che 220 lecture 17 - pennsylvania state university
TRANSCRIPT
Topics for Today
* Examples with ideal gases* Irreversible processes
* Review of Ch3
Reading : 38,39
EXAMPLE l :
5mo1ofN2@5bar.300K.ismixedwithl0molesofNzac5bar.5o0K.a4abatica11yaeconst.P.
Determine final and enevsy balances.
Assumeidealgasmthq >= 29µmol . K .
Solution.
. Adiabatic : Q=0I const . P : Q=oHtt
€t ,
FTZ
h ,N nz initial A : n ,H(T , ,P)th2H(Tz,P )
final It ( nitric ) HCTE,
P )
OH = ( nitm )Hf - h ,H , -
mHz=O nets
mi ( HF - H ,)+m(He - Hz )=0
in x '
DHFCPCTF - T , ) oHz=cp( TF - Tz )
~D Mcptf - T, ) tnzcp ( Tf - Tz ) - ON >
Tf= mi¥T2=
433.3kNitmz
Example I ( cent 'd ) Method 2 : calculate all H from a ref . state :
choose : HCT , ,P)=0
initial :
Hiftn' HKTBHH 't " " )
} � Hinitiakhvptz 'T )
H÷lz = Hot ( p ( Tz - T , )
final : HE( nitnz ) HCTF,
P )
HLT . ,P)= cpt ,. T , )
) � Htm' - MHM ) soft . t , )
OH =0⇒ mcptz - T, ) + ( nitnz ) ftp.T/=0XTf=niT+mTzmitnz
Exercise : Change the ret . State to : HCTF,PHO
and repeat calculations.
Example 1 ( an 'd ) Methot 3
Them, moles of compartment I change from
th,P ) -w( Tf ,P)
OH 'TEn , got ,= - T )
Then, moles of compartment I change from th,P)to (TIP)
DH ztokmzg > ( TF - T, )
Total Change; t¥oH%oHEr=m , cptf , ) tmcptt - h )=0� Tf =
MT tnztz
mitmz
Example 1 cont 'd Energy balances
Dlttk N , ( ✓ ( Tf - T , ) tnzcv ( TF - Tz ) = 0
Qtr = 0
Wtrkoutot - Q = o
Horris this possible ?
* Compartment l expands and produces work.
But
compartment Z contracts and ansumes work .
The two amount 1 cancel each other.
Before
mixingV, = RpI= 4.988×15331 m
: NNHNZVZ = 0.108 m3Vz= 12¥.
8.314.16 Mymo , Mno volume change
trAfter mixing : Vf = My = 7.205.503%1
,-7 Cnithz )VF= 0.108ms
EXAMPLE Z : Repeat Example 1- with Cp= function of T.
92/12=90+9 ,T+9#+a5P+qyT4 at 3.53g9 , = - 0.000261
Solution 92=7.0×10893=1.3053×159
Working equation is same at before : 94=9.9×10-13TF
oHt±o=n,§FpHdT+nzJcptHT=oTz
MR :{ not . - T . )+ai( II . TI )+ . . . }+ nz{ao( TF - Tz ) tai ( T¥ . Tzl )+ . . . }=0
Sdvefov TF ~D TF = 434k
Resultnomucw different,
but that's only bkthegssot Nzdoes mol change much with 't
.
EXAMPLE 3 : Compartment 1 : Nz @ T , = 500k , P,= 5bar
,VE 0.25 m3
Compartment 2 : Vaccum V2= 0 . I m3
Remove partition and allow to equilibrate .
Determine final state & perform balances.
Assume ideal gas with g , = 29J And . K.
Solution
Hit?! vac. Before
on 't ¢¥wFri=o� Tz =Tl
Iii.FI?if . AterFinal Pressure :
R¥, = Pz¥k) � Pz =P ,
I= 3.57 bar
/ # Vitvz
* Process is irreversible
* Gas expands w/o producing work .
Example 3 ( cont 'd ) Perform the expansion reversibly andcalculate the energy balances
.
It.si?eiiiExpand against opposing pressure .
Reversible Adiabatic Process :
Final Pressure
P, .VE Pzckth )t� Pz= Pi (¥-3.12 barV it Vz
f- Cpkv = li 4
Final Temperature
Tz=Ti ( P⇒PkP= 436.8k
Work:W= on . ¢°= cvtz- T, ) = - 1308 Jlmol } � Wttk
39,33g JN= RVYPT
,= 30 .
1 mol