heat and thermodynamics by brijlal n subrahmanyam
DESCRIPTION
A book about Thermodynamics and heat in Physics.TRANSCRIPT
Brij Lal & Submhnranyam
HEAT & THERMOOYNAMICSPFOPEFNES OF MATTERATOI'rc ANO NUCLEAR PHYSICSNUMERICAL PHOBLEMS I\I PHYSICSA TEXTBOOK OF OPTICSTHEBMAL AND STANSTrcAL PHYSICSHT,TTERICAL ET.AMPLES N PHYSICSK.K. Ttwari
ELECTRIC]TY AND MAGNETSil WTTII
EIECIROilGD.N.Vasu&mFUNDAMENTALS OF iitAGNETtsM ANO
Fl FCTmTYD. Chattopadhyay & P.C. Rakshi
OUANTUM MECHANICS, STAMICALMECHANICS AND SOUO STATE PHYSIC:;
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A TEXII'U)K OI APPLIED PHYSICS FORPOIYTICTI\|G;
Rs. 95.00 I{. SUBRAHMANYAM
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BRU LAt
B)' the Same Authors
l. Optics for B.Sc. (Twenty-Third Edition)
2. Properties of Matter (Fifth Edition)
3. Atomic and NuclearPhysics (Sixth Edition)
4. Numerical Problems in Physics for B'Sc'
5. PrinciPles of PhYsics
6. I.l.T. Physics for Engineering Entrance Examination'
t* Ii
RAM NAGAR, NEW DELHtr-llo 055
HEAT
THERMODYNAMICSlFor B.Sc. (Pass, Gendral Suhsidiary), B.Sc. (Hons. and Engineering) and
Medical and Engineering colleges Entrance, IAS Examinationsl
BRIJLAL,M.Sc.Reader in Physics
Hindu College, University of DelhiDelhi-ll0 007
N. SUBRAHMANYAM, M.sc., Ph.D.
Department of Physics, Kirori Mal CollegeUniversity of Delhi
Delhi-l l0 007
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,.(c, ( t ),) l,:5".i'tl-l',1{' q"or <z-.i'- )2{l <lJ I r
1,12 Heat and Thermodynarnics
5 l1 t{lo.tic Theory of Gasec
The continuous collision of the molecules of thc gas with thervalls of the containing vessel and their reflection from thc u,allsresults in the change of momentum of the molecules. According toNewton's second law of motion, the rate of change of momentumper unit area of the $'all surface corresponds to the force exerted bythe gas per unit arca. The force per unit area measures t-he pressureof the gas."/
loetulatcs of the trinetic theory of gaseq{r'
Nature o! Eeat{ 15t
The component .of the ^velqcity with which the molecule P
*itt stri[i it" lppotitt"fag'c BCFG is ur^and the momentum of the
*.i.."ti i" -rr.' ' Tr., molecule is reflected back with the samc
momentum nltll \r ah opposite directign and after traversing a
;i;;;; I rvill itrike thi"oirpoiitC face ADEE'
,r4P.z(d1 The gas is composed of small indivisible particles called
'mol ecules. LThggripqrtier, gljhejrrdlvi d ual molecul es are t he sam e"(1 rr,. g., i
--G|Fil distance between the molecules is large as comparedto thai of a solid or liquid and hence[_ghc forces of inter-molicu]arlattragtio.n are negl
46;ryfnr- molecules arc continuously in. .motion _with varyingvelocities
-hnd the molecules move in straight lines betweer, 'ur,!
two consecutive collisionsf The collisions do not alter the moleculardensity of the gas, i.e.,lo.n the average the number of molecules pre.sent in a unit volume reinains the same.{ AIso, the molEdulEs?oio-ti-ccur;iilaG;f@me of the gas. t
W^rl6rl The size of the molecules is infinitesimally small ancompared to the average. distance.trav.ersed by. a molecule betweesany 1wo consecutive collisions3 The distance between any two 6q1:nsecutive collisions is called free path and the average distance i:-called the mean free path. The mean free path is depindent on tt ispressure of the gas. If the pressure is high the mean free path is le:leind if the pressure is low the mcan free path is more. -
'ss
..!q IE_ *gtSe gtrurs-p-er&slll nju{.gq$j9_!phsr9! 4 n d t h ewhole ol therr energy ls Elnellc.
t. srir s'p".;io;;-t .hli."",nt. of a Gae
,r$ " A. ct of the moleculesu Y on the walls of the containing vessel accounc ior the pressurc of th,s.gar.. na:fr ne
(Co4;idcr a cub-rqallLesg!-,4 B-}\\!OE of side I cm conrainin.thc g)s il.'ig. 5.7). The volume of tEe veffiT)Cfas
-is 13 cc. Let n and ,zt represent- th_e very large number he
ioleculcs present in the vessel and the mass ofeach moleculf
ern
resPcctively.
Consider a molecule P movi.ng-.in a random direction witl avelocity Cr. Thggggllylan be reiiilvEd
Fig. 5'7
The change in momentum produced due to the impact is
mu1-(-mut\ :2mur
As the velocity of the molecule iszr, the time interval between
cwo successive impacts on the wall BCIG is
-- 2l seconds
ur
.'. No. of imPacts Per second
I2t
",ur:2r
Change in momentum produced in one second due to the im-pact of thiJmolecule is
z*u'x# = ry
The force Fx due to the impact of all the n molecules in one
second
Is
comPon.llt-s 1Ll, ?'t
Therefore, : +luf*uzz*"'+"J I
Ig9"!o t!1e_-Unle
C12 : u1,1-vr.2*wrz
axes respecti--lq---
lU Heat and Thermoilynamict
t.',rrce l,er unit area on the walil BCIO or ADEE is equal to the1rr errure Py
Px : #r(urr .| zrrfurr+...... +z,r)Similarly the pressurc Py on the watls CDEF and lBO.& is
given by
,, : # (ur21arr;......+r"2)
and the prcssure P7 on the walls ABCD and EFGE is given by
Pz - fr {rrr+rrz+......+wnzla" $qpresrurc of a gas is the same in all directions, the mean
pressure P is given by
p _ Px*Pv*Pz3
tt, f: 5F L
(urz { x rz 4 ut') + (ur' +u"'. + r'r"l
1- (aaz + o s2 + wzz) + ......
* (u,2+r,"r*.,r) ]
: #[ cf +c22+cJ+......c", ] ...(d),
But volume, Y _: lr. Let C be the root.mean-square velocity olthe molecules (R.M.S. velocity).
C, : C r' *Cr' *Cr'*.-....CJ7L
Then
or nCz - Cr2+C22+Ct21.......Cn2
Substituting this value in equation (i), v".e get
D _ m.n0z'- 3v- ...(ii)Brrt .4f : rntl rvhere .0f is the mass of the gas of volume V, m is
the mass o[ each molecule and z is the number of molecules in avolume [/.
L;From
Noture ol deot
or e: 13tr- Vp
155
...lirr'
tNotc. R.M.S. volocity C ia tho Equaro root of tbo moan of tbe squ-aree
"f rh"!;;;'id;;i"if"iifr-.*ind it,ienot'oqual to tbe meenvolooily of tJre
moloculea.]
TABLEMolecular Velociticr et OoC
Molccvlar WciihtRoot mcan cqtare wlocitY
in cm/s
Eydrogen
Ilelium
Nitrogen
Orygon
ArSOn
Carbon dioxide
Chlorine
2.016
1
?a
20
.10
41
7L
18.4 x 10r
13.l x l0'4.95 x lO'
a.ut * ro.
.1.14 x l0r
3'95 x lo'
3.11 x l0r
<: X.l*S. vclocitY of hYdrogen-'-4r/
The densiw of hydrosen at N.T.P. is 0'000089 glcc' There'&/ forc C for hydrolen can bJcalculated as follor'r's :
c- rT/ 3tzox lr6te8l: V ----T-OoO0-89-_
: l'B4x 105 cm/s
(b) For oxygen the density it N'T'P': l6x0'000089
f sx%;T56tggrC for oxYgen : fJ -e;-O
oOOOeO-
: 4'6x l0{ cm/s
(c) For air the densitl' at N'T'P' is 0'001293 g/cc
I 3 x 76;l3G;(e8tc for air : { _____o:ooizoz__
/ -- 4'850 x 10' cm/s
a#xarople 5.4 calculote the number of moleculee in one c.c. of
orgjri "'t
W?f .p- using the lollowing ilata :-
t5(;
157\ r! rt , <rJ Eeat
Let the mzuls of each molbcule be rz and Avogadro's number be.rY.
M:mXNDcnsity of merarry : 1J.6 glcmtR.M.S. uelocity of ozygen molcculn ot 0.C
: 4.62 y 10t cmftMass o! one tuolecule o! orygen - SZ.8:r.l0-u f
P: J-og,3
Let the toass of each molecule be m and the number ofmolecules in one cc - z
IP : i- rnnC2
J3P?L:
-mc2
P : 76 x 13.6xgB0 dynes/cm2m : 52'B X l0-2{ gC : 4.62 x lff cm/s
tctt - i iq L,l<,*l /-lt.r)t /<i ,t,.,,
H e at an it ( o*- ̂"
or rlr**,
tu,2N
* o,
-f, *No
| '"o
Here & is c^llcd Boltzmann,s .r"ryp yO bO
...(dr)
,{r".{ C
.".(i)
...(i0
...(dir)
Herc
h:
n : t'S$J 1l$tru:4,-oJrxluur(insfi6 Energy per Unit Volume of a Gas
Thus, from equation (;i), thc mean kinetic cnerg'y of a moleculeis directly proportional to the absolute tcmperature of a gas. Whenthe tenpe;ture of the gas ir increased, the mean kinetic energy oflthe moleculcs increases. When heat is withdrawn from a gas, themean kinctic energy of the moleculer decreases- Thus, at absolutezero temperature, the kinetic energy should be zero. It means atabsolute zero tempcrature, the molecules are in a perfect statL of restand have no kinetic energ'y. But beforc the absolute zero tempcra-ture is reached, all gases change their state to liquids and solids.
Also from equation (ii1, G a, T
It means that the root mean square vclocity of the molecules isalso directly proportional to the square root of the absolutetemPerature
-culeshtoelbeaome root meon s$urrctoelocitg ae tlut olhgibqetmoleculee at -10trC ?
The enerry of a gas molecule ir,
i ,,a: +oFor hydrogen molecules
f, "'rcf : ]*r,
For oxygen molecules
J, *Ps : {vr,Dividing (d) by (di)
mr0f fr@7:T
Here Cy : Ca
* :;,r1,""r,,
'\
oe, .y',rr Kn(q\,o%o*
L;( Kv*?: J PCzI wt 3
: T'* r"2: i-E
fi'
w"hgrc E : l.pC! and is equal to the Linetic energy perunit volumeot the gas. p is.the mz*s per unit volume. Hencil'the pressure of a. -C_T_i:
nymerically .cgual to tu'o-thirds of the ..uriEir,"iic energy of
" jiru':.$.'ilH"l[:::|;;ffi:'#J.' b c q'<'"' t'^ r*S
Dhe pressure of a gas, according to the kinetic theory, is
\\'' I
*r".rg* P:i- Pcz
r\c"o P- +!fPV:*'"
Consider I gram molecule of the gas at a temperature ? KPY:RT
J" uo: RT
iro * *o,
3x76x13.6x980
. .. (t)
T.,)1.2
lf,t
From equation (iii1
?2
T2
: 173 K:?:16
- mrOrtT,mtCt'
: 16xl73: 2768 K:2768-273: 2{950c
H eat ond T her modynamic,, Nature oJ Heat
t'- t )I nt3vI
Py:;MC2
Consider one gram molecule of-a gas at anrure ?. The meanlnersy of the molecules
-lMc..2
: )- N'nc'
rv : \' Nmc'
: I tt.-f,'.*c'
Mean kinetic energy of a molecule
: LmCz /
2
:3kT2
PY:-|rv.*r,PY - NkT
But /Vx& :.8
absolute tempera'
T2
E*.r-plq.s,&dculcs al 27"C.
Culculate rle RMS velocity o! the oxygen rnole-
First calculate the n.dfS velocity of oxygcn at N.T.p.
n - l-gP"- V p
Here P : 76X 13.6 X980 dynesTcmr
P : 16x0'000089,:JmrC : 4'6x 10. cm/s
Let the 8.tr48 velocity of the molecules at 27oC be C1
+:'[Tc1 :'cy J T
Here C : 4.6X l0r cm/sT :273 K?r :27oC
: 27 *273:300K
Ct : 4'6t,* Js-. C, : {.8*Xl0t cm/t5,17 Derivatioo of Gas Equation
' F.o* kinetic theory,
PV:frT ...(d,
Note on the Gas Equation.
In the gas cquationPV:RTP is in dynes/sq cmI : B'31X I0? ergs/g mol'K?isinK/ is the volume in cc per gram molecule.
Eraople 5'7 Calculale the volume occupied by 3'2 gtams ofoaygetu at ?6 cm o! Hg and 27"C,
Here P : 76x 13'6 x9B0 dYnes/sq cm
r :27,+273:300K
B : 8'31 x l0? ergs/g mol-KPV:Bf
8'1lxl0?x300l : % xT36x98o cc Per g-mor
I/ - 2{610 cc per g molP: + pc2
,ao
Heat and Thennoily namiu
Volume for 3'2 * "t "at.rX;rr* r.,32
: 2461 cc
Noture of fleof IOZ
(iii1 Begnault'a LauFrom equation (dd)
PY - NKT
When 7 is constant,
Pc., lIIt means that for a given mass of gas, the presurc is dircc4y
proportional to its absolute temPerature provided the volume remainrconstant.
/r'.'ample 1.8 Shoo tha,t n, the nuwber of nolcoula pr u;nit
ooluma of an iileal gaa ie gfuen by
PNO: -ffiwhera N io Aaogadro's number.
For an ideal gas, for one gram molecule of a gas,
?V:RTR: NK
PY : NKT.
Let z be the number of moleculcs per cc. fn that case,
constant, rn.r"ro.. "r(']
akolute temperature
...(d)
But
But
P:nkTPo: -Tr
-n*:-frPN
fl, : nT
where lV is theconstant.
If
Erample 5'9. Calatbte the nunfrer of molcculca in one ctliamelre of an iileol gaa ot N.T.P.
Let the number of molecules per cc be to.
PAIO: 'ffiNumber of molecules in one cubic mcue volumc
,: nXlSPIV x l6cu: -m-
ilere P :76x l3'6x980 dynes/cmr
.l[ * 6.023x lOa
I : B'31 x l0? crgig rnol'K
Ir means that for a given mass of gas,pr<>g;rt ional to the absolutc temperaturercrnains c()nstant. This i: Charles' Law. 9*JI
.)
the volume is directlyprovided the pressure
fi'l{tDerivation of Gas Laws(t*) (i) ggP\Iw-v According to the kinetic theory,
':igP* *#*.
Pr- **oAt a constant tempcratureT, O is
constant tempcraturetJ MC': constant
Hence PV - const. -at constant tetnperature.Li;l -gMk*I&;D-Accoidirrg io the kinetic theory
G9 ": *.,P- *#*
PY:**oConsider oue gram molccule of a gas at
T.M:mNpv: Iy,lr.r.J
The mean Linetic energy of a moleculeI - 3 --Z *C': ;AI
mCn :3bTSubstituting this value in equation (i),
PY : NHI ...(rr)Avogadro's uumber and & is the Boltzmann's
P is constant,Yc.T
la .L e,v*A
762 Eeat anil Thermdynaniu
T :273K76x 13'6x980i6'023x lOax 1064-t:
- : r.Utt*t,,uErample 5'10 Calculate the number of moleaules in one like
oJ an iileal gae al 136'd Q temperoture onil 3 armoepheres prellure.
Let the number of molecules Per cc : 7t
Plvn-wNumber of molecules in one litre,
c:7,X103P][x lOEt: --w-
Hcre P:3x76x13'6x980 dYnes/cmr
-l[:6'023x104E - 8'31xl07 ergr/g mol-KI' : (273+ 136'5)
: 409.5 K3 x 76 x l3'6x980 x6'023 x lOax l0r
B'31 x l0?x409'5/, --t','U*'0"
,Zfzlcrote 5'11 The number of moleculea per cc ol o gae ie
'/., *r*, at-N.t.P. Calcul,ate the number of rnoleculee per ce of lhe
gu.' () of 0C anil 10'6 mm pressu?e of mercury and
(id) al 39oC ond f0-6 mm pressure 'nercuty.
For a unit volume of a gas
P:LmnCN5
(i) At 0"C, R.M.S. velocity is equal to C
At N.T.P.I
P, : -g- mn1 Cr
At OoC and l0-'mm Pressure
Pt : i mnror
From (d) "td (dd)p,
: n,Pr n,
nrxPlor ,rt :
-FI
... (i)
.. . (ii)
Naturc of Ecot
Hcrc n1 : 2'7 X l0rePt : 76x l3'6 x 980 dYnes/curP1 : lQ-e mm of Hg
- l0-? cm of Hg& : l0-z x l3'6x980 dYnes/cmt
2'7x l01e x l0-?x 13'6x980fi1 : ------7il13i-;580--z1 :3'553 ll0ro
(ddt Let thc R.M.S. velocity at OoC be C1 and at 39oC bc Ot'Numb'cr of molecules Per cc at OaC and l0-0 mm of Hg pressure
e tl,tr
and at 39oC and l0-c mm of Hg pressure: fL,
P:!mnC'J
Here presure is the same in both the cases
ImnP;:Imn,cg,n1?1t : tus0gs
C,'7lg : tl2
W
But C e' t/-TC,, T,
rl2xTrtt': -Tl-' Hcre r\: 3'553 x l01o
f1 :0o C:273K
fs:39oC:273+39:312K
3'553x1010x273ms: _*____m_
ro ,, \: 3'109 x loloI 5't9 Avoga&o's HYPothesistJ-L, " konrider two gases A and B at a pressure P and each having a
1 vo)ume I/'r C Z Mass of each molecule of the first gas : m1
\V Number of molecules of the first gas : rLL
Mean square velocitl of the molecules of the fint gas _ qtFor thc 6nt gar
. p: I e,e.
: + '# ...(,)Similarly for the second gas
P: $ r,4,. _+ryL ...(,,)
where zn. ,spres@l the ,nass of each molecule,. z, the :rumber ofmolecules atia crr tue meu,, rq""i"".r.1iry ot.the morecules.From (d) and (dr),
l **C!__ I .mrnr1szT ---7-: 3or n*t7f _ ,irnrCrr'
,.,_^^,lf the twogasesn g at- the same temperaturc ?, the;:::f,rnettc energy of the morecures oi [.iiiirr" gases is the same
+m,c,,-|wc,"From (r:di) and (iu),- " h \ 6i\-8 i ...(du)
.--- r-:'""' "qpd ":lfn :,: :.,#= ::r.;-; conditions or;:ffifi :lX"U:J,;ff f$;91.";;;l,u*u..;;.i;il.il'tni,5.20 Grahan's La,w of Difueion of GeeegAccording to the trinetic tfr"orf
P:*r*or '- /F
Cn l
^c - It means *, *:311,,:J[r"re vcrocity of the morecures3:1,ffr .,"#,x,mr:::lt{;t,:1T..d,",?::Lb,}"Tu
C,_ T P,c'-Vp,
Nalarc of Eeal
Here, r, and r, represent the ratesg?ret.
,*-,;ff;?l;,u;;;::1"#;ti::'tototranitomhineticenersvo!oncTotal random kinetic energ.y per gram-molecule of oxygen
I
' , m@xNe
= ; kT.N
: +. #,,c
- J-R?2
. : f *e sx r0?x300
: 3'735 X I0ro ergs
- S7Sijoules _/
ffi ill"tlgdH'"".H#:o,:r1ffi ff :.?"i:'#:t-'H:ffi ;1'"'""",*",#I
cvle
@
,ffff':;:',!a!,;];*:;'!;ov.'*'kineticenersvo!omote-Average K.E. of a molecule.---=---r
=j-m@:Here & is Boltzuann,s constant.
& : I.38x l0-rc erg/molecule-degAveragc K.E. of a molecule : f *l.3Bx l0_rcx300
Noto. The aven : 6'2r x ro-r' ergs
m:*ild&,'H?fl ,YirlT*""i,""9,"k.:;';ffi '"#"TI.iT"#;EaTplc 5.14. Colculnte the mean bandationol bhetic cnery!gw mol,eculc of a gaa at |Zt"e, i;;r:; n":';.JZ joulcs/motc_K
Aoogailro'e nurfier, (Dcki' 1974)
N - 6.06x10u
rrerc i::?*:ii-il"1*"iI:6.06xIOD
L_!:l 8.32 \,T - (606;10-"/ jouler/molecule'K
Mean tranrlational tinctic cnergy per molecul " - |Ul
165
of diffusion of the two
*o* -
+-/f ,-/
!ii:-l166 Eeat anil llhermollynamice
3x8'32x1000: T;TT6;TT: 2.059 a l0-ro joule
Erampte 5'15- Colculate the lotal ranilom kinetic energy oJ onegram of nitrogen at 300 K.
Total random energy for one gram molecule of nitrogenq
: ;RTTotal random energy for I gram ofnitrogen
3Rr: -2Af
whcre the molecular weight of nitrogen y : 28 g
3RTn:-
-2M
3x8'3x107x300- --ETE-: 133.4 x 107 crgs
: 133.4 JoulcNotc. Tbc total resdom kinotio energy for I gram of a gar h difierun!
for diferent gaaes at the gamo temporature.
-Eranple 1.16. Colculote tlu I'otal random hinctic cnugg o! 2 gof helium at 200 K.
Energy for I g of helium3RT: -zfr-
Energy for 2 g of helium2x3xRT: ----mi-3RT:__fr*3x8.3x107x200:____l-
: 1245x 107 ergs
:1245 joulee
_ 1'.n- ple 5.17. Calailate the root mcan aquo?e oelocjty o! amolecule of mercury aqpoltr at 300 K.
Mean Li.ctic encrgy of one molecule of.mercurytc:+mO-;W
Nalure oJ HuAV
Let il bc the Avogadro's number.IeThen j nNCs : ;- kNf
I --- 3--v MCz : ;Rr
c:r{ryHere the molecular rveight of mercu\ M - 221W
C:.T@v -ezi-/ : 1.93 a 10r crn/e/
Eramply'5 18. Wilh what speed woulil one grom molecul,e otorygen at 300 K be moaing in oril,er that the translational, kinetiaenegry of its cenlre of mass is equol to the totol ranilom binetic enetggoJ all, ila molecul,es 2. Molecular wei,ght of ozygen (M) : 32.
Total random kinetic energy tf I gru--*olecule of olygen
- f,i'nr : +xB'3xlo?x3oo- 3.735 x l01o ergs
Kinetic energy of .ilf grams of oxygen moving with a velocity o
- iMas
|*r,: g'Jggy lQro
Ij xSZxu2 : 3'735 x l01o
. 3'735 x l01o__--]6-r : 4.8 X 10r crn/s
. Exarople 5'19. Calculale the temperature at whiah the r.m.e.tselocity of o hyilrogen mol,ecule will be equal to the opeeil oJ the eorth'e
first aatellite (i.e. o : 8 km/s).Energy for I gram molecule of hydrogen
:!uor:1ar- 2"-'- 2
m Moz,:-572x(8x105)r: T;3:3;-i3r*
./. : 5'14x1$ KExaople.520. At what temperalure, ptessure remaining e;a.tu,
tant, wil,l the r.m.s, oelocitg oJ a gas be halJ its talue oA OoC IlDel,hi lEone.) 1975J
?t: ?
Tr:273 Kr _ r-r;T: 1, E5',:o,
,\ + 273
"r:+:68'25K*
o' .
?z: -2(/r'75'c '/R, 5.21 Degreee of Freedom and r![arwett's Law of Fquiparti-AJ tion of Eaergy --- - --:"6i -'
ffiffi#Hfl,.m#;d =;iffi "re
nas ?ffi ;e -a.ffi;*s*m **[,].T:l*mol ec uI e" h as thr ee d-egr ees
-"i ir;l";;r td* I. t i * "na;;fl;Hof freedom of rotationl It has in au nve e;;;;;;';'i;;;d:B -Z;$ ;Y^*:*:x r,:ffi:;":H??jffi:];he mean,,^,oi d(n
)*o': * o,But C, - *qoe{td
*",#"?*Xli#f all equivalent, mean square velocities arong
Here
ur-oE-rdlm(ur):Imgf):lm(wr)I mC' : 3 [l m(ur)i: 3 B rn(u!)]
: 3 [] m(w')7q:; KT
lmut:lbTI moa : lkT*^rd: *lil
. Thgreforc, the average- -linctic encrgy associateddegree of freedom J yf
Nature of Eeat rcg
Tn,rs(he energy associated with- each dqgree of frcedom(whether trinsratory 3i -t"i.ryf i, I iA:) Ew u?trEE .,r rrcec
,,
This represents the theorem of equipartition of energy .WE \^el'?2 Atom.icity of Gasec / -
fe\(l Mono'-otomic- ga.E. A mono-atohic gar molecule has oneAtom'-tach molecure has thrcc d.gr;;r;i?..do- a* io-i."rif,tory
Encrgy associated with each degree of freedom _ | WEnergy associated with three degrees offrcedom
:; bT
Consider one gram molecule of a gas.,"r:eryyg:
-rYx |r*:{ g*4r
JI7x,t : .BJ
a:$ar
I:##*ii::"fl'' " :*; nil * $:: *l:g.H, il;:*f i1$*
c, lTC, E.V Tncr"r: T
Heal anil Thermoilynamia
..-(')
...(i0...(ddi)
...(iu)with each
(# - the increase in internar energy per
,"-R".",r"" )
But . Cp-Cv:BCp : Cv*n
: t**n: f,nFor a mono-atomic gas
Coa-9vq
TR \: -a-- : 167 )]a -/
[But,
c,-#:3raunit dqpec rise of
or
170 Eeat anil Thermoilynamia
The value of T is found to be uuc experimentally for mono-atomic gases like argon and helium.
\fi 6 Oiot*r;" g*. A diatomic gas molecule has two atoms.SuchY molecule has three degrees of freedom of translation anC twodegrees of freedom of rotation.
Energy associated with each degree of freedom
: l*r2
Energy associated with 5 degrees of freedom : $ Vt
Consider one gram molecule of gas.
Energy associated with I gram molecule of a diatomic gas(5: il*i kT :; RT
.(a:;Rr^dUwv: _EF
q:;R
But Cp-Cy : RCy: CvlR
: i.o** - tp n
u-C'r- Cv1+BA: +-: 1.40 \rp )2'- -/
The value of 1 : l'40 has been found to be true experi'mentally for diatomic gases like hydrogen, oxygen, nitrogen etc.
1ii$ Triatomia gae. (a) A triatornic gas h-aving 6 degrees offreedom has an cnergy associated with I gram molecule
- .Irx $*r : *ra :3RTc":-!fi:sn
, But Cp-Cy: RCp - Cy+R
: 3E*8 - 4.B
v- cPr- Cv
: #:r.33(b) A triatomic gas having 7 degrees of freedom has an energy
asociated with I gram molecule : N x -2 kT : ; B
7u:-;-Rra.,: !! : 1- ndT2
But C?-Cv : R
Cp: CylR7(l: 'R+B:
; R
v- cPr_ Cv
: ffi:1.28Thus the value of 1, Ce and Cv can be calculated depending
upon the degrees offreedom of a gas molecule'
5.2t Marwell's Law of Dirtribution of Velocity
At a particular temperature, a gas molecule has a fixed meankinetic eneiry. It does not mean that the molecule is moving with,n. r.*" spJid th.oughout its movement. After each encounter thc."."d of ihc molecule changes and due to a large number oli=oitirio*, the speed is different at different instants. But the root*""r, rouur" vilocity (r.m.s.) C remains the same at a fixed tempera-
i"r.. ei iny instant, all the molecules are no-t.moving wit! the same
.r"io"iw. So*".t" moving with a vclocity highcr than C and the;tL*'*ittr "
velocity lowei than C- But the mean kinetic energy ofall thc molecules remains constant at a given temperaturc'
I)erivation of Maxwell'e law of Distribution of Molecularvelocitiee
mean slluare velocity of molecules is defined by the
c,: |J] * ar
Here dIV is the number of molecules having velocities betweenc and oadc. If the total number of rnolecules is iv, then a fracticn
$wiff have the comPonents of velocities in r direction in the range
Nalurc of Ecal lft
\o\ v$
Theequation
r76 EeaC and Thermodynomico - n.',cLi.:,-.f-o 'lr><-i't+<(n-\tr1 "')(,\de())l)(Naturc * r':frv)-\c,r-', i', eru(.i'c( +r)(- crrt J rtc . 1.o'(1.end of the plate. After-a sbort time, sufficicnt quantity of silvcr irdeporited on the plate p. Gid;;p.IirJ-pr,",.'-.t"i,'tr,". r"r"iir.
t-
-!ts__!t_I
*i*A cnr.r Ua-_u
Fig.6.0intensity
-of silver- on the- prate p is studied and this represents thevelocity di_strilutioa of tht molecules. --
Th; graph repie...rii's tlrenumber of nolecules and verocity agre* *itn'Ml**Lil iiriau"rri""of velocity.
In 1947, Estermann, Simpson and Stern desisned a moreprecise apparatus to study the veiocity aistribution.
Cesium atoms from the oven emerge from the opening z{(Fig. 5'10).. B is-a slit and D is a hot tuigstcn wire.
- rie *-riore
_appbratus is enclosed in an evacuated chamb?.i;r;;; fb:i** ofHJ). The opcning 4-and the slit s.* t*i^"ifu.--i" ti.. uilr."""ol a gravitational 6eld, cesium atoms will strike the wire at D.But due to the graviational field, the path ir;-p;;rili".'.il.
"io*
i:--\\
Fig. 6.10
f,.:g:lTg-11" qa-ttr p_ao not rcach the wire. Thr atoms soinsalong thc pathr I and 2 reach at_ D, and D3.respectivity. ft "i,ito-.city of the atoms in path I is highei tr,"nifii pitfr i.
- ,
3 \ \ \ \ \ \
S25 Mean Free Peth5v ^. . Iq de-riving the expression for the pressure of a gas on thc basir
ot'kinetic theory, it was assumed that the molecules aie of negligible
When cesium atoms strike the wire they get ionized and re-evaporate. They.are collected by a negatively-charged detectingcyllnder. su-rroundrng- the tungsten wire. The magnitude oI thecurrent indicates the intensity of the atorns at variouipositions. Thcdetector can be ooved to differeot positions of the wiri. The atomsreaching.at Dl have higher velocitf than rhose reaching at Dr. Thcverticai height of the derecLor iepresenu the magiitude- of thevelocity and the ionization current indicates the- nu-mber of atomrstriking. the wire at a particular point. A graph is drawn betweeathe ronlzation current along the y-axis and the vertical height (spceCof the atolps) of the detector along the c-axis. The velocity iistri.bution is found to be in ag.eemerit with the Maxwellian disiributionlalof velocity.
sizc. They were asumed to- be geometrical points. A geornetri.rlpoint has no cimensions and hence inter-molecular colli.eiins will notbe possible. But, a molecule has a frnite size (though small) andmover in the space of the -vlisel containing it. tt colliies *itfi iG*molecules and the walls of the containing vcssel. (Thc path iourr"aby a moleculo.betwten-any tw.? ..consecutive colliiionsis
" ,r."ightline, and is called the frcc p.tL, The direction of the moleculS Lchanged afrer every collision. After a number of collisions, the totrlpa-tF app_?B to bc zig-za1 and the free .path is not constant 1Fig.5.lly. -Therefore, a.tero(ucrn.frcc prth is ured to indicateihcmean distance traveiled bi a llolecule between two coltisions. It'ihctotal distance travelled aftcr JV collisions is 8, then ttr" *."n
-fi."pathtrirgivenby
r_ B )^-T f "'t;lLct thc molccules he assumed-ry'be spheres of diaarcter d. A
collision betweeo two molecules will'take place if the djstanr:e bet-ween the centres of the- two molecules is d.-, coilirion wiii also ociurif the colliding molecule has a diameter 2dland the other moleculcis simpl'7 a geornetricat point. 'I'hrir, assuritine all ot!:er moleculcsto be gromerrlc.rl p,'ints and.the c()lli1ri1q n:oiec,;lr c:'C .rr::crrr ! jthls molccr.r je wili '."'ir.'cr a volume rd?t tn c'r:.* rr,:?: I j r"rrr.]:i-px .,,!r tr., ih,c vr.l 'rirl,i :.l'a cylinder of diamrrr:: it i *:
ft,I
hI
yE Eul otil TlurmodYnnmiat
(l.t ,t bc the numbcr of molecules Per cc.
ihco, the number of molecules Present is a volume ndlu: ril,tvXn
This value also represents the number of collisions made by the molc.cule in one second.
The distance moved iir one second : u and the number ofcbllisionr in one second : rd\t v.n.
.
llolurc o! Ecal 179
Detcrmiaation of Ereea free 1nth. As dlacussed in article5.27, the relation berween coefficient of tiscosity and the mean free
pattr of a molecule is given bY .l
r : i mnC)t
But 4* crilc : du ihs dw
dlv : .u/s "-bo'
du ilv ilwAlso 6t * .y'+-ts2+.toz
...(;i)
dN : NAs e-J (rrr+rs+otl du & dw ...(ddd)
Equation (irid) has to be modified in case the gas as a wholc
POSSeSses ma$ mouon.
Let uo, uo and rao be the components of the mass vclocity'Therefore,
-the actual vilocity of a molicule consists of two- Parts :
(i) the mass velocity comPonentS oe, us and rao
ldi) the random thermal velocity conPonente
u-uo, u-us and w-ll)scorresponding to thermal motion without mass motion, simi-
lar quantitiis with lnass motion arcA - u-uo| - a-as
and W - $-woFrom equation (ddi;, Maxwell's law of distribution of velocitl
can be wiitten as
d,N : IiAsc-6(u8+vt+'')d(l dY dw...qixyEquation 1io) holds good only if ao, zo, tto, T and 'Y are cons-
taDt throughout the gas.
If the gas is not in an equilibrium state, there are three possi'
bitiiies occurring singly or jointly.
For unit volume,mt: P
r: Ieclr- 3't
pC..'0L//
The root mean square velocity C of a molecule can becalculat'.ed knowing pre$ure, d=ensity and iemperature. The coefficient of,ir".ti,, .itire gas is determined expeiimentally' Hcncc the value ofthe mein free pith ol a molecule can be calculated lrom equation(d).
5'26 TranePort Phenomear
According to Maxwell's law of distribution of velocity
ilN - +*NAa "-4ct
,26, ...(i)
,-\ .'.1 u. rfree path ^ : #r*:#)
This equation was deduced by Clausius'
...(i)
...(i0
..{o)
of thc law of diltri-
Then' 'llllll : P
ml--
7.o"P...(did)
The mean frce path is inversely proportional to the denrity ofthe gas.-- 'Th"
expression for the mean path according to Boltzmann ir
^:h ...(iu)
He essumed that all molecules have the samc average rpecd.
Maxwell derived the exPression,
Thc mean &cc path is inversely proportional to the sqrrare ofthc diameter of the molecules.
Let m bc the mass of each molecule'
,*#
^-T.r* )He calculated the value of I on the basis
bution of velocitiesTAsLE
Meaa Srcc Petb (tr)
?'l1x lf cna
&6Ox l0-t om
2.18 x l0-8 om
3.ag X10-{ sDo
l.t!x lO-. om
0.S{lxlH oro
2.85xlflom0.009 x l0-! ors
I n",
l- Hrurogon
/,\J Eeal anil Thumodyramict
: /VxgkT:gR?U :3RI,duaH: -m
: 3.8 : 5'96 cals/g.mole KThus, the atomic heat of solids (.4s) is 38 and this value agrcct
with the Dulong and Petit's Lav.,.
E=arople 5'21. In an experiment, lhc aiacceilg o! the gc,a wa,Jou,,d to l,e 1'04x70-t d1'nssi(:r-rt per unit oelocity gradieat. The.B.ff.S. rclocity of the moleculea ie 4'5 xJ0trcmis. The denaity ollhe gaa ie 1'25 grams ,?e/ litre.
Calcalate 1i1 the nean lree path oJ thc molecu,lcs o! the gac, (ifit,equeneg of collisiott anil (iii) molecular diameter of the gd* malc-cules.
Ifere
Jd : 3;10-8 ca
Eraaple 5.22. Im an experiment the oiacooitg of tie got woa
tound to be 2.25 x 10 -t CGS uzrts. The BMB aelocity of the moleculeai.s 4'5x l0r ernlt. The density of the guis .l gram per litre, Colcu-lcte the mean Jrce path oJ the moleculee .
Nature of Eeot
d:
Here
(r)
! : l'56x l0-'unitsC : 4'5 x 10. cm/s
p : 1.25x l0.r g/cc
. grln- @
. 3xl'66x10-.^ : rU5;TOTIZE TO.
i : 9110-c c-n
(ii) Frequen", "t.",I'iAi."IlS.;f coilisions per second
= Ti6-freemX:9
,t
Ir : gllg'I x lO-c
l{ - 5xlffliid; Avogadro's numlru.o23
x loBNurnber of moieculeg p;;"r;r;"
22+00
According to Maxwell's relatioa-l^*Jm,l--l- r'|4iffix'fiI
C : 4.5 x lS cm/rp : I g/titre
. 3ztn- ec
, _ 3_x! 4>< lr'" - l0-rx4'5x l0'< ,/, l-15 1[0-ccm'> *4"-etc 5'2t. \alaulale the meon lree ytath of a gaa moleouk,'
giuen lllo;t ibe molcaubr diameter ie 2 x l0-s cm ond the tumbcr ofmoleautre pet cc ie 3x10t.
IA:-mI.:@
l:3yl(}rcrnNoto. Thc moco froc peth ir lor than iho wlvclcngth of lighf in tbc
viaiblc rpootrum
Eramptc 5'24. Calculote the meon lree path of gaa molecvleein o c,tnnrbei of 10'o mm o.f rbercury 1)reEsure, uauming thc molcculorilianckr to be 2L. Ane ryam molecule. of the gos occupica 22'4litres olN.T;P. Toke thc lemperolure of thc chamber to bc 27 3 K. (Agro 19? 5l
At 760 mm Hg pre$ure and 273 K temperalure, the numberof molecules in 22'4 litres of a gas
- 6.023 x lOsTherefore, the number of molecules per cmt in the chamber at
r0-! mm pressure and 273 "J:rTiXi;:ll,o_,
22400 x 760
r - 3'538 x l0ro molecules/cmld - 2L: 2x l0-8 cm
Mean frce Path,-l^-r,dro
I
I :2'25 x l0-' CGS units
3'l,t x (2 x l0-r1t* 3'538 x I01s == 2'25 X l0 cm
194
and
Eeat ord Thanoilynamia
Below the Boyle temperature, the gases are highly compressiblethis suggests the eristence of inter-rnolecular attrattions. Beyond
I
Natlrre o! Ecat'-'\'i '1 ' '-'o,uh
(
c-i>fttoz t'ltia (
llere c is a constant and I/ is the vol '*c of the gar
Hence correct pressure
: (P+il: (r*# )wherc P I ihe observed pressure.
Q)f Corr""tion for Volume. The fact that the molcc.rleshave 6hite size shows that the acrual space for the movenrent of thcmolecules is less than the volume of tb6 vessel. The morecules have
the gas : (f-b). .11 a--Dtk, the radius6tdne'motecule be r.
(J- ,'.,u ,-t-
L
T "''
196
BoYLEr-l-- / TEMPERATURS
the_1p!9le_-ol-rl_9q9l"q aro-und the'.n- and duC-o t6,is-Ectoi rtx, thei6rrection -foi-vilirEre ls 6 rr&ere 6li-a imateiy four times thesorrcL'rton ror volume
-tS o lr'nere O rS approamateiy lOUr ttmes the
actual volume or the molecuies. Thercfoie the corrected volumc of
Thc volume of the rnolecule
the Bgyle temperature, Boyle's law is obeyed and intermolecular(". a,.t7&,tions are less significant.
l\ 5/36 Van der \flaals Equation of StateA 5t36 Van der \flaals Equation of Statey
^- (lVntt. deriving the prefect gas equation PV : RT on the basis
(S'\ oi kjnetic theory, it was assumed that (r) the size olthe molecule of/AY-/ the cas is nesliEible and (ii) the forces ol inter-molecular attraction/)\'-/ the qas is neeligible and (ii1 the forces ol inter-molecular attractiont are absent. But in actual practice, at high pre$ture, the size of the
molecules of the gas becomes significant and cannot be neglected incomparison with the volume of the gas. Also, at high pressure, thernolecules come closer and the forces of intermolecular attracti'lc areappreciable. Therefore, correct;";* hculd be applied to the ga!e(luatlon;l
1i) Correction for Presgure. A molecule in the intcrior of aqas expericnces forces ofattraction in all directions and the resultantIohesive force is zero, A molecule near the walls of the containerexperi enc es a . resu I ran t fo@helaattF=D ue to.thi!'...'[email protected]*r.dThe :orrecliotr foi prestute p depJnds upon (d) the numberbf molectrTes striking unit aiea of the walls of the container Persecond and (id) the n"umber of molecules Present in a giveri volume.Both these far:tors depend on the density of tha gas.
. .'. Correction fcrr pressure p * pt -+
The centre of auy two moleculo ".o.pp$,.ch
cach othcr onlyby a 'rinim"m distance of 2r ri.c., the dia-meter of cach moleculi.The volume of the sphere of influence of each molecule,
AB-in(2r)!:Brcoarider a container of volume 7. If thc molecules are a[ouied
to enter one by one,
The volume available for first molecule:Y
Volume available for second noolecule: Z-B
Volume available for third molecule: F_25
Volume a,railable .rjt}:#:il i
Average spacc available for each molecule
:n
: v- $l+z*3......(n-rIti: ,'-g (n-'l)n-i' ---T-
OT NB B- 2fn
Fig. 6.18.
op:-7,
P-------+
t- 'l- ,O* the Vau dcr Waals equation of state for a gas is)- t-' (r**)(r-6) : Rr ...(i) ) ,.wherc c and b are Van der Waals constants.
From the Van der Waals equation of statq
(;;+F;;:;)?;"^ $rt, : ,+u_;, I
-/-
196 Eeal anil Thermoilyrnmict
Ar thc number of molcculca ir vcry tu.g" $ ca. be neglected-
.'. Average rpace available for cach molecule
: f-g (ButS-Bc)'2r--__ Y_n(ul'2: Y-4(ns): y-b
b : 4{tt*l - tH Jl?iL,*.
actual volume ol
Nalurc of Eeat Dldecrease in oressure. It is not possible in actual practicc. The rtater18 and I'D, though ,r*t^ti.,'".r, be ,"at;'.d i_n practice by carcfulexpcrimentalion. - At higher tempcraturg, the theoretiial andcxperimenLal isothermals ire similai.
until now "s
qany as 56 differcnt equations of state bave beeosuggested. But no single equation satisfie;aII the observed fact!.Dicterice (1901) has suggested an equation
P(y -b) - Rr "-#Berthelot has suggeEted an equation
(r+t:*)v-b) : Rr
5.37 GriticelConstante
-f6\" (rn.
";ti.ul temperature and the corresponding ,ulo". . Jf''/Jprerrure.and volume at the critical point are catied tajcritical.cons.rants. At the crrtrcal point, the rate of chnngc of pressurc withvolume (#)is zero. This point is calledthc point of inflexion.
r-/-Accorriing to Van der Waals equation
( ,*#) rr-,r : Rr
,:l#)-+Differentiating P with respect to 7
iIP _RT7V : (7:[]r *
At the critical ooint * : ndyv-ToY -Y"
-RT, 2o.
-
I - -
tt(7r-D)8T Vr' - v
2o R?oor W : O;D;-.),
Differentiating equarion lrrir;
dzP 2Rr@ : ip=rf
_
...(rt)
Ug:l{lUl
E
Ir.r t:r.q ol
2a-v-
...(0
...(j0
...(di0
V0LUME ------+Fig. 5.19
Graphs betwcen Pressure and volume at various temperaturcs are<irawn using equation (d). The graphs are as shown in Fig. 5'19.In the graph, rhe horizontal portion is absent. But in its place, thecurve ABCDll is obtained. This docs not agree u'ith the experimentalisothermals {or C0, as obtained by Andreu's. Houever, the portion1J3 has been explairrcd a.s due to supercooling of the vapouts anci theprrrtion .ED due to super-heating of thc liquid. IJut the portion BCDc rnnot be explained because it shows decrease in volume with tu
'vi
...(iu)
,os
zRro(7"-D)t
Dividing (tu) by (u)
Substituting the value of
ry_rvV:Y"
6a7] : t-t
6o 2RT"w: v;6)e
Y, V "-b32zVo - 3l'r-3bYo:3b
Vo : 3b in equation (ic)
2a RT"fis: @z
mBat":i_Rb
Eeat onil Thermollynamice
...(u)
.. .(r'i)
...(uir)
At thc critical noint -jS- : o,
Noture of Hedl l
TASLE
Critical Temperature and Pressure of Comraon Gasee
id..
IIelium
[[ydrogon
Nitrogen
Ait
Argon
Oxygen
Carbon dioxide
Ammonie
Chlorino
Wator
Stfrata,rnoe Criticollfcnpnraturc t0
-2Bgoc
--24t0"c
-146.C_140.c
-118.C3I'C
130'c
I4b-U
Critical Preacr.rr.c(atn?p{phcrro)
2.26
t2.80
33.60
39.00
48.00
60'00
73.00
15.00
76.00
218.00
Substituting these values of 7" and ?" in equation (ii)D ExBa a'o: TEW - 56r-
^ \
P": h'.
'lt t$ ')"'o""
/V5.38 Corresponding stater f
Two gases are said-to be in corresponding rtates if the rarios oltheir actual pressurc, vqlprg and temperatura and cridcal pressure,critical volume and critical temperature have the same vilue. IiIDCatrg
Pr Pz::-Ed Po,
Yr Y,W: N;
cnd +: #lgl lal
...(i)
...(rd)
,.,1,1i1)
...(iu)
5.94 C<refficient of Van der Waals Constants
^-J, t r,r* \(\^, (,"/ t',:3b'qg P":;o
msa'" : D-Ei-Fr'.,m s*r.,ions (rioi; and (ii)
T,, 61 az 27t'2P, (27 )' il':bz n
@
Dividing (t'di) by (dr')
6*a:TIR'o1 1)24 2Ll 1t t Co .: "6+ -F:
f , 8o 27ba
P; : TEi X-a-B6: lE-
r, - Ri"-- gP" ...(u)
?10 Eeal and Thcrmodynomica
...(ui)
Nahrc ol Ecat
(;i)
Notc, Eoro V " -
moloouler w0 x apecif,c volumo.
The experimental value of the critical coefficient of all gases isgreater than thc theoretical value of2'67.
- -E,-gple 5'2{. Calculate the Yan iler Waols constanta lor dryoir, gioen that
?6 : 732 K, P" - 37'2 atmosplwreo,
B per mole : 82'07 cml atmos K-1.Here Pr:37'2 atmospheres
llo : 132 KR : 82'07 cm! atmos K-r(0 ":#ryo: /11\(t3'ozl'032)'- \3rl-r7E-a : lt.3l x l(F etuog clqr
@elhi 1975)
^- RT,
" - -8lE, 82'07 x I32O:--
8 x37'2or D:36'41 crnr5.+0 Reduced Equation of State
- -Taking the pressure, volume and tcmpcrature of a gas in tcrmr.of reduced prcssure, volume and temperature,
PVT-F:-: A, *O- : F, 7,- : Tt c f 2 14
P:a.P",T-PTo,T:\ToAccording to Van der Waals eguation
(r*#) rr-rr : Rr
( -r"*&)rF r,"-u) - Ry r"
But P, :;i,Y":3b
and T": h. ly.:,+)re.3a_6t:y#
127 b'rPz t
[.* p', ] [3B-r] : sy
This is the reduced equation of state ior a gas. If two gaseshave the same val rres o[ a, B and 1 they are said to ie in corresfrnd-ing states.
5 4l Properties of Matter Near Critical poiat
Based on the experiments of Andreu,s, Amagat and others, thestate of matrer near the critical pcint can bi srrmmarised zrsfollows :--
(l) The-dersities of the vapour and, the liquid gradualyapproach each other and their ciensities "r" .qrril at t$'c criticalpoint.
(21 At the. critical point or just aear the critical point, theline of demarcation between the liquid and the vapcur d-lsappears.Consequentiy, there must exist riutual <iiEusion'bet*r.cn thl r*ophases and the surlhce tension mrrgt tre zero. This alro meaus rhatthe forces of inter-molecular atrraction fur the liquid and vapourstater must be equal at the critical point-
AkoRT, B (tu) .27U-m - Tfre@R?, 8-P};: -t )
The quantity i-i-rt*T * calted the critical coe$.cient of a gas.
Its calculated value : 3 and it i3
is the same for all gases. l. L.,) \,+-:r.,,aC!O(.
The experimental values of the critical coefficient for differentgases is given below.
0,TABI,E
F.rperimeatal Values of Gttical GoefEcient -$#
Pain atm.
Spccit oolulmadn cmlg
RTo7Vo
Srbel;l;ncc
Ilelium
Eydrogen
Nitrogoa
Orygen
Carbon dioxido
Water
T"inK
6l
33.1
t26.9
164.2
so4
u7
2.26
r2.8
33'6
19.7
7Ztg
2r8
I6.4
2.)-o
3.21
2.52
2.17
3.181
3'lt8.28
8.12
g.a
3.{r
4.30
...(d)
...(i;) .
2tl Eeat anil Thermoilgnamict
Write short notes on :
Mean free pathjoule-Thomson EffectCo;rtinuity of state
Rowland's experimcnt.for finding JVan der W"als equation of state
Pressure exerted bY an ideal gas
Critical constantsDegrees of freedomAtomicity of gases
Maxwell's iaw of distribution of velocity. (Delhi 1975'S
Andrelr's' experimentsAmagat's expcrimentsHalborn's experimentsBehaviour ofgases at high PrcssureCritical Pointftrresponding states
Intermolecular attractionTcopcraturc of invcrsionReduced equation of state for a gas lDelii lfiorta.) 19761
Porous plug cxPerimcnt"
Thermodynamics
6'l Therraodynam,ic Systern
A thermodynamic sysi em is one which can be described interms of the thermodynam,c co-ordinates. The co.ordinates of athermodynamic system'can be specified by any pair of quantities ufz.,pressure (P), volume (Z), temperature (f) and entropy (B). Thethermodynamic systems in engineering are gas, vapour, steam, mix-ture of gasoline vapour and air, ammonia vapo'-:rs and its liquid. InPhysics, thermodynamics includes besides the abovc, systems likcstretched rvires, thermocouples, magnetic materials, e]ectrical con-denser, electrical cells, solids and surface films.
Examples : 1. Stretched wire. In a stretched rvire, to findthe Young's modulus of a wire by stretching, the complete thermo-dynamic co-ordinates are
(o) thc stretching force /(D) the length of the stretching wire and(o) lhe temperature of the wire.
The pressure and volume are considered to be constant.
2. Surface Fihas. For liquid films, in the study of surfacetension, the thermodynamic co-ordinates are
(@) the surface tension(b) the area of the film and(c) the temperature.
3. Revcreible Cells. The thermodynamic coordinates tocompletely describe a reversible cell are
(a) thc E.M.F. of the cell(b) the charge that flows and(c) thc temPeraturc.
its centre of mass is equal to the total- random kinetic energy of allits nrolecules ? (N'lolicular weight of hydrogen : 2l-
^^-[Ao". l'93 x 106 cm/s]
61. Calculate the temPerature at which the r.m.s' velocitv ofa helium molecule will be equal to the speed of the earth's 6rstsatellite i.e., 7, : B km/s. [Ans. 10'28 x lOp K]
62. Calculate the mean kinetic energy of a molecuie of a gas
at 1,000'C. Given,E : B'31 x l0? ergs/gram mol-KN : 6.02x lOa
(Delhi 1969) lltta. 2'07x l0-rr ergpl
63. If the density of nitrogen is l'25 g/litre at N'T'P', calcu-
late the R.M.S. velocity of its molecules.
lDelhi 1972 ;'Delhi (Eons.\ 19731 lLne. 4'95xlS cm/sl
6{. At what temperature is the R.M.S. speed of oxygen mole'cules twice their R.M.S. speed at27"C?' (Delhi 1973) [Ane. 927'C]
65. Calculate the R.M.S. velocity of the molecules of hydro'gen at 0oC. It4olecular weight of hydrogen : 2'016 and- R: 8'31 x I0? ergs/gram mole oC
(Dith; 197 1\ lLtr. I 8'4 x I S cm/rl
66. Calculate the R.M.S. velocity of the hydrogen moleculesat room t€mperature, given that one^ Iitre of the gas at room temPc'rature and normal pre-ssure weiglrs-0'086 g-'
(Oetni /9761 1Aoe. l'BB x lS cm/s)1
67.(i)
(ii)(dii)(iu)(r)
(oi)(oii1
luiii\(ir)(r)(ril
(rii1(riii\{ria)(lcu)
\rl,il(roiil
(x',tiii)lair\@r\
lAgro 1962 ; Delhi (826.) 1966)
(Agro 1962 ; Delhi 1974, 7 5\,
215
216 lleat and Thermoirlnamica
62 Therrnal equitibriurn and Conceot of Temperature
.\ thernrodl'narnic s)'stem is said to lle in rherrhal equilibrirtnif arry trvo c,l its independent tlrermrclvnamic co.ordinatei ,t and I:remain c()nstant as long as the e\lernal conditions remain unaltered.Consider a 3as enclosed in a cylinder fittcd rtirh a piston. If thepressure and volume of the enclosed mass of gas are P and 11 at thetempcrature o[ the surroundings, rhese values o[ P and I/ u,illremain constant as long as the external conditions t'iz. temperatureand pressure remain unaltered. The gas is said to be in thermalequilibrium with tlie surroundings.
The zeroth law of thermodynamics u,as formulated alter thefirst and the second laws of thermodynamics have been enunciated.This lau, helps to define the term temperuture of a system.
This law states that if , of three syslems, A, B ar;d, C, A and Bore separately in thernol, equilibrium utith C, then A anil B are olsoin thermal equilibrium with one a.nother:
Conversely the law can be stated as follows :
4f ,hrt.y' or more systems are in thermal contact, each to each,by roeans of diathermal walls and are all in thermal "equilibriumtogether, then any two systems taken separately are in thermalequilibrium with one another.
Consider three fluids .4, B and C. Let Pa, Pa represent thepressure and volume of -4, Ps, /s, the l)ressure and volume of B,ind Pc, Ysare the pressure and volume of C.
If .4 and B are in thermal equilibrium, thenSr(P*Vn) : 6r(Ps,Ysl
or Ir[Pl, Y7., Ps,7a] : 0
Expression (d) can be solved, andPs:/r[Pa, V*,Ys)
If B and C are in thermal equilibriumflz(Ps, Ps) : #slPc,Ycl
or frtPr, Ya, Pc,Zcl : 0
AIso Ps: f2l?s, Pc, I'c]
From equations (zi; and (diz) for .r{ and C to beequilibrium separately,
.fr(Pe, Ye,Vs) : lzlra, Pc,Yc)
...(d)
...(ir)
. . . (iir)in thermal
...(iu)If d and C are in thermal equilibrium with I separately. then
according to the zeroth law, A and C are also in thermil equiiibriumwith one another.
.'. .F'r[Pl, Ve,, Pc,7c] : 0.
Equation (it') conta;ns a variable 7s,does not contain the variable 7s. It means
$r(P* V s) : ps(Pc, I'c)
. ...(t )
whereas eguation (r)
...(ui)
rntrrne-tc4 ecl -_ it.z$2ruzr,-Thirmoilynam;cs 217
In general,
d,(Ie, I'a) - cr(,I'o, I'n) : dr(Pc, I'c) ....r,ii)These three functions have the same numerical valrre thougi,
the parameters (P, [/1 of each are diffcrent. This nurnerical valireis termed as tentperatur': (T) af the body.
...(riii;This is called the equation o[ stare o[ the nuid. -,Q'Therefore, the temperature of a system can be defined as
the property that de termines ,rvhether or not the body is in thermalequilibrium with the neighborrring sy!rems. If a nuntl,rer. oIs,vstemsare in therm"] tquiiibrium, this cor,:rnon property of the system canbe represented by a single numerical value called the temperature,It means that if two systems are rrot in thermal equilibrium, t{reyare at different temperatur r.
_ Example. fn a nr"rcury in glass thermometer, the pressureabove the mercury column is zero and volume of mercury measuresthe temperature. If a thermometer shorvs a constant reading.in twosystems. A and_B separately,. it will show the same reading evenwhen .r{ and B are brought in contact.
6'3 Concept of Heat
Heat is defined as energy in transit. As it is not possible tospeak of work in a body, it is also not possible to speak oi heat in abody. Work is either done on a body or by a body. Similarly, heatcan flow from a body or to a body. If a body is at a constant t"m-perature, it has both mechanical and therrnal energies dqe to themolecular agitations and it is not possible to separaie them. So, inthis case, we cannot talk of heat energy. It mtans, if the flovr ofheat stops, the word heat cannot be used.' It is only used whenthere is transfer of energy between two or more systems.
Consider two systems.r{ and B in thermal contact $,ith oneanother and surrounded by adialatic walls.
For the sJstem /,fl : Ar-arlW ...(r)
where II is the heat energy transferred, U1 is the initial internal€nergy, U2 is the frnal internal energy and W is the work done.
Similarly for the sPtem B, '
E', - U2',-Ur',*W',
Adding (i) and (ii)E + E' : (Ur-Ur)+ril + (U 2' -At') +W'fl +E' - llU t*a r')-(Ur*Ur'))+ (f +W')
...{r4
... (idr)
Thc total change in the internal energy of the composite system
[(u.*ur,)_ (4*u i))
2t8
The work Lrne by the cor^rposite system : W +W'It means that the heat transferred by the composite s1'stem
: E +H'. But the composite system is surrounded by adiabaticwalls and the net heat ransferred is zero.
u*'; jo-u, ...(iu)
Thus, for two systems A arrd B in thermal contact with each
other, and the composite system sur-rounded b-y ad.iaba.tic walls, the
heat gained by one-system' is equal to the heat lost by the othersystem,
6'4 Quasietatic Proceee
A system in thermodynamical equilibrium must satisfy thefollowing requirements strictly :-
(i) Mechanical Equilibriuro. For a system to be in mecha-
nical equilibrium, there should be no unbalanced forces acting on
".ry p.it of the system or the system as a whole'
(di) Thermal Equilibrium. For a system to be in thermalequiliLrium, rhere should be no-temperature difference between thepirts of the system or between the system and the surroundings.
(iid; Cheaical Equilibrium. {or- a system to be in chemicalequiliLrium, there should be no chemical reaction r.r'ithin the systero.rid ulro no movemerlt of any chemical constituent from one part ofthe system to the other
When a system is in thermodynamic equiiibrium and thesurroundings are kept unchanged, there will be no motion and alsooo *c.k wi'il be dorie. On theother hand, if the sum of the exter'nal forces is altered, resulting in a finite unbalanced force actir:3 onthe system, the condition for mechanical eguilibrium wiil not besatisfied any longer. This results in thc following :-
(i) Due to unbalanced forces within the -system, turbulence,waves'etc. may be set uP. Thesystem as a whole may possess anaccelerated motion-
(dt) Due to turbulence, acceleration etc' the temperature dis-uibution within the system may not be uniform. There may alsoexist a finite temperature difference between the systelo and thesurrouncl'ngs.
(iii.y Due to the presence of unbalanced fprces and differencein temperature, chemical reaction may take place or there may bemovement of a chemical constituent.
From this discussion, it is clear that a finite unbalanced forcemay cause the system to pass through non'eguilibrdurn states. Ifduring a thermodynamic process, it is desired-to describe every stateof a system by thermodynamic co_ordinates_referred to-the system rua whole, the process should not be brought about by a finite un-balanced force.
Thermodynamicc 21'1
A quasistatic process is definei as the Pr99c: in which -thedeviation from theimodynamic equilibrium is infinitesimal and aU
the states through which the system passes during a quasistaticprocess can be considered as equilibrium states.
In actual practice' many proc$s.e-s :losely approach a quasista'tic orocess and mav be trcated as such with no signihcant eror.Consider the exparsion ofa gas in-a closed-cylinder fitted withapiston. Initially'weights are on the piston and the pressure of the gas
inside the cylinder ii trigtrer than the atmospheric pressure. If thewershts are'small and aie taken off slowly one by one, the process
"^rii" considered quasistatic. If, however, all the -weights are re'
'moved at once, expinsion takes place-suddenly and it-will be a non'equilibrium p.oceis. The system will not be in equilibrium at anytiiee during this Process.
A quasistatic process is an idcal concept that is applicable toall thermodynamic iystems includin_g electric and magnetic systems.
It should bi noted that conditions for such a process can never bcsatisfied rigorouslY in Practice.
6'5 IIeat-A Path Function
Heat is a path function. When a syst€m chalSes from a stateI to state 2, thiquantity of heat transferred will depend upon thcintermediate stagls through which the- systern passes i;e., its path.Hence heat is an ineract differential and is written as 8II.
On integrating, we get
[*m:E'olla la
Here, 1EI, represents the heat transferred -during thc givenprocss bctwien the states I and 2 along a pardcular Path A.
6'6 \f,orlr-A Path FunctionSuppose that a systen is taken from an- initial equilibrium
state I to'a final equilibrium state 2 by- two different paths A and Btfig. O't). The prbcesses are quasistatic.
H eat anil ThcrmoilYnamics
tI
P
V _---+a[. 0.1
250 Heat and Thermodynomica
l'lrc;.r'els rrndcr these curves are different and hence thelrrrntitics of rvor k clorte .rle :rlso different.
For the p;ulr .\,r2AI pdt'J r.{
r28I PdVJIB
It is customary to represent, \.vork done by the st'stem as {"r'.,work done on the syste:-o as -ve, heat flowing into tl,e s,'stem ras
^1-ve, and heat flowing out of the system as -ve.
frr)rt First Law of Thermodynarnicsx' vJ
heat produced. It is true rvhen the whole ol the rvork done is usedin producing heat or t'ice oersa. Here,W : JE rvhere J is theJoule'smechanical equivalent o[ heat. But in practice, rvfen a certainquantity of heat is supplied to a svstem the whole of ttr-e heat energymay not be converted into 'a'ork. Part of the heat may bc used indoing exrernal rvork and the rest ()[ the heat :nighr be used inincreasing the internal enerqv of the molecule(Qflet the quantityof heat supplied to a system be 8I/, the amorrntYF-external rvorkdonp be 87 and the increase in internaI energy of the molecules berlu)/Ine te rm U represenrs the internal energy"of a gas due to mole-cylar agitation as well as due to the forces of inter-molecular attrac-tion. lflathematically
\ w : aa 7m') ...(d)
@
condition. Therefore for a cyclic process $aU : O
$ r': fa'
Thermodynarnics
and
22t
r 3'\ll'A:l 6lf:J l.{
For the path BrlB
;r-e - .l re 8rr/ :
...(i)
. . .(i,)
The values of II'e and Il'6 are not equal. Therefore workcannct be expressed as a difference between the values of someproperty of the system in the two states. ThereJore, it is rtot cotectlo represettt
Ir, : []l:, BtF : ty, _ wlJW'
... i;ii)It may be pointed out rhat it is meaningless to say "rvork in a
system or rvork of a svstem". Work cann^t be interpreted similar totemperature or pressure of a system.
In terms of calculus 817 is an inexact differential. It means
that F is not a property of the systen, una j 817 cannot be express.
ed as the difference between two quantities that depend entirelyon the initial arrd the 6rral states.
_ Hence, heat and work are path functioas and they dependonly on the process They are not point functions such ai preisureor temperature. Work done in taking ttre slstem from state I tostate 2 will be different for differenr paths.
6'7 Gomparisoo of Heat and l{ortThere are many similarities between heat and work. These
.are :-I. Heat and work are both transient phenomena. Systems do
not possess heat or work.2. When a system tindereoes a change, heat transfer or work
done may occur.3. Heat and work are boundary phenomena. They are obser-
ved at the boundary of the sysiem.
4. Heat and work represent the energy crossing the boundaryof the system.
5. Heat and work are path functions and hence they are inexactdifferentials. They are wriuin as 8,8 and 8I7.
6. (a) Eeut.is defined as the form of energy that is traruf€rredacross a boundary by virtue of difference of iimpcrature or tem-perature gradient. -
(b) Work is said to be done by a system if the sole effect onthings external to the system could be the raising of a weight.
...(14lBc,th arc erpressed in l'reat urrits].'I-lris erluation represents Joule's law.
For a svsrem carried through a cyclic process, its initial andlinal internal' energies are cqual.' From the first law of ti,.rmodv"a-mics, for a system undergoing any number of complete cvcles
ur-ut : sl
" D,H : 16 rrrY:rH : llt [Both are in heat units]
6.9 First Law of Thermodynarnice for a Change ia State ofa Closed System
, Fo.r a clo.sed. system during a complete cycle, the first lau. ofthermodynamrcs rs lvrrttell as
" aa : rli st;,Yr
229
In practice,rathcr than a cYcle.
Heat and Thermodyrumiu
however, we are alco conccrned with a PJocclBLet thi system undergo a cycle, changing its
Tlxrmodgnamict
(8.e-8If) depends only on the initial and thc final states of thesvsrem and is independent of the path followed between the twostates,
Let d.E : (8H-Etr)From the above logic, it can be seen that
2
I an : constant and is indepcndent of the path.JI
This naturally suggests that E is a point function and dE rr an.cxact differential.
The point function .O is a iroperty of the system.
Here dD is the derivative of E and. it is an eract differential.8H-8W : dE ...(du)
8H : d0qtW ...(r,)
Integtating equation (o), from the initial state I to the finalstate 2
tflt - (E,_E)+LW,[Notc. 1EI1 carmot be writton ae (EI1-II1), bocsueo it dependa upou
the pathl.
Similarly, ,}[. cannot be written as (W1-W1), because it alsodcpends upon the path.
Here lEIs represents the heat transferred,
1fl1 represents the work done,
E, represents the total encrgy of the system instate 2,
E, represints the total energy of the system instate l.
At this point, it is worthwhile discussing what this E c^npossibly mean. With reference to the system, the^ energies_crossingi:he boundaries are all taken care of in the form of E and W. Foi.dimensional stability of Eq. (u), this f mustbe energy and this mustbelong to the system. Therefore,
E2 represents the energy of the system in state 2
E, represents the energy of the system in state I
This energy E acquires a value at arry given equilibrium con-dition by virtue of its thermodyna-mic state. The working substance,for example a gas, has molecules moving in all randbm fashion.The moleiules have energy associated by virtue of mutual attractionand this part is similar to the potential energy ofa body in macro.scopic terms. _ They also have ve locities and hence kinetic energy.'fhis energy E therefore can be visualised as comprisine ctf moleculirpotential and kinetic energies in addition- to maLioscopic potentialind. kinetic energies. The first part, which owes its rxrstence ro the
223
H;-state from I to 2 along the path :{ and from ? to I aiong-the
-path B'
This cyclic Process is iepresented in the P-Y diagcam (Fig' 6'2)'
According to the first law of thermodyaamicl
ftr-{twFor the comPletc cYclic Process
2A r8 2A 18
I,rt'* J ta - {,t'* I,:'
Now, consider the second cycle-in lvhich the-rystemchangesfrom state'l to statc 2 along the paih /, and returns from state 2 to
rtate I along the Path C. For this cyclic process
2A. lo 2a lo
l,** [:': [,'** I;*Subtracting (id) from (;)
IB LO 18
I*'"- [:'- L''-l8 LO
or Ittr-tr):l(DE-Dr)l, zo
Ilere I and C represent arbitrary processe between the statesI and 2. ?herefore, it-can be concluded that thc quantity (88*8W)ig the same for all processes between the stdtes I and 2. The quantity
I
ro
[,,)ro
...(0
...(dr)
... (iii )
224
lhermodynel,,tc r'l:l,rrre is often cailcd tl,c internal enercycomFletely deper,derrt on thc ihermodynamic stare. and'trvo depend o,t mecllarric;rl or phvsicaI surte oI thc system
E : U yKE + PE +Othcls whiclr depenrl uponnature etc.
For a closed s),stem inon-chemical) rhe clranges inexcept U are insignificant and
dE:dUFrom equation (r)
8H : dU +EWFlere all thq quantities are in consisrenr unirs
Era:nple 6 l. ll,hen o .ey.\tem is taken, lrom the state A to tLestalc B, alonT the Vath ACB,80"ioules oJ heat f.ows ilto thr systrm,anil the system dois S0joules o/Lortc 1i,;g. A.S\.
(a) y7w much heat fl.owa into the syslem along llrc ltullt ADB,iJ the work ilone is J0 jouies.
- (6) The -system ia returned, lrom the state B lo th,c stute A along
lhe curaeil poth. The work d,one-ott. lhe syslem is 20 joules. Docs thesystenx absorb or liberale heat anil how mich ? -
(c) l! Ua --C, UD: 40 joules, finit the ltcat absurbed in the?roeess AD anil DB.
Tkrmodytwrtiu
Erpr : AB-U^+WE - 50*10 : 60 jouler
(D) For the curvcd path frcm B to A,W : -2}joules
: -50-20 : -I0.Jouler(-ve sign shows that heat is liberated by thc syrteur)(c) Dr : 0, Up - 40 joulca0s-U1 : 56
Or : 50 jouler
For DB
6.10
a:fiY,r)Differentiating equation (r)
In the Drocess /?A, l0 joules of wort is done. WorL danefrom.r{, to D is J-10 joules";d};;; J';; is zero.For A-D,
.EIrp: (Up_Ag{W
- 40+I0 - 50 joules
trIpa - Ac-Ao*W: 50-40+0 : l0 joutee
II eal and lhei.modynamics
which isthe other
chemical
all othcrs
. ..("i)
Ot! ta&d
ti
P
Applications of Firat Law of Thermodynqnic! q
Speei6c lleat of a Gae (T aod, v Independent)Thc internal energy of a system is a single valued function ofthe state variables or:2., pressure, volume.
case of a gas, any two
*ifffr,,,*.,,, *,o,li'Si*:# i.h-*T#.1'JHl *;
oo: (#), rr* (#),* ...(,')ff anamountof heat gEI is supplied to. a thcrmodyaamicallyrtem' say an idcar gas and if the'iorume lncrer*e! by ily at a
ffXo", pressure P, theu according to the 6rst law;i-6il;;.-
...(d)
Fig. 0.3
AJong the parh ACB,Ilacs : Uy*U1!W
Ilere A:+80joulesP : $30 joules
.s. +80 : as-Ua*3OU s-t t: 80-30 : 50 jouler
(a) Along the Path -r{DB,
W : *l0joules
Here8E: da+EW
8W : P.d,V
8E : dU+P.i|YSubstituting the value of dU from equation (d0
,u - (#),ur*{#\,rr*r* .. . (r:r'd)
^ -l
()r
Dividing both sidcs bY dT
#: (3.r)..(#),#*#(# ) : G+)" "[ '*(sur),){,
iI the gas is heated at constant volume,, EH\[77 .)u
: t'dv :odT
(# ),: (-37;":,.\\Ihen the gas is heated at constant Pressure,
(#),:",
H cot ond Thennodgnumict
...idr)
..'("t)
...(utO
?hermodynaniu
Herc Cp, Cy and I are expressed in the same units.
From cquation ldir;
'r: ( #)"rr*lr* (#) ,)uoFor a process at constant temPeraturc
ilT:g
(a^a)r - P(irltr+ (#),ruo, ...(t)
...(d")
...(rid)
"..(ri;i)
...qrio|
From Joule's experiment, for an ideal gas on opcning thc stop'*n.k, i-ro *o"tk u'as done and no heat transfer took placc'
So 8.8 : 0 : dU +0. Thercfore, dA : g. Even thougt thco,.'-rlrrml- cnu'gea while the tcmperature is coostant, there is no
cirange in internal energ'Y'
(ao_\:0\ aF-1, - "
lirora ttre ideal gas eguationPY ':' R?
i:r P (*F;,:ocp-.c1 : r, l# ),*(p), (# ),
This equation rePresents the amount of heat energy -suppliedto a system ii an isothlrmal reversible process and is equal
- to thc
s,'.m of the work done by the systeE aird the increase in its internalenergy.
For a reversible adiabatic process
8E :9,Therefore, from equation (fr),
o: cv ur+lr. (#) ,T,cydr :- [r* (#),y,
Dividing throughout by dV,
,,(#) :-[r*(*#),]The isobaric volume coefficient of expansion
": +(#),",: (#),
ce-.cv -' (-# ),Cp-Cv D-
-:t
aV
'* (*f ),: o * F--F
(#), * ("#')-,*(e,#'):-[on(iF;,1
...(u)
From equation (iu)'
,,: (#)"*[ r*(#),](#),c" : c"*f ,*(#),1(#),
c,-cv: [ ,*( #),](*f ),
/:lL \F'r,t i '^,, 1
l(t T
L'P* {)!
'. l,y *'- ,
=,.( ;i)u:o,* "H .. . t.ilii i
or
228
or
From equations (rid) and (rio)
"(#) :-(#)
/ a" \ Cv-Qt\aTJ:-W
Thir expression holds good for an adiabatic reversible Proccss.
6'11 lrotLcrmal Procege
ftf " "rrt.- is pcrfectly conducting to the surroundings and
tl" t.g-p.rli"r. *-"i* cotirtarrt throughout the process, it is called
Heal ond ThcntoilY*amio
V--+
frg. 0'a
anisotherma!process)_9'yidf 1,3i"i*,c--':b::1ll'"::";;fi;""*ir;-l.o4t"t. u"a nt"i"g a vol-ume represented
hrrroundings and itl temPeratureof the
substance, there is rise in temPerature because the extcrnal worltdone on the working substance increases its internal erre-rgy. \{}rr:nwork is done by thJworking substance, it is done at the cr:st of itsinternal energy. As the system is perfectly insulated frorn thesurroundings, there is fall in temPerature.
[Ttrrr, during an adiabatic process, the working substance r:rrerlettly insulated from the surroundings. AII along the process,
ihere is change in temperatureJ A curve between pressure andvolume during the adiabatic pro"cess is called an adiabatic curve oran adiabatic.
Examples. l. The compression of the mi,xture of oil vapoureld air during compression stroke of ao internal combustioa is anadiabatic proCess and there is rige in temPeralure.
2. The expansion of the combustion product! during theworking stroLe of an engine is an adiabatic process and there is failin temperafitre.
3. The sudden bursting of a cycle tube is an adiabatic
Processr.
Apply the first law of thcrnodyuamics to an adiabatic Process,$.EI - 0,
Tlwmodynomir"r 229
8.8: dU+Etr0 -du+8IF ...(d)
The procesd that takc place edlilenly or quickly are adiabatic
,..(rt)
certainby the
inint a (Fig. 6'a).
Pressure is decreased and work is done_ by the w.orking sub'
.*r,".^"[f.-";;;i-i" i"t"t"tl tt'"tgy and *eie shor{d be farl instance at the cost oI- its rnt€rnal energy an<t ulerc Juuutu uE 14'
F;+F{{ts.*::l::r*?:f*.*,,r:d::'f '-':,**::x1:f, ;:-;;;;'#J[3f i;;Ptheiurroundilq-3:d-::'::::::::::ilit t.-p"totr... Thus from A to B the temperature remalns cons'--- --;i^-
^.,^,^ a Il io -ollarl t}'e iaotLcrmal curve or uothermal,[",.*Ttl,:ai,-. t a.a it Lrr.a the ieotbermal c-ur've or iaothermai'
Proccsses.6.13 Iroc.horicProcesr
fif ,U" working substance is takea in a non'expanding cham.ber,
the h-eat supplied riill increase thc presstrre and teryperature. Thevolume of thc substance will rcmain coultant. Such a process iscrllsd at iaochoria groceee.l The work done ir zero because there irnochange in voluml. Th"c whole of the. heat .supplied ineeasesthc inteil,al energy. Therefore, during thc isochoric Process 8F :0.
Consi<ler the working substance -at-the point B and let the
oror,rif-. i""r*"a. Ex"ternal work is done on the workiog sub'
5;:;;Jtu.t. tuo"rd te rise in temperatlT",', !i13",-1v.t::Ti{:*:E *iis::, x :H:::"ff:J?*k, ji"1';r,'ffi }? : "'
8E:dAThe heat transferred in guch a Process
Efl : A#rcrilT - da
...(0
"..(id)
/fnrrt. durinc the isothermd process, tle temperature ol the
*orfio g t.rLsta nc i rem ai ns ronltant. . It.-?l- ?-1Y:jS:iSi:il;;;.d;-r;;J[8r. ru. eguation for an isothermal procers is
PY : BT : conltant [For oae gram molecule of a 88s']
For r gram moleculer of a gar PY : *nf )6.12 Adiabrrtc Procc!'
Hence C, is the specific heat ficr one grarn'molecule of a gas arconstant volume.
6.f.1 Ieobtrlc Procerfif O"workingsubstanceis tahen ia aq crpanding chamber
t.ptLJ. constant !r.rrrrr", the pocesr is called an is.rir-^" prt''c:s)Hire, the temperaturc and voluml change. If an amount of heafEf,I is-civen to-the working substance' it is partly used in increasilgthe teilperature of the wolking substaace by ilI and 'r-'t ^ 'in doin-g external work. Considering one gra* - arrrt'ul[substance,
2E0 Eeol ond Thermodgnamice
.. . (i)
...(ii)
... (iii)
. ..(dd)
?hermodynarnic*"
C,.P.dV +C,.Y 'dPlCr.PdV -CoPdl: : o
Ce.P.dV+Co.V.dP -'0Dividing by C,PY,
c) ilv dP
i'-r-+-P:uBut
8EI - I xlrdz+P$But 8E - OydT
P.dY : r'dT
Caill - C..d7+ +o'-cn: +
Here C, and O, rePrescnt the specific heats for I gmm of a gas
and r is the ordinarY gas constanL
If C, and Crtte the gram-molecular specifrc heatu of gas' then
o,-c, : + ...(io)
Hffiis the universal gas corutant'
-ttS/ Gas Equation Duriog rn Adlebedc Processv
Corrrid.r I gr-am cf the work-ing substance (ideal gas) perfectly
ioruf"iJ-iro* tfrE r"r."""ai"gt. Lei the external work done by the
gas be 87.Applying the 6rst law of thermodynamics
8.E[ : da +8wBut 8E: g
and 8W : P'dY
where P is the preslure of thc gas and dP is the change in volume'
o : da+!-*!
co : ,C,
dP dVJY --:: {)P II Y
Integrating, log P1Y log Y - const.
log PYt : coost.
PY't - const. VT--'--"This is the equation ccnnecting pressure and volume tl 'rr:rr:
,W+\.fu)"*r
adiabatic process.
Taking PY:rTo-'?'-v
,"1 ";;#, d;;;i' i"u it temPeratdrc bY??'dA - lx?,xilT
c,dr+!+:9' r
Also
,t
Thu)during an adiabatic Process
'$ ,, PPI : const'
$rY TY-'."- const. and
t .ft'tlit\/b: const.
i,rzrrot" 6'2. A rtotor oar tYre
"niult tln-room ternperature of 27oC'
furd the reaulting tertPerolure.
/rT\(-r-,) ' i'Y: const'
But r is const.fTYr-t : Colutt.
rYl-r : const.
r?.-_{: CCflSt.
: COIrSt,
: COISI.
...(4
As the external work is done by the gas at thc cost of its inter'Y
rT 1'r_lPJflr'fPFTPf .L
T
haa o preslure of 2 ar,r;If the tgre audilenig bu'-
| : c'-c,
For an idcal gas- Pv:q ^ "'(;;;1Differcntiating,
P.dYqv.dP : r'itT -'.1? -B#Substituting thc valuc of dT it equation (di)'
a1%*ff=ocolP.&Y+Y;d4+r.'i' : o
But,
-a
( f,)":0'4log (0.5) :
-0'1204 :l'4 log T, :
log ?. -
T,::
.,2-m:aple 6.3. A ou.antitg_o!.air at Z\"C anil atmoapheric pret-., .euilenly ccmpre.iaeil rc-naiy ;ta
-or;a1i:not ool.ume. Iind thc..i preasvre anil (ii) temperatire.
td) Pr : I atarosphere ; ps : ?, 7 : 1,4
Yt:Yi Yr- +Duriag sudden compression, the progess is adiabatic
P,Y,, : P,Y,T
Pr: Pr[ t)': I[2]1..: 2.636 atmoqpheres
iii)P1 -P; yr: +rr-t:
Trl7r;t-r -,r=
Eeal and ?he.rmodyaamict
2 atmospberes27? +27300 KI atmospheret1.4
Prr -r-a-a-rt
( ?,\'\T" J
/ 7', 1t'r\300/l'4 fiog
"1-log 300]
I'4 los fi-3.40803.4680-0.12043.34763.3+76-lT-
2.39t I2.16.1 K-26.9.C
300 K i?t: I1.4Ir lYryt-t?r121t.t-t30q2y.t39s.9 Kl2i.9.C i
J
Tltcrrnodgnamiu giE
. Erlmple 6 {. 4i, it. compreased adiabatically to ha$ itet'olume. Calculate the clmnge in ita lemperature. -
@eth; lbAglLet the initial temperature be ?, K anC the 6nal tenperatureT,K,Initial volume : VtFinal volumc : V2
:vl2
During an adiabatic processTrYrt-r : Trl/rr-t
m ^f l', -17-rr2: r,Ly, J
T, : Tr121t-t
_7 for air : l'40?, : Tr121r'ro-tT1 : Tr121o.to
?t : l'319 ?tChange in temperature
* Tr-Tt: l.3lg rr_T,: 0.319 Tr K
n- Exampl"- 6.9. I g_ra1n molecule o! a monoatomic 1y : 5l3lperfect gas ot 2)"C ie adiabatically compieaaeil in a ,"r"rr;bi D"rx,costrom an initiol pressure o! 1 atmoaltierc to a final grcssr;c of d0otrnos7lherea,Calculatethe.resultingdifferenceintimp'eiotutc.
LDcthi (Eo*.11978)In a reversible adiabatic process
Ptz-r P't-r-Tr't : V;-or (+)'-':(+)'Here, Pr: 50,
&- l,Tt: 273*27
:300KTt:?Y- 5'3
(50;an_(#-)'"q
Iog (50) : -i- ft.S fl-log 3001
?r - 1,{3{ K: 11161r
Pr:,v_rl
-
DIt:qt_-t
-,:Prr-,-TJ-
(+)'-'
But
lltll,'
2
T-
234 Heat and ThermodYnamirs
rXlanople 6'6. A quantity o! dry air at 27'C- is compresacd(il elowiuanld G;l auddenli to 113 r,7 its rolume. Finil the clnnge iniifrTature in each c,aae, a.*su.n1.ing 7 to be i'4 lor dry air. -- . -
lAgra 1969 ; Delhi 71,7 i)(l) When the process is slovu'. the tenrperature of thejlslgm
@ thCre is ng:1,"$-.:-.,temoerature.
L@;fifnermodynamia // z--V ufirlopee of Adiabatics and Isothermrls
^ ,Y In an isothermal Process(v PV:const.Differentiating, " vJt =-?A{PdV+VdP : O =)
or #: -+In an adiaba""
!17it .or,r,.
..(t(2) When the compressiou is sudden, tire process is aCiabatic.
.Here Y1 :V, nr:I?1 :300K, ?r-?'t : l'4 ^'
Tr lrrlt'r - Trlrrlv-r
12:,1;l;1,,?2 : 3oo I gll' '--- LV J
: 300 [3]t'r-t
^ =
i8i:i"tThe temperature of air increases by
192.5-27 : 165'5'C or f65'5 K
''Xi"*T|1)u, : o .=7v ? w =Px v{*t A{- dP ''1P
-dT : -,n-T'herefore, the slope of an adiabatic is Y
isothermal.
.. .(tr)
times the slope of the
/Eranple 6'7. A aertain mass oJ gas cf NTP is -ea9tanded,
lo
-threc times-ita t)olwme uniler adiabatio conditions. Calculote the
"e;ffiiig femperoture onil preature. '( for the gas ia t. 40.
lDelhi (Eons.) i5!Here, Y1o 7, -V, : 3l'
Tr.: 273 K T2 * ?
?rrrt-r : IlYrt'r12: 11 [+]' '
rz: 27s[+]"-'Ts*176K:-97'C
Here, Y1 : Y, l't : 3V
Pr: latmosphere, Pr:'!PrVrt - P1V1l
P2: P1[+]'
,Pz: | (+I'P, : 0.21{8 atmosphere J
V*Fig. 6'6
Hence, tlre adiabatic curve is steeper- than the isothermal curve(Fig. 6$, utL poi"t where the two curves intersect each other' \ *.L7 Work Done During an Isotherrral P:eggss
When a gas is allowed to expand isothermally, work is done
bv ir.k I t.o, the initial and final volumes be 71 and 73 respectively. Inri;.Y'6, ih. .t". of the shade{.s-1rip represents the work done for
"'i"^fi'"to"g" in volume d7. When the volume changes from 71
to Yr,
Work dcne - [:t P . dY J arca aBba ...(r)^ )vr
Fis. 6'6 represents the indicator diagram' Considering one
gram m6lecule of the rii : A,oRTorv
(l)
(2)
W-
V ----+Fig. 6'6
- RI log"
W = RTx 2.3026 log,,,
Also l)1v1 = Pllt'\'1 I'lor r\= h
.olume dV = P.dVby,
ptw = RTx2.3026* togro d \__--...(i\,) t
. Here, the change in the internal energy of the system is zero (because
the temperature remains constant). So the heat transferred is equal to the work
...(id0
. Here, heat transferred is zero because the systen 5 thernallyinsulated fiom the surroundings. The decrease ir the internal*.rgy of the system (due to,fall in temperature) ir equal to the
> work done by thc system ^Dd
a$e !Qr8o.
@ Irnrg. Relrtroa Bstween Adiabetlc end rrothcrrorl Etgtlctttcrl. IeotherrnrlElesticitYDuring an irotherual Procers
PP : const
Thcdytonkt
During an adiabatic Proc€:ts'
'o:: r': for ,__VT I
w : Kli:#-l-J. ! l- -l: r-716-F:rJ
Since / and B Iie on ttre same adiabaticP1Y1t: PrYra:K
w:1+l#^-#,=]I r PrVrY &717'1n : r=7-LVp=r--7rFi- J
: * [''n-'r']Taking'f1 and ?ras the temPeratures u! q.-poinc
repectively;nd considering one Sram molecule of the gas
P1Y1 : RI1and PlYs: BT7
Substituting these values in equation (di)
w : #[nn-ar, ]
Diffeentiating,PitY+vitP - o Q v ol'g = ? Av
y.ilP D-=aJF- : 'From the definition of elasticity of a gar
?dpEi,- _
=trfV:#
Ecat ond llffiYrrr;mia
u\i"+
P
V.t
V,.. 1ii )
. (i,,)
...(d)
...(ii)
AandB
lr2
T
done.nl) d.uwork Done During an Adiabatic Process
shown by the indicator diagram (Fig. 6.7) the work done for an increase tn
,--tFig. 6.7. c(
Work done when the gas expands fioln V1 to l/: rs uiven
l,a':
tY=l PdV=AreaAlllta" I',
-..(i)
238
From (d) and (di) lE6 n Pr/2. Adlebedc EleetlcttyDuring an adiabatic proces!
P77 : const
Differentiating, P'(Yt'rdV +Yt dP : O
YdP4V-1P
From the definition of elasticity of a gas
Eeitt: #_,:#From (du) and (o),
fr.4:7PComparing (iii) and (oi)
Er41 :7E1,sThus, the adiabatic elasticity of a gas is T times the
T oermodynamica 239
rlre atmospheric pressure be Po. Tbe;ressure of air inside the verselis Pr.
The stopcock B is suddenly opened and closed just at themoment when the levels of the liquid on the two gid$ of the mano'tDeter are the sarne. Some quantiiy .of air escapes to the atm.osphere.The air insidc the vessel exfands adiabatically. The tenPerature ofair inside the vessel falls due to adiabatic expansion. The air inridethe vessel is alloli,ed to gain beat from the surroundingp and it finallyattains the temperaturelf the surr oundings. Let the pressure at thecnd be Pr and ihe diffe.ence in levels on the two sides of the mano'meter be [.
Theory. Consider a fixed mass of air left in the vessel in tlccnd. l'his mass of air has expanded from volume 71 (less than thevolume of the vessel) at preisure P, to volumeT3 at pressure P3.
The process is adiabitic aishown by the curve /B (Fig. 6'9).
PrYrr - PoYrt
. *i: (*)'Finally thc poipt C is reached. The points A and C are at thc
room tempelaturi. Thcrefore AC can be 'considercd as an isother-rnal.
Eeol and Thcrmoilynamiae
...(di0
...(iu)
...(o)
...(ui)
isothermal---
...(0
...(r0
elasticity. \.\_7-6'20 Clement aud Desorpes Method-Determinatioa of 1
, Clement and Desormes in 1Bl9 designed an experiment to find'f, the ratio between the two specifc heats of a gas.
P1Y1 : P2V2
Y2 Plv;: -4
c(4,v2)
B (8,v2)
Fig.6.0
Substituting the value d + in equation (d),Y1
Taking Iogarithms,
log P1-log .l3r : Y[ioq Pr.- log Pg]
.. iog /', *iot i-'o| * losT;1G"""
rig.0.8
The vessel .d has a capacity of 20 to 30litres and is fitteC ina iro:r containing cotton and wool. At the top end, three tubesare fitted as shown in Fig. 6'8. Through 8r, dry air is forced intothe vrssel d. Ttre stop cock B1 is closed when the pressure inside.4 i: ;li6htiy greater thau the atmospheric pressure. Let the,.{ifierence i* }evr;l arr the two sides of, the ruanometer be .& and
t:(*)'
'---:1 rj:--1 r-:--j.l E:-l L:
:::I r-----:I l:
Ihcrmoilynamice 241E e,tt and ?hermodynont:t
But P, : Po1 H and Ps - Po*lr
. v_ _tggt&+_{l-_l9gj._-' - Iog(Po1I/1-lcg (Po*t), /Po*E\
* -'tg \-7; /,_,o, (t#)
r.s (r. +)-;4;H-)t
Approximately, ,:&-#u_P;
Hence y:=.8, ...(iu)- E-hSimilarly, 1 for any gas can be determined by this method.
Ilrawbacks. When the stop-cock is opened, a series of oscilla-tions are set up. This is shown by the up and down movement ofthe liquid in the manometer. Therefore, the exact momentwhen the stopcock should be closed is nor known. The pressure maynbt be equal to the atmospheric pressure when the stop-cock isclosed. It may be higher or iess than the atmospheric pressure.Thus the result obtained rvill not be accurate.
6'2f Pertington'cMethodLummer, Pringsheim and Partington designed an apparatus
to determine the value of 1. In this method, the pressure andtemperature are measured accurately beforc and after the adiabaticerParuron.
Fis.6.10
Tne apparatus consisl of a vessel / hav{tg a capqcity between130 and 150 litrcs. The valve .t can be opened and closed suCdenly.
It is controll..d. !y " rpfuS.arrange'n-ent (Fig. 6.10). Dry air (or gas)at a pressurc higher than the arn cspheric plessure is loiced into'ihevessel ,{ and the srop-cock I is closed.
- The oil manometer .ll{ is
used to measure the pressure of air inside the vessel ;t. '.lhe bclo-T.Le_l .B (a platinum u,ire) and a sensitive galvanomerer are used inthe Wheatstone's bridgc arrangement
The vessel is surrounded by a constant temperature bath. Letthe -initial pressure -and temperature be P1 and'?1 (room tcmpera-ture). The bridge is kept slightly disturbed from the barancedposition. ,The valve .L is suddenly opened and closed. The wheat-stones bridge is at once adjusted for balanced position. Tire remrlera-ture of air inside ,{ has decreased due to adiibatic expansion oiair.Let the remperarure inside be ?o- and the atmospheric p.ess,.re Fo. rrthe ap-pararus is allowed to rcmain as such for some time. it will iai.nheat from the surroundings and the balance point gets aLtr.u"ll Inorder that the balance point 'emains undistuibed, iome piec.s oiiceare added into the watcr surrounding the vessel ,tl. whtn the icm-peraturc of water-bath is the same as that of air just
"rt.r rJi"b"ticexparuion, thc bridge will rcmain balanccd.
. ^The-tempgTtu_re ?j of the bath represents the temperature of
air aftcr the adiabatic expansion
&, . Por-,T:-T;r
1 p, \r-r_ ( T, \,\Tl : \7;/(r:lXlog P1-logPq) : y [og f1-log ?o]
v _ _ log P1-log Pot-
. As Pp P9, ", :$- T, arc known, y can lre calculated. The
value of l for air at l7"C is found to Uc i.iO:+4dqot$:s. - (l) Due to thc large volume of the vessel, the
expansion is adiabatic.
- r tl) .TA. ,,gTpT"tur€s are measured accurately just beforeano arter tne adlaDatrc expansion.
Drewbechs. This method cannot be used to find the varueof 1 at .$sh.. temperatur* because it is not possiblc t" a.tr-'-"the cooling correction accurately.
6'22 Ruclherdt's Experiodentr_ In 1929, Ruchhardt designgd an apparatus to find the value ofy. It is based on tbe principlJof mechan-ics. Air loiglj i, .""r"*ain a. big -jar - (Fig. 6'i I ). h tube of uniform
".." of ils. .""tio, isfitted and a ball of mass ry 6ts. in thS tube just like a pirt"n.-in tf,.equilibrium position, the baU h at the poin:t j. th.';;;;;.'p "fair inside the vessel, is given by
P : Poq !f-
AIR
244 /q3/ krenersible Process
E eat tn iI Thernoilgn amict
The rhermodynamical state of a system can be defined rvirhthe help of the thermodynamical coordinates of rhe s)'srem. Theslate of a system can be changed by altering the thermodynamicalcoordinates. Changing from one state to the other by changing thethermodynamical coordinates is called a ptocess.
Consider two states of a system ie., state Aand state B.Change of state fiorn z{ to B or r:icc ocrsa is a process and the direc-tion of the process u'ill depend upon a new thermodynamical coor-dinatc called entropy. All processes arc not possible in the universe.
Consider the following processes :
_(l) _Le1 two blocks .z{ and I at different tcmperatures ?, andTr(Tr;Tr\ be kept in contact but the system as a whole is insulat-ed from the surroundings. Conduction of heat takes place berweenthe blocks, the temperaiure of I falls and rhe temperatur'e of Brises and thermodynamical equilibrium will be reached.
(2) Consider a flywheel rotati.ng with an angular velocitv -.Its initial kinetic energy is |1<.,r. After some timJthe wheel comesto rest and kinetic energy is utilised in overcoming friction at thebearings. The temperature of the wheel and the blarings rises andthe increasc in their internal energy is equal to the origiial kineticenergy of the fly wheel.
(3) Consider two flasks r{ and 3 connecrcd by a glass tubep_rovided with a stop cock. Let / contain air at high pres-sure andB is evacuated. The system is isolated from the surroundinss. Ilthe stop cock-is-opencd, air rushes from .d to 8, the prorrr.i i., .4decpeases and the volume of air incrcases.
All rhe above three examples though different, are thermody-namical processes involving change in thermodynamical coordinares_,Also, in accordance wirh the fint law of thermodynamics. the princi_pl^e-of c-onservation of ene-rgy is no-t violated-becaurg the total inergyof the system is conservcd. rtis also clcar that, with the inirial co*-ditions described above, the three processes will take place.
Let us consider thc possibility of the above rhree processestaking place in the reverse direction. rn the first case, if the reverseprocess is possible, the block I should transfcr heat to ,{ and initialconditions should bc restored. rn the second case, if rhe revcrseprocess is possible, the -heat energy must again change to kineticenerg'y and the fly wheel should stait rotating with the initial angu-lar velocity_ar. In,the third case, if the revcrse process is possiblethe air in B must flow back to / and the initial corrdition shluld beobtained.
- But, it is a matter of common experiencc, that none of theabove conditions for the reverse processei are riached. rt -."n,that the direction of the pr-ojess carxngrbe determi"ia Uy L.*i"gthe thermodynamical coordinates in the two end states. 'r" a.t.r-mine the direction of the proccss a sew thcrmodynamical coordinatehrs been devised by Cliusius and this ir c"tfe-a'Ge .-..Ji.T,n.sysrem. similar to internal cDGrg[r entropy is also a functi6n of rhe
rrate of a syst:.rn. . For any possible process, the entropy of an isolat-rd lystem should increase or remain constant. The piocess in r.r'hichthere is a possibility ofdecrease in entropy cannot take place.
- If th-e enrropy of an isolated system is_ maximum, any changcof statc will mean decrcase in entropy and hence that change6frtate will not take place.
To conclude, procases in wi;ch the enlrol;y oJ an iaolateilsudem ilecreaaes ilo nol lttke place or tor oll processea toking placein an isolateil syatem the entropu oJ the system shoulil increase orremrin constont. It means a _process is irreversible if thc entropydecreases when the direction of rhe process is reversed. A processi!qid to be irreversiblc if it cannot be retraced bggL3fi.e"flI1x_iE-.-o'ffigin! an lrrt-vCisibic pro.efi hJat-iiiffy isal ways used to overcomE-fiiEtion. -Encrgyft also 6i$ipa iEfi n:flreform-of-gonrdu-c1ion and'i-aaliatioil. This loss of ;iieiEy alw-efi'=tekesp l'ate w.liiihir-ttre-cnginE'woiTilin ,one . *di.recrion or ihe revirse di-rection, Such energy cannot be regained. In actual practiie allth? engines are irreversible. If electric current is passdd through arvire, heat is produced. If the direction of the current - is reversed,heat is again produced. This is also an example of an irreversible'process. All chemical reacrions are irreversible. In general, allnatural processes are irreversible. ,
6'21 Revcrsibte Proceeg
Consider a cylinder, containing a gas at a certain pressure andtemperature. The cylinder is fitted with a frictlonless piston. Ifthe pressure is decreased, the gas expands slowly and maintains'aconstant temperature (isothermal process). The energy required forth-is,expansion is continuously drawn from the sourcc (surroundings).If the presrure on the piston is increased-, the -gas contracts sloi[yand maintains constant temperature (isothermal process). The energyliberatecl during compression is given to the sink (surroundin$).This is also true for an adiabatic process provided the process takesplace infinitely slorvly.
The process rvill not be reversible if there is any loss of heatdue to friction, radiation or conduction. If the changes take place
Thermoilynamice 2t5
rapidly, the procgss wp-not be revcrsible. The energy used in over-6fbe retraced.
any heat engine or process
he pressure and tcmperature of thc working
ldi.tlons of reversibility foras fsllows :-
or. r1$t rgl,. .-Ig+is -pro€esr, the inirial conditions of.the workingsubSTanEE c5_ be obtainedl
Thecan Dg-stat
substanec
246
must not dih-er appreciably from those of thestage of the_cycle of operation.
Jq)rAii p/ pro"or.s taking place in themust bi inEditely slor.'.
rLe-1c,11-\/ I o n- - z)(' (4 6 r4.,1-<taJt<-Wr
Ecot ond ?hermoilYnamico
surrotmdings at any
cycle of o.peration
be completely free
d.re to conduction
Thcrmoilynamiu 24?
Firet Part. According to Kelvin, the second law can also bestated as follows : '-L'#'
'/ "lt is imppSible to-_ge!_p_ c_qe!n!_oU!_ llpply of work from apody, !r eeoling_ft to a temperature_lower than that of its surround-
-l!gy--In a heat engine the working substance does some work and
rejects the remaining heat to the sink. The temperature of thesource must be higher than the surroundings and the engine will notwork when the temperatures of the source ind the sink ire the same.Take the case of a steam engine. The steam (working substance) attigh pressure is introduced into the cylinder' of theingine. Steamexpands, and it,does external work. The contents remaining behindafter doing work are rejected to the surroundings. The teriperatureof the woiking substanie rejected to the surrouidingr is higier thanthe temperature of the surroundings.
If this working substance rejectedby the first engine is used inanother enCil-e,_ it can do work and the temperature oi the workingsubstance will fall further.
I It means that the working substance can do work only if itstemperarure is higher than thai of the surroundings. --- I -
Second Part. Accordil1to _C_!""riu. :
"It is !-mpq,q!ible-to_m_a_Le b_eat_{q!y from a body at a lowerte-rnp:I?lirle ro a. body at a higher temperature withbdt doing ex-tirnal work on the working substance." _V' This part is applicable in the case-of ice plants and refrisera-tors. Heat itself cannot flow from a body at a lower temperatuie toa body ata higher temperature. But, it ii possible, if some externalwork is done on the working substance. Take the case of ammoniaice plaut. Ammonia is the working substance. Liquid ammonia atlow pressure takes heat from the br-ine solution in the brine tank andis converted to low pressure vapour. External work is done to com-press the-ammonia vapous to_ high pressure. This ammonia at highpressure is pased through coils over which water at room tempera-ture is poured. Ammooia vapour gives heat to water at room tempe-I3tulg arrd gets itself converted inio liquid again. This high pressureIiquid ammonia is throttled to low prdsure liquid ammon'ia. In thewhole process ammonia (the working substairce) takes heat frombrine solution (at a lower temperature) and gives heit to water atroom temperature (at a-higher temperature). This is possible onlydue to the external work done on ammonia by the piston in comp-ressing it.. The only work of electricity in thi ammonia ice plant'isto.move the piston to do external work on ammonia. If the exter-nal work is not done, no ice plant or refrigerator will r.r'ork. Hence,it is possible to make heat flow from a body at a lower temperatureto a. body at a higher temperature by doing exterilal work on theworklng substance.
_Thus, the second law of thermodynamics plays an imporrantpart for prlctical devices e,g.,heat engines and-refrigerators'. Thefrnt law of thermodynamics only gives the rclation- between the
*orking parts of the engine mustfrom
or radiaiion during the cycle of operation
It should be remembered that the complete reversible Processor cycle ofoperation is only an id-eal -case. In an. actual Process,therc is alwiys loss of heat-due to friction, conduction o,r radiation-
. The temperaiure and pressure of the working substance differ apprc-ciably from those of the surroundings.
,61-laS Second Lew of TheraodynaaicetDr "
A heat engine is chiefly concerned with the conversion ofheat'cnersv into mec-hanical work. A refrigerator is a device to cool acerta"i; space below the temperature ot'its surroundings' - The firstlaw of thermodynamics is a qualitative statement which does notoreclude the poisibility of the-existence of either a heat engine or a'rcfriserator. 'Thc firsi law does not contradict the Cxistence of_ai60"7;ffi;i.nt hlar-ngine or a self'acting refrigerator'
In practice, these two arc not attainable. These phenomenaare recognized and this lcd to the formulation -of a law governingthese tw6 dcvices. It is called second law of thermodynamics.
A new tcrm reservoir is used to explain the second law. Areservoir is a device having infinite thermal capacity and whichcan absorb, retain or reject inlimited qiiantity of heat without any:
. f:SlylS-Plalgk statemcnt pf the second law is as follows ;
' "It is impossible to get a continuous-supply-of-work from abodv (or ensine) which can transfer heat with a single heat reservoir.T''Th( ii a nelgative statement. According to this statcment, a single
, reservoir at i single temperature cannot continuously transfer heatinto work. It me-ans thal there should be two reservoirs for any heat
' engine. Onc-rcser+reir{callcd -thc s9.ur-cc! it t"F:l at a lrigher tem'peiutu.e and the oiher reservoir (called the sink) is taken at a Iowertemp6 --l**--tccordlfrg-ia,
this statemen t, zero degree atsolute t"-p.."tur"is not attainablE because no heat is rejected to the sink at zero deg'ree Kelvin. If an engine works between any temPeraturc higherthan zero degree Kelvin and zero degrec Eelvin, it means it uses asinsle reservoir which contradicts Belvin'Plancl('s statement of thesec"ond l,aw. Similarly, no engine can bc 100% efficient.
. In a heat engine, the engirle draws heat from the source attdafter doing some external work, it rejegts thc remaining-hcat to thesinE Thi source and sink arc of infinite thermal capacity and theyrraintain constant tcmP€rature.
l\'.-)rk done(.1\ namics
Heat atd Thermodunamica
But the second larv of thermo-v''hich heat can be converted
COI.IDUCTING;
T-----'amAT Tt
ffiATT2
Fig. 8.12.
A pcrfect non-conducting and frictionless piston is fitted intothe cylinder. The- working substance undergoes a complete cyclicoperation (Fig. 6'12).
A perfectly non.conducting stand is also provided so thatthe working substance can undergo adiabatic operation.
cf the working substance increases. lVork is done by -the workingsubstance. As the bottom is perfectly conducting to the source atternperature 7r, it absorbs heat. The process is completely isother-m;^1. The temperature remains constant. Let the amount of heat
76d the heat produced.ives the conditions undcr
Thetmoilynamics
absorbe.d by rhe rvorking substance be I/1 at theThe point J? is obtained.
Considcr one gram molecule of the workingWork done from A to B (isothermal process)
219
tcmperaturc ?,',.I
sub!tance.
having ""' irrr,ur.a',o[lsubstance. The volume
/cgr-g!-B-er.er.ibl"&et..--.-' Heat cngines are used to convert heat into mechanical u'ork.Sadi Carnot (Frcnch) conceived a theoretical engine which is freefrom all the dclects of practical engines. Its e{Eciency is maximumand rt is an ideal heat engine.
For any e'rgine, there are three essential requisites :
(l) Source. The source should bc at a fixed high temperature?1 from u'Ii-rch tlie heat engine can draw heat. It has in6nite thermalcapacity and any amount of heat can be drawn from it at constanttemperature ?r.
(2) Sink. The sink should be at a fixed lower temperatureZ, to which any amount of heat can be rejected. It also has infinitetirermal capacity and its temPerature remains constant at fr.
(3) I{orking Substancc. A cylinder with non-conductingsides and conducting bottom contains the perfect gas as the wotkingsttbstanae.
CYLINDER
WORKINGSUBSTANCE
I
I
P
(B , V3)
t,rEFG
.V.-..----.-.-}Fig.6.l3
increases. The process is completely adiabatic. Work is done bythe working substance at the cost of its ,:aternal energy. The tem-perat Te fa1ls. . Tle.working substance^undergoes- adiabatic changefrom I ro C. At C the temperature is
"1 (Fig. 6.13).
Work done from B to C (adiabatic process)
- \Yi': \i: P . dv 1 But PV't : constant : 'K
tvtd; i" Pzv'-Rra
:ll
l.u_r_ Y' I Prv, - RT,: 9"-'-RV't-r-
1-- I Ptr"r:PzYzr-RE I+I{a
| -"t,RVr-Tr} _ RLTr-rzl: __T=r__ : *_ "/_l
ff1 : Area B0flG , ...(di)
aY,: \i',' dv : Rr, ',c,+: arca ABGD
(2) Place the engine on the standDecrease the pressure on the working
ffi C.a.leo.'T(rycle
,Y\g-b tr{ Plaee the engine containing the workin} substance over-2 D' fth" source at temperature Tr. . The working sub.starice is also at a
temperature, 1r. _]J. pressure is Pt and _volume is 71 as shown bythe point d in Fig. 6'13. Decrease the pressure. The volume
250 Ecd and Thcrmodynamia
(3) Place the engine on the sink at temperature ?;. Increarethe pressure. The s'ork is done on the working substance. Ar thebase is cgnducting to the sink, the proces is isothermal. A quantityof heat IIs is rejecled to the sink at temperature ?r. Finilly thtpoint D is reached.
Work dogc from C to D (isothermal process)
\il: ff,'
,u,_ RTr,,s :+vt
- -RTrbS+
Thermodynamia
The points I and C arc on the same adiabatic
. TrIrl-r - rzyrY'rrr / 7, 1r-rT: \7;/
From (ui) and 1uid1
(+ )": (#:)"Yr l',v;: Y"
V, Ys
T:N;nr,bslL-n?,log
*l**I,,-,,Hr-H,Useful outDut W, :
----------------', : ---l;pui- T;Heat is supplicd from the source from L to B only.
E1: RTIdC +W H,_8.rl E E;:-tr;-RLr,-r,).-(+
): "rlGG)-
,H212 : l- -E;
T"rl: !-T
261
...(uit)
r : area CEID(The -ve sign indicates that work is done on
substance.)
(4) Place the engine on the insulating stand. fncreace thepressure. The volume decreases. The process is completelyadiabatic. The temperature rises and finally the point d is reached.
Work done from D to u4 (adiabatic process).
WI: ,f,, 'OU
._ _ R(Tt-T]l/'- Y-l n.ufrr: ATIaDIEA ...(?).^ b:i"
[]71and Waare equal and opposite and-cancel each othe..f 9HThe net work done by the working substance in one
"o*-pl.t.cycle ,: Area ABGEIAreaBCEO-A., O*!-OOrea DIEA
: Area ABCD
The net amount of heat absorbed by the working substance
- fl1-flgNet work : WL+Wy+W"*W,
- RrL br+*W-Rr,bs +,-^T={,,w : nr, rcsft-ar,tor h ...(o)
The points A atd D are on the same adiabatic
The Carnot's engine is perfectly reversiblc. It can be operatedin the revcrse dircction also. Then it works as a refrigcrator.The heat IIs is taken from the sink and external work is donJon theworking subitance and heat II1 is given to the sourcc at a highertemlrcrature.
The isothermal process will take place only when the pistonmcves very slowly to iive enough time for the heat transfer ti akcplacc. The aaiabatiC Prcccrs will take placc whcn the piston tnovet
...(rdd)
thc working
...(t i)
From equation (u)
W:
Efficiency
w:W-
yr
Tl
-1, - T u'-
: *-T-2 \-Z
tI
...(adrT)
TrYr:r'r : TrYr'r-r?, / 7t 1r'tE: \-%-/
extremely fast to avoid heat transfer. Any practical engine cannotsatis[y t]rese conditions.
. AII practical engines have an efficicncy less than the Carnot'sengrne.
6'27 Caroot's Engine and Refrigerator'Carnot's cycle is perfectly reversible, It can r,r'ork as a heat
engine and also as a refrigerator. When it works as a heat engine,it absorbs a quantit,v of heat .t/, from the source at a temperaturi 7r,does an amciunt of work lY and rejects an amount of heat .&, to thesink a_t temperature ?r. Wtren it rvorks as a refrigerator, it absorbsheat E2 from the sink at temperature Tr. W amount of work is doneon it by some external means and rejects'heat Hr to the source at atemperature 7r (FiS.6'14). Ir, the iecond case heat flows from abody at a lower temperature to a body at a higher temperature,with the help of external work done on the vvorking substance and itworks as a refrigerator. This will not be possible if the cycle is notcompletely reversible.
Coefficient of Perfotmance. The amount of heat absorbedat the lower temperature is 112. The amount of work done by theexternal process (input energy) : W and the amount of heat rejected: Hr Here f/, is the desired refrigerating effect.
o<0 H eot anil Thermodynamice
(iil nernce narrcn
Fig. 6.14
Coefficient of performance.ErH, : =W' : Er-r,
Suppose 200joules of energy is absorbed at the lower tempcr-ature and 100 joules of work is done with external help. Then200+1001: 300 joules are rejected at the higher temperatuie.
The coefficient of preformanceE2: -ly:
Thcrmoilynamics Z5A
E2H r-H,
,* :,300 - 200
Therefore the coefHcient of performance of a refrigerato r : 2.
In the case of a heat engine, the efficiency cannot be more.than l000/6 bu-t in'the-case of a refrigeiator, the
'coefficient of per-formance can be much higher than 1OOo1.
/'.r-ig+ D ample 6'8. Iinil the efi.ciency of the Carnot't etgine aorb-T ing detween the ateam point anil the ice point.
Tr :273*100 : 373 KI \ '::',':t:273K
Tr
i 27g t00' :t-zlT: Wg
o/eefficiency: $xfOO
: 26.91%
Eraaple 6.9. .?inil the effwieacy oJ o Carnot'c cnginc worlcingbetween 127"C onil 27"C.
It:273*127 : 46911Tt :273+27 : 300 K
I:L+: l-jP - 0.25
400
/o cfficieucy : ?5.o/o//
_ lxlmple 6'f0. A Cornot'e ctgirtc whoae tcmperalurc of tlrcaarice is 400 K takee 200 caloriea oJ hcal at lhil tcmpqraturc andrejecls 150 coloriea ol hcot to thc sjn(. What ia lhe lemperottre oflhe sinb I Also ulculole tl* eficieocy oJ the enginc. " i
JIr:200cal; trI:150calTr -*00 K ; Ir: Itt E,-7i: T,
rt : _f,i "*,
?r: ffi*O* : 3{X) K
(I) HEATENGINE
26t Ecal and Thcrmoilgnanice {lrlrnodyramia 246
,-r{+: 1-# :0.25
0/6 efficiency :25%z' Exraple 6'11. A Carnot'a engine- !9 oyg'qted belween two
necntoirr oilemperattrea of 150 K anil J50 K. IJ the cngine reccilrtce
1000 caloriee of heat troh the source in e-och cycle, c,alculate thaomou,nt of fuat- rejecteil to the aink in eoch- cgcle. Calculate lheeficiency-ot the engine ond the work ilone bg thc engine in eoch cycle,(I calorie - 4'2 joules).
*450K; fr:350K: 1000 cal'i - E1 : 1
Trll
,r* #,-tg# :272.77 Lrs
.71'rr_ 3s0 100l-m-86
:0'22220/6 efrciency :22'22o/o
WorL done in each cYcle
: Er-flt: t000-777.77
- 222'23 cal:222'23x4'2 joules
:9l?'rgjoutceErrnplc 6.12. A Carnot'a enCill working aa- a refrigerator
bclwacn "ZN K and 800 K receioee ,500 calorics of luot lrotn theruscruoir dl thc bwer lemperatute. Calculate thc amount of heat rclicctcdAo tln rceatoir at the higher tempetalure, Oolculate aho thc amoualof worlc donc hr eaa,h cycle to operate thc rc7'rigeralq. -- - -lDelhi lflona.l 197aJ
E1: I E|: 5(X) cal
fr - 300 E Ir: 260 KEr lrrT- n;;
&: Er. +z, 500x300ra,:
-
:576.92ca1' 260
lf : Er-Et: 76'92 cal
76'92x4.2 joula: 323.08joulu
Era_mple-6.!3. A Carnol'a_retrigerator lahea heot lrom ualq al0'C anil iliacarda it to a room ot 22"C. 1 kg of woter at-A"C h to bcclwngeil into ice ot.0'q. Eow mony caloriTe of heat ore diacorilcil tothc room I what is the uork-ilone by the retrigirator in thie proceaa IWhat is the coeficienl oJ perlormonie ol thi machine ? '
[Dclhi 19741
Et:?IIr : l000x80 : 80,000 car
?r:300K?t :273K
flr TL
T:41E, - E#'
Jl
80,000 x 300- ---2
-Er _ S7,9oo g$Work done by the refrigerator
:W:J1E\_ESW : 4'2 (87,900-80,000)W - 4.2x79A0
F : 3.I8ilxl0.Joutc.
CoeGcient of perforu.ance,
90,000-- 97,900-80,000
80,000- 7900
: 10.13
rrEr
E,E;
Ez*
!:(l)
(2)
(3)
Er: E;g-
,tS ,/- Eeat onit ?hermoitynamica
'".)Unt"T_pIe 6I{. A carnot engine uhose lou temperalure rescr-coir ia ot 7"c t,,s on efi,ciency ,I 50%. It is desired ii ;iiir.rf, tn,:{-r!r:?v.to 7 0o/o; ns-how i""y.a"[]iu-_rmrW tii ir*i*i",i' "tthe hqh temperatutc reaqroir be increased t (Oetht lgliy
In the first case
Thennodynamiu
Ii(Eciency of the engine d
-1:Er-8,--E;-
or
or
:? .- 50o/o : g'5, Tr : 273*2 : 2g0 K.It-?
" : t-TJ
tl
0'5.: I-- 37r : 560 'K
In the se@nd case
1'- 70!s: U7,Tt:280 K'rr':!,' : l-\ m,
0.? : r- '?g-. T,,'
T'' : 840 EIncreale in temperaturca Stl0-560 - 280 K
Fig.6.t6.
Efficicncy of the engine B
\ - 71': E"-E{ -\ flr,
Since I>rl';E{)ErAlso, 'W : Er-Ht: E{-E{
w:4
WEl
Consider two reversible engines d aud B, working between thetemperature limie ?, and ?r (Fig.6.15)- .d and A"are coupled-Suppose.r{-is more etcient than 8. The engine d workr ., . t..telglne 11rd^8 as a refrigerator. Thc engine? absorb,s an amouatoI hcat Irr ,rom the source at a teEperature [. It does exterua]work P and transfers it to 8. The heat r-jected to the sink is E. ata temperature &. The engine B absorbs Eeat Ea, from the sink' attemperature fi and. I[ amount of work is done on the workinf subs-tance. The heat given to the source at tempcraturc ?1 is .E1,."
Supposc tbe engine / is more efficient than B.
Thr.rs, for the two cngines A arrd lr vs tem, (4 +,i tr t L..q,. "'ti
t i;r r.i*fr#'i*t rfl },,i:oJ. :tcmperature ?2 and (Er,-Hrl i, tt. o"""i;
rytltl**i,"-.r,*.1$;ffi?i;[:":J".1_"#;i#:ffirll* lilthp', i::: i H:#,.x,# r*rH.Ifl{done on the svstem. This is
"orrto.V'toihe secgnd law of thermo-dynamics. Thus, a ""r;"1 L;;;,'..*,iL n) The two ensines(reversibre) working between *,J*-.'t*o-tempe.ature timits iavethe same efficiencyf fl\{;r;;;;;;t" 1';.';"r;r h "; i, ; ;' i.,,'
" d' r, 1", a ".
-,l - - i.i"ii;" ;""! i"I,tf TJ''*:lSi.;l'ff ETi: " efit? :$ ;:,* glr$ :,.e-.- i:# #:LT;the efficiency depends ."1'y ;p";ih; r-*I i._p"rarre ri,nits.
. In a practical engine there is alwavstricticil;ft ,";ffi :i"La;",i"".r".:;;;.-*"':l.Jd:,."Hr_o"l;r;:I.wer thar. that of a Carnot's ."*,"r=f '";' 11G_) \-n .*',u'r' Gt1''r\
'o
H eat aail Thermodynumicd
?h;rd Fourth
Thctmoilytlzmico
flt H,-T: Trl l)Er Tr:Er: r;
Er_E, Tr_T,-Z;: -T_Here, flr - L+ilL, E1 : L,
T1 : T{iII, T1 - lIEr-flr: LailL-L : ilLTr-?r-T+/IT_T:ilT
. dll itf_T-TThe area of thc figure
ABOD-fl1_82-ill_ dP(Yr_Vrl
. dP (vt-vi dr
L:--T (VL-V)
2f3
For the same compression ratio, the efficiency of an Ottoengine_ is mor_e than a 'diesel
engine. fn practice, th. "ornpr.rrio.ratio for an otto engine is from i to g.rrh fo.. ai.r.r."sl"."it i.from l5 to 20. Due to-theJ-righer
"o*p..*ror ratro, an actual dieseren,gine has higher efficiency lhan the' Otto (petrcrfl .nein.. -fhecyrrncter must bc strong enough to withstand very high p.6ss,r...6'37 Multicylin6.l Enginea -
With an engine,havin.g one cylinder, the engine works onlyd.uring the u'orking stroke. The piston movcs a,r.ing'ihe ,"rioi,t,.tlrree strokes due ro the momentum of the shaft. i""" -"iti""t-ina..engine (say 4-c_ylinder.engine) the four cyri*ders
".. "o"11.J1
--tir.working of each cylinder is given below i-
- Thc cycle AND reprcscnts a comptetc cycle and Carnot'stheorem can be
^ applied. Suppose the volume it the point z{ is p1
and temperarure h f 1d?. '[he pressure is just below iu raturationprcssure and- the liquid begins ro evaporate and at the point B thevolume_ is 72. .The substance is in the uapour state. Suppose themass of the liquid at I is one gram. The amount of heat a-bsorbedis IIr, Here E1:L1d!, rvhere L+dL is the latent heat of thc liquidat temperaturc (T lit?1,
At the point B, the prcssure is decreased by dP. The vapourwill-expand and its tempcrature falls. The tcmperature at C it f .Ac this pressure and temperature f, the sas bceins to condense andis convcrted into the liquid stare. At thc-point D, the substance is inthe liquid srate. From-c to D, rhe amount of hlat reiected (civenout) is II1. Here II1 : .t where Z is the latcnt heat .i t".ociit,r."'/'. By in-crealtlS_t1r9 pressure a little, the original point z{. is iestored.The cycle ABCDA is co-mpletely reversible. -Applying thc principleof the Carnot's reversible
-cycle
272
Eirst quartor
Ssooad quBrtor
Third quarter
Sourtb quartor
fdret
Workilg
ErhausL
Charging :
Compreesion
8@nd
Erhaust
Charging
Comprossion
YPorking
Yz
Compreasion
T9orking
ExheuaC
Charging
Charging
Coopreesioo
Working
Erheust
In this wa-y, -the poyqr of the engine increases and the shaftgets Eomentum during each qgarter cycl-e.
[,9#Clap_evroor,eteq-1H,9a!-F-.{gi-tio.ntT)-*""ider thei*ott..*"ii-iaen ., 'i.ilr.o,ure f+d? andauafl at temperature ?. }Ierc EA and.ED show the liquid state
P i,P@ ...(,
v'------;Fig.6.23
ol the substance- At I and D the substance is purely in the Iiqtridstate (Fig. 6'23). From I to I or:D to c trre subitance is in tra'nsi-IIT fr.T the liq.uid to-the gaseous state and o;cic ieiii.--ai'Cl"au the substance is purely in the gaseous state. From B to .F or Cto u the substance is in the gaseo,s state. Join I to D and B to cby dotted lines.
This is called thc Clapeyron's latent heat equation.
. . Applicetloar. $ eficct of clwnge o! gtreaeure on thc mclting
point.When a solid is convertcd into a liquid, therc is change in
volume.(i) If % is greatcr than 71dPfiV ls a positive quantity. It means that thc rate of change of
274 E eal and Thermodynaniet
pressure with respect lo temperatrrre is positive." In such cases, rhemelting point of the substance rvill increase rvjth increase in pressureand lice lersc.
(ii)' If l'o is iess than Ir,.dP,6,Vis
a negative quantity. It me ans ihat the rare of chenge
of prcssure u'ith respect to temperature is neqative. In such cases,the me lting point of tlre substance u'ill decrease rvith increase inpresture and ricc t,er,sn. ln the case of meltinq ice, the volrrme ofrvater formed is less than the volume of ice taken. Ilence 1tr q l"r.
Therefr:re, the melting pcint ol ice decreases rvilh increase inpressure. I{ence ice u'ill rnelt at a lemperature lorver than zertrdegree centigrade at a pressure higlrer than the normal pressure.
Ice melts at 0"C only at a pressure of 76 cm of Hg.
, 12) Eflect o! clange tf yre.ssure on the boiling point.' When a liquid is cortverted into a gaseous state, the tr.,lrr:re I's
of tlie gas is aln'ays qrearer rhan the corresponding volume I'1 ofthe liquid r.e. I', > l'r.
dP'Iherefrrre, j7 is a {ve quentity'.
lVith increase in pressure, the boiling point of a substance in.creases and tice uer.-ra, Tire liquid n'ill Lroil at a lorver temperatLrreunder reduced pressure. In the case of water, the boiling poirrt irr-creases w'ith inclease in pressure and z.lce uersa. Water boils at 100'Conly at 76 cm of IJq pressure. In the laboratories, rvhile preparingsteam, the boiling point is less than 100'C because the atmosphericpressure is less than 76 cm ol I{g. In pressure cookers, the liquidbr.rils at a higher te mperature. irecause the pressure inside is morethan the atmospheric pressure.
-, Example 6'17. Aolculatc the depreeeion. in the melting point oJice Ttroiluced by one otmosTshere increase of prensure. Gitten latenlhe.ct t{ ice : 80 cal ner gram rt.nd the specific uolurtues o! 1 gram ofice anil wq,ter at 0'C are l'09 I cnts and l'000 cmxrespecliuelg.
\Panjab 19{)S)
Here L: B0 cal : B0x4 2 x l0? ergs
1' :273KdP : I atmosPhere
: 76 x I3'6 x 980 dynesfcmt
I'r : l'091 cms
l', : l'00C cmg
dPLdr : ?:(|:Y,)o dP.f.(l'r- I'1)*r .= __
T_
?lvr'modynamice
76 x 13.6 x 980 x 273(t - I.09t )B0x4'2xl0z
: _0.007{ KTherefore, the d.ecrease in the ntclrinq point of ice wirh an
increase in pressure of one atmosphere
:0.0074K-0.007{"c* - -!,rerrplc 6 18. Find lhc incrcase in the boiling ltoid"o! uoter
lt 1!0:C when the ple"lure ia inueated by one atmiipherc." Lalentheat o! tuporisation lf.1team ia t40 callgrim and I gram o! eteamc,ccupieE a talrrme of 16?7 cms.
dP : 76 x t3.6 x 980 dynes/cmr
T : 100*273:373K
, : 5{0x4'2x l0? ergs
l'. : l'000 cms
I/r : 1677 cmo
dPL-a7 : TW_I1
ur_ dPxT(Y'-Vi..-_-_-Z-76x I3.6x980x373x 1676
540x4.2x10?:27'91"C
Tlrerefore, theincreuae in the boiling point of water u,ith anincrease in pressure of one atmosphere
: 27.92.C
: 27.92 KErample 6.19. C.alculate lhe-ehange.-itt temperalure o! boiting
water whe n tlee pressure i,s inqcrceil W 27.12 mm o/ Hg. Thi normilboiling point o! waler at utmocpheric pressure ;s l00.Cl
Latent heat o! ateam
anrl specifc volume o! ateatn
: 537 callg
- IA74 cmt (Delhi 197 4)
dP : Z.7l2x 13.6x980 dynes/cmr? : 100+273 :373KL : 537x4'2x l0? ergs
Irr : l'000 cm8
['r : 1674 cm8
dPL, d:f : 16;-11
n8 Ecal aad Thermod;yta,amict
:2.792K:2792"C
. Therefore, the increase in thc boiling pint of r+,ater rvith anincrease'of 0. I atmosphere pressurc
:2.792,(:2.792C
. -"- P:!-ple 6.23. Calculol;e the c,h.,a,ruge in lhe mclring point ol icewhen il h wbjectcd lo a pressurcof 100 otmosphere,e
Densitg of ice : 0.917 gfcms and
Lalent heat of ice : 336 Jlg(Delhi 1972)
dPLm : rv;ndP : ld0- I
: 99 atmospheres
iIP : 99 x 76 x 13.6 x 980 dynes/cm2
L : 336Jlg: 336 x l0z ergs/g
T _273K(vt-vi: ,- #i7
0'083: -- 0.917
* -0.091 cm3
d? _ r dPvr_yrl':--J-
Thermodynamicc
d7: -l K'T :273 K
I'.-I', : -0'091 cmiL : 79.6 cal/g
- 79.6 x 4.l B x l0? ergs/g
)p _ L. dT"' - ru;wdP :'gq#Haf dynes/cmr
79.6x4.18x 107d.p:. _______ o atmospheres273 x0 091 x I 013 x l0
dP : 135.2 atrnospheres
Pressure required: 135.2+ I
, Ll : f36.2 atrnoepberesWl,L*ple 6'25. lf ater boils a! a tem1teroture o! 101"C ur a
pressure ol 787 mm o[ Hg. i gram o! water occupties 7,60 I cm3 oneuaporution, Calculate the lotenl huct d steam. J : 4.2X 107 ergs/cal.
lDelhi (Hons.) 197 tldPLm : fv;TtdP : 787-760
:27mmofHg- 2'7 cm of Hg: 2'7 x l3'6 x 980 dynes/cmr
dT :1'g: lKT :373K
' Yr-Vr: 1,601- I : 1,600 cms
L:1, T dP 17r-Vryu : ----m-
373 x2'7 X l3'6 x 980 x 1,600
dT-dr*
273 x99x76x 13.6 x 9B0x (-0.091)336 x l0?
-0.7326 K
-0.7326"CThe decrease in the melting point olice rvith a pressure of 100
atmospheres..t : 0'7326"C
'- Erary.ple-6 2{- Calculate the pTessure requireil to l<twcmelling point o!_icc W l"C.
.\L = 79.6 c?llg, apecific aolume o! wuter al 0"C - 1.000 cm.
specqrc_aolq?ne o[ ice ar 0"c : r 09r cm3 and I armosTthere pressare: 1'013 xl0c dyn.r7.*rr. ' lOrin;-iiiA1dPLii: ro;q
tr : 511'3 cal/g
Exarnple 6'26. lYhen. leail is melteil at atmoepheric preasurc,
Ithe melting gtoittl is A00 K\ lhe density d,ecreases lrom 1.1'01lo 10'65glcms and the latent heot of lusion is 2* 5 J,rg. tllwr is the mellingTroint nt a pressure oJ 100 atmospheres ? iDelhi (Hons.l 1972)
L-
r._I
373x2.7 x I3'6x9B0x 1,6004'2 x l0?
ergs/g
callg
.Eed aill lf&rlrrpitrrtrj.t
Il1 - -{-l(XX) joulerE: : -8(X)joules (since hcet is rejected!fr:500Kf3:300KH rofl) -800
-- -.1- -
T 5m 300
Thrrmoilyr.omict 0tg
Now consider the reversible cycle from state I to state 2 aloogthc path d and from state 2 to state I along the path ()'
For this reversible cyclic process
o: -; joulc/degrec
(3) Conridcr a C.arnots reversible engine working betwcen thetenpcratura 500 K and 300 IL Suppose 1000joules ofheat energyis drewn from the high temperature rescrvoir.
flr E,T,: T
l0o0 _ E!500 3m
^E : 600joulcstrfl flr. H,
\ _-_-4 __4? _ T, ' T,Er : -il0C0joules
Er :-600jouler?l:sCgi;?r:300K
\1:4.1',;$ :oFrom equations (i) and (ii)
fla sE rlc 8Elzn-T : Jzo-T-
Herc
?
...(d0
... (d;i)
thermodynamicd
...(is)
...(o)
fi l&lJ, . ,-6001T 500 ' '3m
ET:O
,.lr.:';:This shows tn"t t# ,* the sane value for atl rhe reversible
paths from state 2 to state 1. Thc quarttity I S- tt iniieper.jent of
thc path and is a function of the end states only, thereforc it is aproPcrty.
This property is callcd entrpP!' Entropy is apnoperty ana ii dc6ned by the relation
,r: +or 81-81 : f +
The quantity Br-Br rePrescnts the change in entropy of the
system whc'n it is 6hangcd from state I to statc 2'
,. &f:, Eatropy changee of r Clorcd Syrtco Duriog aBIrrcvcrdblc Procccc
Consider a rgversible cycle wherc the stat! is clr'ngeC from Ito Z "f""ltfrip.tn
aand 2 to I r.long thc path A tFiS' 6'26)'
For a reversible cYclic Process
3[ar:o
This example showr f# : o, only in the limiting
T
casc and in no casc ># is grcater thaa zero.
t-'6.*2. Eotropy ud tLe Sccoad llw of Thcrmody-a-icrConsidcra cloacd systcm undcrgoing a rcversible proccss from
statc I to rtatc 2 along thc path .{ and from state 2 to state I along.thc path I (Flg. 6'25). As thir is a reverrible cyclic process
{4:o2t EE frD &a,r7+ lrr'7: o ...(0
80 Eedt and llhermodgna*iu
lza EE rrB 8EJu TT)za -?-: o
Frorn equations (d) and (ir:)rta 8E' tro 8I1
Je" --:F- -lzo -7- 7 o
,$ince path B is reversiblc and entropy is a property
['B _qL: Jj| as - ll] asJ*r f
ur> +r Br-o , i: D.R
To conclude,
For a revereible procers
U-rr: J; #ad for an irreversible process
8,,-..e. :- f jg-
...(iio
...(ttu)
Thermoilynamiu 291
Equation (ia) shows that the eflect of irreversibility is alwaysto increase the entropy of a system.
...(r)
fu** Entropy W-.".'Consider adiabatics L and,.df on the P-7 indicator diagram
iFig. 6'27). All along the adiabatic f,, with change in pressure
V____+F.ig.6.27
there is change in volume and temperature. This shorvs that allalong the adiabatics L or M, there is change of temperature" Consi-der the isothermals at tcmperatures f1, T, and, Ts. ABCD representsthe Carnot's reversible cycle. From ,4 to 8, heat energy .&1 isabsorbed at ternperature f1. From C to D, heat energy Il, isrejected at temperature ?s.
-fl'- : !!^:-aI t7
Sirnilarly considering the cycle DCEIEr:HxTz TN
*:"*r: !l': *constant
From one adiabatic to the other adiabatic, heat enerey iseither absorbrd or rejecteri. The qr"rantity of hc'at absorbed orrejected is not constant but it depends upon the temperature. Fligherthe temperature, more is ihe heat energv absorbed or rejected andoiccoerso. The quantity H,'T betrteen nvo adiabatics is conslanrand this is callcd the change in cntropy. Let the entrrlpy lirr thcediahatics L and. M be .s1 and S1 rcspectivcly.
Here ,S, and 8, are arbiirary quantities.
,Ss*.S,: i.{'-.,,nrron,. "
P
"-.-._.-Fig. 6.20
Now considcr an irreversibre path o frorn state 2 to state r.Applying Clausius inequality for the cycle of processe! A aad OrgJz<o. t'o 8^E , fro
Irt f - )zo ...(j08^A
7"-< o
H.\-.-^ 12
ooo
If the adiabatics,..1." t.a ;,la ;; Hili.,Y,?rit",
' Change in entropy
dB: I{1l
In general, the change in entropy
Eeat anil Thermoilgnamict
and the heat absorbed or
...( I )
Thermodynamice
the total gain in entroPyABCDA
293
by the working substance in thc cycle
Er E,lt rt
tB 8E rsr :
J:: d's: &-Br : I: Lg ...12',
lo -f : J"rUU
represents the thermodynamic co.ordinate of aEystcm' This integrar refers to the value of the function at the final:f,[#:::tf I3]: ut tt. Liii""r ',1'",". This runcrion is carred'ti.i'ii,iJiif ,i,:ol'ifixl:,1lf
"il.Jffi :i.*"r.f*:o"i'.11,'iir",.ntAll alons tt
rWfriWmyeeytt*rxx;t*Yrf
!r":r".i'ia Entropv in a Rcversibre proceee (qryl
6'*. ffi o Tff r,,l.8 i.":.I:?::#tr:i: IHHI r,t jil?f
But for a complete reversible procdsEr Ezr;: 4
Hence the total change in entropy of the working substance ina complete reversible Procesg
r.-. : f'r - F: -ff : o'
w6,*O Ghange in Eatropy in an frreversible Procesg
In an irreversible process like conduction or radiation, heat is'lost by . Uoay at a hightr temperaturc f1 and is gained by the bodyat a lower teriperatrr.:e ?s. Hire Zlr is greater than ?2'
Let the quantity of heat given out by a body at a tern?eratrtreTl be E and the heat gained by the body at a temPerature 'I'g be l' 'Consider the hot and the cold bodics as one system'
ELoqs in
SntroPY of thc hot bodY - 71
Gain in entroPy of the qgld bodY - tTherefore, the total increase in entropy of the system
Hq: r;-NIt is a positive quantity because- 71 is less than fr' Thus the
entropy of tlie system increases in all irreversfble processes'
( 6fi Thtrd Lew of TteroodYnaraics-V/ tV \ U all heat engines, there is always loss of heat in the form o,f
"o"i""iion, radiatiin and friction. Therefore, in actual heat engines
El -. ----^, .^ E, 1
7; is not equar to T '" )
*-*ir,,o, /.- but it is a positive quar^titv' when
cvcle after cycle is repeatcd, the entropy of the system increases. and
;:ili;. tir^*i-"ti val'te. When -the
system has attained themaxi*rr- value, a 3tage of stagnancy is reached and no work can
t. aorr. by the engine-at thisltage. 'In this universe the entroPy
irl"..."ri"g and u-ltimately thc rlniverse will also reach a maximum
""i". "f "niropy when no ivork will be possible. With the increase
i" ."t-rr. thi disorder of the molccules of a substance increases'
T-h;;;#dy is also a tucarure of the disorder of the syttem' With
i l!i:,:'.rt:T T:rr3 Ei fll : ffi d,"il,,t-.r,,o* or, h e work ins
ill:{i:lh##ri:{r:#t::rl#i,.il# ;i:".. fins substance ar a t.*i*jr,,ll, Tir"rBL!, rs reiected by the worl-
ii,u'i,u*#i+,i"fi f{f li'jJtt$:,i}*ii;iitemperature Zr). From D ," a ,r,.rllr.nl,.rrurrg. ro enrropy. Thus
294 Eeal anil Tl*rmodrynamia Thermodgtomia
Here Br:ur:+-+-ffi.-. Area ABID : ,',4SfU - E,-n,
Therefore, the area ABCD represents the encrgy convcrted toworL
: +? -t-'*:r-+
29t
decrease in entropy, tbc disorder d,ecreasant/At absorure zero tem.r)etot*te, the entropg lettih lo zero and the moleculee oy a subsianie or?^'-y-t::!..:::ye!ect order (welt arrensedl. Thia is'the tiird ii,tet*f moa!tuatntcE.rt
Erample. The molecules are more free to move in thegaseous.state than in the liquid state. The enropy ;, mo..-'in ttregaseous state thaa in the Iiquid state. The molecutt u.l---lr.'rr..to movc in thc liq-uid stare lhan in the solid state. rir.
""i.oo, i,more in tjre liquid stare than in the sorid. Thus when ^
,"u.iJ;i" i,converted from a solid to a hquid and.then from the i;q"ia']o",t,.solid state, the entropy increajes and oice ueraa. Wh;"-i;;l;
"i"nr.r-ted into water and the.n into steam, the entropy and disordcr "i:ir,.molecules increase. when steam is converted'i"to *"t.. l"a"it".,into ice,.the entropy and disorder of the molecut"r a.".IrrJ.""E"n",
entropy ia o mcasure of the ilisorder of the moleculio ijlir-iisii.'_L , By any ideal procedure,-it is impossiblc to bring any systern toaDsotute 1:rg !:*p.: atur.e performing a finite numbi,
"f opl"utio"r.r nrs rs.calre.l thc _prrnciple ol unattainability of absolute zeio. Thusaccording to Fowrer an.r,Guggenheim, the
"".tt i".uiiiry'pii"iipr"is called the third law of theiirodynamics.6'48 Temperature-Entropy piegram ---
*="fi;4'rd:x',lJ:-.'iff ,.ti,li",l[*r;'"i:.1rrfi
ifi ,?.1illfu ?,j,1From .d to Il, heat energy .H1 js absorbed at templrature- ?r. i.f,.increase in entropv Bs ti[* place from z to d-ffi-o'igi;ili'r*-
Fig.6.20B to C, there is no change in^ entr-opy. Thc tcmpetature decreasesat conltant entropy. From c to D, -:here is decreise in ent.oo,
-7,3,J.
i:.":::T:.temp^e_iature fi. Irom D to'd, thcre is "; ;H;g; i;entropy but the terDperaturc increases.
.o""Jlixffi jjg?#":H'try:t?jiH:;s:Hffif au,.ir;f
**The area ABCD : Sr (\-!I.) - '8, (Tr_Ttl
Efficiency
r
Here Hs is the unavailable energY.
H,: !^' x1t:s1 x?r|"
The unavailable energy depends on the change in entropy attemperature ?1 and the temperature 7r'
6;{9 EntroPY of Perfect Gas
Consider one gram of a perfect gas-at a prcssure P, volumeP
""a t.-p.."turc ?."Lct the quantity of heat given to the gas bc t;1.
8H : da+8w
8E: lxcuxdr+P# ...(d)
gH : TdS
rdB :cvar+ ff ...(i0
Also ,\ = ,lo
or P: n-7:dl : C"df +ryds:cy#*3+
rntesratins, Ix: * :," !iis*+li"+
B,-s, - cy log, -**i
be + ...(idJ)
Br-8, : cnx2'3026 loglo ft+|"r'3026'logr ft ...(ig)
The changc in entropy can be calculaied in terms of presurc
alro' PY - rT
DifierentiatingPiIYlYilP - ril?
PdY - rilT-YdP
P
c=H
But Br:+andB1 :t
296 Ecat anil Thermodyzamhe
Substituting the value of PdY ia equation (ii1
f ,iS : CyxdT4J!! - ry*) n'- '*
lknwitynomico 29?
E:emple 6'30. Calculatc lhe chong-c itt cntropry vfur 5 kg olualer ot 100'C b anuertcd inlo deom al the aaru lcmlreroh.rc.
Heat absorbed by 5 kg of water at l00oC whcn it is convcftedinto steam at l(X)oC
: 5000x540: 2700000 cal
8II : 2700000 cal
The gain irr entroPY
.- 8^Eoo: _F-
2700000: + :?24A c.t/E373
n dT vdP", -T- ----iTrTf
Tas:cr$*! #
"ff:+-+l:;gcptoso #-+r"&+.
Br-8, : cpx2.3o26xrogroft - l*z.zo2orog,.f ...1ur)
Notc. r ir tho onflinary gae co-net1a,nt and -hae to be takon ia unite ofwort, o" rclnrento tbo apooiflo heat for I gram of e gas rt oon"teoi-piJerr"e.
If Cp reprcsenls-gpm 'nolecular specific heat of a Aas atcoastant pressure and B the universal gas constant, then
Br-Br - ctx2-3026 logp *-+x2.3026rosro+ ...(udd)
of ice at fC h ootoqleil itu ualer at the eame teipiroture. 6'-(punjah 1963, Delhi lg75l
Heat abaorbed by l0 g^of icc at 0"c when it is converted intowttcr at OoC : 10X80 : 800 cal
8trI : ggtg ""1?: OoC :273K
The gain in entropy
,r: #= # :2.e3 crryB
\..o"Ersmptc 6 31. Calcdate tlrr- increuc in entropg-wt*n f Eamof iJc d -ly1 ia conoedeil i*to dteom ar 100'C. Spca',{'a l*at of icc2-0:\ latenl heal oJ ice - 80 callg, latcnt -heat.ot etun :-.510<xrlle. - lBonboY 1971; Delhi 1973t
(l) Iucrease in entropy when the tcnPeraturc of I gram oficc iniriass from -10"C to OoC
u:l;"+t* il?: *lrrT
T,: tB tOSo -flf
: ,rtE x2.3026 k"ft: I x0'5x2'3026 t"et#: 0'01865 callK
(2) Increaec in entropy when I gram of ice at fC is
into waier at 0"C.
fU: +: ,tr : o.re3 cal,rK
(3) Incrcase in enboPy vrhen the teraperature of Iis rsis;d from OoC to 100'C.
ou:1;:+
But Cv*
dB:Also PY -
Y_T:
Integrating
ras: (e"4.
f:t,
["'ds :Iar
B3-81 : ...(u)
converted
g of water
298 Eeal anil Thermodgnamiw
: msx2.3026 log, f,o7;. : Ixtx2.3026loCr.#
: 0.312 cal/K(4) Increase in entropy when I g watcr at 100"C is. convertedinto steam at 100"C
,":+540: ffi :1.4*T callK
Total increase in enfopy: 0.01865+0.293+0.3 l2 + t.+47
/ : 2'07065 cal/K
- '/ ETanple 6'32. one grammorecule oJ a gas expan*a ieothermar-ty to tourtimea it.g aoturie. catc"iiii;i;;h";;;t;-liri"#ffi, ;,termc oj the gaa conslant.
: [o' Pav)Vr
PV:R?DRr
w - Rr Ii:$- R? r"r"#
fi:rW : RT x2.3026log1s (4)
H.erc, W anct I are in the units of work
Gain in entropy : t#
:#:ys#tu3
Thermodyrnmice
(d)rrtr:50g;?r:273Kmr:50gi?t:353K
Let the final temperature of the mixture be ? Kmr sX (T-?rl - m, s(Tr-T).
50x I x (T -273) : 50x I x(353-7;?:3lgK ^ il
(ii) Change in entropy by 50 g of water when itsrises from 273K to 313 K.
299
temperaturc.
8r/-T-iT dT*t .l*rT
lYcrk done
But
Here
v Erampte 6.33.qual maas oJ water atcrotroptg.
: I.387 I caUKJ
50 grams o! water at 0"C ic- ,nixeil uilh an83'C. Calculate the resultant ;"criiie -;n
(punjab 19681
.. :50xlXtof" #- 50x2'302Oxf"Srr{l}
: f6.829 callK
Herc, thc fve sign indicates gain in entropy.(ddd) Changc in entropy by 50 g of water when its temperatur(
falls from 353 K to 313 K8.8 tr dT: T_: -r l rrT
:50xtxlos"{}f,
- 50x2.3026xlogro ffi: _6.023 cal/K
Here, the -ve sign indicates loss in entropy.
Therefore, the total gain in entropy of the system
: 6.829-6.023: 0'806 cal/K
J Example 6'34. Calculate the change in entropg wh,en 50 graynrof uater al 15e, ia mixeil with 60 grams o/ water at 40"C. Bpecifiaheat ol woler may be aasumeil to be equol to 1. (Ilajosthan 19611
(t) mr:50s?r : 15+273 : 288 Kmr : l0 grams
. Tt: 40+273: 313 K
300 Eeat ond
Let the fiual temperature bc f K.rrlrxdx (T-"r) - mrxE x(?r-f)
50x I x ("-288) : 80x I x(313-?)? : 303.4 K
Themwilynamice
(ii) Change in entropyrises from 288 K to 303.4 K
:
when the tcmperature of 50 g of water
:50xlx2.3026xtoSroffi: +2.602 cal/K.
- (ri{ phan-ge- in_ entropy w-hen the remperature of 80 g of waterdecreases from 313 K to 303:4 K
8E tr iy|: _T : _, J r,-T_
: Boxlx2'302- 303'4b y logroTT3-
: _2.497 caUKTherefore, the net change in the entropy of the system
: +2.602-2.4at: +0.llc cal/K
Hcnce the net increasc in the entropy ol'the system
, : 0.ff5 caliK :-:
" f'.-e'nplg 6.35. 10 g o! eteam at 700"Cis bloun hdo g0 srdmr-of watu ! (PC, cotzhineil in a enlorimeter of uater equivatint I0grarns. llhe uhok o! thc ateam ia conilenceil. ialcu,late tie increoaein.the cr*ropry of thc eyatem. tDelhi (Eo*t) filq
(i) mr - l0gfr : l00oC : 373 K,rh : 90+10 : 100 g
?t:273KLet the final temperatue be ? K
l0x5110*lOi.?n-fi : rciif -273)? : 331'2 K(di) Change in entropy rvhen the temperature of water and
calorimeter rigcs from 273 K to 331.2 K8fl t? dTT : ^, I4.Trcolx''q
: l0o x 2'3026 X log,o (#+): a 19.32 cal/K
(iii) Change in entropy uhen l0 gram! of steam at 373K ir
Thermoilynamiet 301
8E t? itT-T : ^t !*, ,-
condensed to water at 373 K/ 8E \ l0x5,t0
_t-.t:_-\ r ) 273
- -14't7 caUK
(-ve sign indicates decrease in entropy)'
(iu) Change in entropy when l0 grams of water at 373 K iscooled to water at 331'2 K
8E tr dr-T : *t )rr-:F_r!
: l0x2'3026 log,^ f9-^!! \r^u \ 373 ): -l'188 cals/K
Net change in entroPY: t9.32-14.47-l.l88: a3'662 callK
Ifence thc net increase in the cntroPy of the system
3'E62 cal/K
1d/ Erample 6.36. 1 g oI water at 20'c is conaerteil into ice at
-i|'C ot Zonslant. preasure. Heat c.rytocity for I g o! water ia 4'2lls-R anil that of ice ia Z'l.Ile'R. Heat of tusion of ice at"'03C - 335 J[g.
'Colaulate the total changa in the entropy of the
sgstem.
(l) Changc in entropy when the temPerature of I g of water at293 K falls to 273 K.
ds: + :*tl;:+t273 dT_ I x u., ]rr,
.r_
: 4'2x2'3026 logro (#): -0'2969 J/K
(ir) Change in entropy whcn I g of warer at 273 K is convert'ed into ice at 273 K
gH -t ;z ??rd^s : -? : --ffi : -1.227 JiK
d'''-&
- I x2.l x2.3026 .r,r(#i )
: _0.07834J/K
Total change in entropy of the system
: _0.2969 _ 1.227 _0.07834_ _r.6022{ J/E
Negative sign shows that there is decrease in entropy of the system.
.., +"-. Ue 6.37. - |!^e_Z! wa.tet ot Z7g K ia brought in contactwith a heat reacnoir at.iTiK (l1what _i_s ti "i"ig"-ii i,irr-rii ,rwater uhen ite temperalure reaiiee Af g K g
,h, ,f)"yr:.at ia the chonge in entropy o! (i) the resen)oi,r ond (ii)
(l) Increase in urtro_py wh_en the temperature of 1000 g of'water is raised from 273 K io 373 K
,u : [?r 8.8
Jr, T,,sx2.gl26rorro #l
:1000x1x2.3026 lo1ro!ffi
- 312 calK(2) (0 Change in entropy of the reservoir,
-8.8'" - -T-
_ -- l000xtxt00373
Negative sign shows decreasc in entropy
: -268.1 caliK
(2) (d, change t':"#1#;universe
* ,{3.9 cal/KTherefc-ire, ihe net increase in entropy of the universe
: 43.S caliK
According to Kinetic theory, the energy of a system at absolutezero should be zero. It means the moleculei of the' system do notn::::..?"y motion. But according ro rhe modern corrc.pt, cven atabsolute zero, rhe morecures are not compretery depriJed of theirmotion and hence possess energy. The eneigy df tli;;;i;J.;';;absolute zero .emplrature is caiied zero pofit i.;;:-.6'5f Negative Teaperatures
The specific heat of a substa'ce decreases with increase intemperature, However, the specific heat does not tend to zero as thetemperature rends to infiniry. This shows that the t.*p"",ii"* ir"ia + ve sign only.
But recentexperimenrs by Ramsey (1956) have shown that apart of a system i.e., the nucleus of a so[id,-can-have a ".gatiu;-t..1perature. This sub-system is considered isolated from the' marn s/s-
tem (f.e-.,.solid lattice), The specific heat of the sub.system tends tozero at high temperatur-e. A imall amount of hcat "rtrgy tcnds to
raise the tcmperature of the system to infinity. It is posiibte to .aJstill pore.energy to the.sub-sysrem at infinity and it f6rces the sublsystem rnto the negarive temperature region. rt has been shown bvmicroscopic statistical anarysii that there-is no distin"tio"-uit*.Iithe tempgrature of {o and -e. -In thermodyna-;"r, tt.
-pu-rl'.
meter l/f is rnore significant than ?.fr. negative temperatures are hotter than the positive temoe.
ratures and milu_s ?:1".
(.-0) is . the hottest t.*p.rlt,.ri ;"4-'H;
zero ({0) is the coldest temperature.
. . The negative temperature is not possible with. the system. as awhole and is only an exception to the iule that only posiiive t.*o.-ratures cxist. The ncgative te-mperarures-are possiutd
""ry lo. iroii.ble.sub-systems. For all normal purposes the temperat*ir'ar. ut*n"!spositive.
q. U9.52 Marwell's Thermodynamical Relarlonr )> "L/ From the two laws of thermodynamics, u"**''"tt was abre toderive six fundamental_thermodynari.rical reiations. The gtate of a'qxsrem c1p.be specified by.gny piir of quantities udz. pressur. ipivolume (lz.)' temperaturc (?) and entropy (s). In solvihg u,y tt.iimodynamical problem, the most suitabte' pair is chosin uird thequantities constituting the pair are takt,n as'inriependent variables.
From the first law of thermodynamics8E: dU1gtrtE : dLt+pdy
or 8U : }E*PLVFrom the second law of thermoCynamics,
809 Ecat otd llhcrmoilynamia
2?s K(tl,:,:"lgf in entropy when the tcmperature of I g of ice at
Thermoilynamict
6'50 Zero Point Encrjy
E0t
dB: 8.4 t4 itrT : ^, I,,T
,t: t#*
8Ii = '?'ri"9
301 Heal ond ?hcrmodYtomi*
Subnrituting this value of 8II in the 6rst equation
iru : Tdg-PdY ...(i)
Considering 8, O and 7 to bc functions of two independcnt
"ari"Ufal-""i y" tt lt" x and y can be any two variables out of P, 7'
7 and Sl,
. iu - (g:), r"*(H), r,t
uo : (#),a,n(Zl).uu* : t*),u,*(1#),uo
Substitutirrg thegc valud in cquation (i)
(#), au+(fi)y: ,[(#),,,*(#) .uo7
-'l(#),u"*(#). * l: [' (H) ,-'(#))*
+.[, (#).-r(#)J*Comparing the cocficicuu of ilr arrd dy, wG get
(*),-n(#),-'(#), "(i+
('#). :'(#). -r (#). "'('ii)Diffcrcntiatiag equation (ii) with respect to y and equatioo
(iid) *'ith resPcct to ,
1#* (#).(#),*,#.-\:#)"(#),-,#
lfhermoilynomirl
ft meaas ilA is z perfect diffcrential
end
?'a : ?tD "odardy azN
(# ) .(:#), ..
#;, )rfUr { .,-ff),(#),
Simplifying,
(#t(H),-(*).(Y"),: (f.),(#).-(H ),(%). ...(do)
Here r and y cao be any two variables out of P, Y, T and B.
I)crlvrtion of Rclrtloar(l) TaLing f and 7 as independent variabler and
s: lIg-Y
#:r,#:r,'#-0, { -'o
Subatituting thcic naluer in equation (io)
(#),:(#),But aS:S
(2) Tating ? and P as independent variablec andz-Tl*P
dt
- :_1.
aroT' : o-as
\ oOtP)" -' oyo"
atBaray
o l'Vardy
t#)"u"*(#).uu
...(o)
...(o0
#:(#),(#).*,#-(t),('#),-,#
The chanse iD iDtcrnal energy brought about by, changing 7and f whether? is changcd W
-dV first and f W df later or t'icarerao is thc same.
APoc -"
VPat
E eal and T hermoilynamice
Derive the lollowing reli.tions :
c,_a.-: _.? ( f+ )"(+# \,
rds : car*r($),arGPlaTls
r^ "-4
Write short llotes on :
n\- ;-{-i) Isothermal Process/ t--@l Adial,ratic Process
(iii') Isochoric Process
/rb) Carnot's enginej$ Carnot's theorem
(ui) Second Larv of thermodynamics' (oii) Clement and l)esormes' rnethod
(uiri) Ruchhardt's experiment for y
p$.l Absolute gas scale
(z) Rankine cycle
1ri; Diesel engine(a.ii) Steam engine
(riiil Otto cycle
lDelhi, 19751
n-.\riul Entropy is a measure of disorder(ru) Entropy tends to a maximum
,---'lroi1 Third Law of thermodynamics(rrii) Absolute zero temPeraturc
y,\*aiiil Entropy of a perfect gas fDelhi (flona.l 19771
(rir) Tempcrature'Entropy diagram(zr) Thermodynamic systcm
UlAf Thcrmal Equilibrium(rzii) Conccpt of Tcmperature
\/.(*oiii) Concept of Heat
1-,@rio) Zeroth Law in Thermodynamics.
(rcu) Phase changes of the second order. [Delhi (Hon's.) 751
' 80. A motor car tyre has a pressure of 3 atmospheres at theroom teinperature of 27"C. If the tyre suddenly,bursts what is
the resulting temPerature ? [Aas. 218'6 K :'-54'4'C]81. A quantity of air (Y : l'4) at 27oC is compressed sud'
denly to I of its original volume. Find the final temperature. - -[Aos. 522'3K-249'3"C]82. A quantity of air at 27'C and atmosPheric pressurc is
suddenly compressed to I of its original volume' Find (i) the finalpressure and (ii) the final temperature.'
[Ans. (i) 8'29 atmospheres (rr) 571'1 K : 298'l'C]
Thermailynamica
83. Find the eflicienq, ol rhebetrveen 150'C and 50"C.
79.
(al
(r)
(c)
311
Carnot's ensline . rvorking[Ans. 23.640";)
Y
r--t [Delhi 1Hons.), 19781
84. Find the efficiency of a Carnot,s engine u.orking betwecn227'C and 27"C. tAns. 402,185. A Carnot's Sngine rvhose temperature of the source is 400K takes 500 calories of,hrat ar rhit terrf'e.uture and ,"j..t, +ti,l'.uro-ries of heat to the sink. What is the' temperaturq of the sink ?Calculate the efficiency of the engine.
- --
[Ans. (r) 320 K, (i;) 20o/L)
86. A Carnot,s _elgine is operated between two reservoirs attemperatures of 500 K and 40b K. If the ."gin. ..*i".", iooocalories of heat from the source i., ea"h cycle, calcul:_te (a) theamount of heat reiected to tlre sink in each ;i;i;; (/,)'";,1'".rf,11.,,.yof the engine and"(c) the .rork ;;;;';y ;; engine in each cycte in(i) joules (ii) kilo-Watt hours. -r -"-
[Ans. (a) I600 catories, (b)2}o/.o, (c) (r) t780joutes,(ir 4.941x lO n-tWfri
87. A Carnot,s. engine^lvorkilg a-s a refrigerator between250 K and 300 K receives"r000 caroriei Jn.ut from the reservoirat the lower temnerature.. (4- cal;;late the -"*"""i'"ii.".,
,.-jected to the reservoir-at the 'highe. *dperature. (if) carcuratealso the amount of work a"ie
-i" -.u"^r,
"y"i.',J'.ip#i" tr,.refrigerator.
/5 ,\ _ [Ans. (4 1200 cal, (irl) 840 joules]( w.' calculate the depression in the melting poi't -of
i".plod[ced by 2 atmospt e.er ii,-".e]*-of fr"rr,r... Given latent heatof ice : B0 cal/g -and the specific u'oiul., ol. I gram of ice andwater at OoC are I.091 cms and 1.000 cm3 iespectire&.
-.-! @ ri'a the increase ," ,lr",o"i,,ti];r::'Jt-:."J::i::,?whenJhe pressure i, il;;*;.;'-6;;;il:rpheres. Latent heat ofHl,tgg. of steam is s+o car/i J"i i.; fr,t""* -.ll'iii"l"ibzz
/6f,-: ^. . [Ane. ss'B4K;;'dir+:cf
"__ @ calcurare ,l:"l11,' r. ,,]. *:i;; *ffi;il:l:
lttii'fff 3#i'i"1,:.,'?"0,,:T'[.i;e,::] j:ti"]1"T**giffi r,&4rs,L near oI lusron is 4563 cal/mol and increase"i".*iu*"on fusion is I8.7 cmr/moi.- -i- iii :"+:i'{ib, ."g..
[Ans. _0 0697Y6 K or _0.06976.C]9r. calculate the temperature at which ice wiil freeze if thepressure is increased by 135.2 ";ilh..*:i,oi"-. ;r,." ; il;. "l_*1,.;' i;;u;;. il" ff :,"148i.",;,:o.B::atmospheric p*rrurr :
^l 06-dynes7"*;.""i.,"nt heat of fusion of ice@..*;t,"0., : : r*rn1';;V;;i. ua'
[Ans. _ l.o.c1
Given that -the "fr"ng"in specific vorume *(." o". ;.il;l'il;, ;';;_::',1lr,li."llX?rli,s 1676-cms. Latent t."t oF-"uo;#;^.i, nr a+o^* e,^ _ ttis 1676 cms. rut""i '[-."i';F'G;il?*l".r'?[l',lt'-1 lxf '"'.",?I,I : 4.2 x l0? ergs/cal .;J;r;;##;i".;;;;.rrr.e _ r os rr,rna",,^_2
J : 4.2x r0? ergs/car .;J;,;;##;i".;;Hii#.3iq;ffi;,:[9.JAne. I50"Cl
