thermodynamics. energy: ability to do work or produce heat. work=force x distance force causes the...
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Thermodynamics
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Energy: Ability to do work or produce heat.
Work=force x distance
force causes the object to move
Gravitational force causes the water to fall. can generate electricity
Energy
kinetic potential energy possessed by an object in virtue of its motion.
Ekin=1/2 mv2
Ekin=3/2 RTNever confuse
T and heat
Heat is the energy transferred from one object to another in virtue of T-difference
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Potential energy:
energy possessed by an object due to its presence in a force field i.e. under the effect of external force. Object attracted/repelled by external force. stored energy!
Epot=mgh
Attraction causes the ball to fall, h smaller, Epot smaller.
Attraction causes the potential energy to decrease.
Repulsion causes the potential energy to increase.
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Law of conservation of Energy (Axiome):
• Energy can neither be created nor destroyed.
• Energy of universe is constant.
• Energy can be converted from one form to another.
Ekin ↔ Epot
Heat ↔ Work
• Thermodynamics: the study of energy transformation
from one form to another.
• First Law of TD.
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System
Part of universe under
investigation.
sys
sys
surroundings
surroundings
sys + surr =
universe
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State Function
Change in state function depends
only on initial and final state.
Irbid
Amman
Sea level
h=650 m
h=900 m
Irbid → Amman
Dh=hfinal-hinitial
Dh=hamman-hirbid
Dh=900 m-650 m=250 m
Initial state
Final state
Change doesn’t depend on path
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• Examples of state functions:– Temperature– Volume– Pressure– Altitude– Mass– Energy– Concentration
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State function
xy
xy
xy
zzz
zdzdzd
dyy
zdx
x
zzd
aldifferentiexact
yxz
yxfz
342
),(
3
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),( 11 yx
),( 21 yx
),( 22 yx
(DU)x
(DU) y
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xy
z
yx
z
x
yz
y
xz
y
z
x
z
yxfz
xy
xy
22
),(
Schwartz theoremEquality of cross-derivatives
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00
46
342
),(
22
2
3
x
yz
xy
z
y
xz
yx
z
y
zx
x
z
yxz
yxfz
yy
xy
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2
2
2
2
2
),(
m
Tm
mmm
V
m
mTmmV
m
m
V
R
T
Vp
TV
p
V
R
V
Tp
VT
p
V
TR
V
p
V
R
T
p
V
TRp
TVfp
m
m
Show that p in the VdW equation is a state function!
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zxy
xzy
zxy
x
y
y
z
x
z
z
y
y
x
x
z
x
y
y
z
x
z
yxfz
1
),(
Circular Rule
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1
1
),(
2
2
RT
Vp
RT
V
R
p
V
R
p
V
V
T
T
p
R
p
V
T
V
TR
V
p
V
R
T
p
R
VpT
p
TRV
V
TRp
TVfp
mm
m
T
m
pmV
pmmTmmV
mm
m
m
m
m
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Internal Energy E
Sum of Ekin and Epot of all particles in the system.
State function
First Law of TD
DE = Q + W
The internal energy of a system can be changed1. by gaining or losing heat, Q
2. Work, W, done on the system or by the System
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Heat• energy flow from one system to another in virtue
of temperature difference.
– Appears only during a change in state
– Flows across the boundary from the a point of higher
temperature to a point of lower temp
– Not a state function
– Path function
QdQ
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Work• Any quantity of energy that “flows” across the
boundary between the system and the surroundings as a result of force acting through a distance.
– Appears only during a change in state
– Completely convertible into lifting a weight in the surr
– Not a state function
– Path function
WdW
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DE = Q + W
electricity (work) can be completely converted into lifting of weight.
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Q and ware path functions (Depend on
path).
full
initial
empty
final
Path 1
111 wQE
Path 2 22 QE
211
21
QwQ
EE
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sys
surroundings
Q
heat transferred from surr to
sys.
Surr loses heat, loses E, Esurr↓
Sys gains heat, gains E, Esys
for Surr: Q < 0 (neg.), DE < 0
for Sys: Q > 0 (pos.), DE > 0
sys
surroundings
Q
heat transferred from sys to
surr.
Sys loses heat, loses E, Esys↓
Surr gains heat, gains E, Esurr
for Sys: Q < 0 (neg.), DE < 0
for Surr: Q > 0 (pos.), DE > 0
DE = Q + W
Qsys > 0 : endothermic process
Qsys < 0 : exothermic process
dQsys > 0
dQsys < 0
dE = dQ + dW
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A
gm
A
F
pp
opposite
oppositeisys
1
m1
sys
surr
m1
sys
surr
m2
oppositeisys
opposite
ppA
gmmp
21
m1 m2
surr
sys
A
gmmp
pp
pV
opposite
oppositefsys
syssys
21
Ep=mgh
h↓, Ep↓
h of m1 and m2 ↓
Ep of m1 and m2 ↓
Esurr ↓
Esys
Work done by surroundings on system
(Esys)f > (Esys)i
DEsys > 0
wsys > 0
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A
gmm
A
F
pp
opposite
oppositeisys
21
m1
sys
surr
oppositeisys
opposite
ppA
gmp
1
m1 m2
surr
sys
A
gmp
pp
pV
opposite
oppositefsys
syssys
1
Ep=mgh
h, Ep
h of m1
Ep of m1
Esurr
Esys ↓
Work done by system on surroundings
(Esys)f < (Esys)i
DEsys < 0
wsys < 0
m1
surr
sys
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• Ex. 6.1
A system undergoes an endothermic process in which 15.6 kJ
of heat flows and where 1.4 kJ work is done on the system.
Calculate the total change in the internal Energy of the
system.
Qsys > 0 Q=+15.6 kJ
wsys > 0 w=+1.4 kJ
DEsys = Qsys + wsys
DEsys = (+15.6 kJ) + (+1.4 kJ) = +17 kJ
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m1
sys
surr
hi
m1
surr
sys
hf
initial final
Vpw
VVpwhAhApw
hhApwrApw
ApFA
Fp
rFw
opp
ifoppifopp
ifoppopp
oppopp
-
m1 m2
surr
sys
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(Vf, p
f)
p
Vm
(Vi, p
i)
popp
=pf
W
Ideal GasOne-stage isothermal Expansion
T=constant
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m1
sys
surr
hi
hf
initial final
-
m1 m2
surr
sys
m1 m2-dm
surr
sys
dVpdhApw
dxFw
oppopp
In each step: - dm is removed - Piston rises by dh - psys = popp
- dw=-poppdV - dw=-psysdV
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Reversible isothermal Expansion
reversible: System in equilibrium with surroundings at each point of process oppsys pp
i
f
V
V
sysopp
opp
V
VnRTw
V
VdnRTdV
V
nRTdww
stepeachofworkofsumworktotal
dVV
nRTdw
V
nRTpp
dVpdw
f
i
ln
Ideal gas
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i
f
V
VnRTw ln
Reversible Expansion
Ideal gasVf > Vi
w < 0
Reversible Compression
Ideal gasVf < Vi
w > 0
Reversible work larger than irreversible work.
But impractical.Requires infinity long time.
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• Ex. 6.2
Calculate the work associated with the expansion of a gas
from 46 L to 64 L at constant external pressure of 15 atm.
46 L 64 L
15 atm 15 atm
o Expansion against the external pressureo External pressure opposes the expansion
o popp=15 atm = constant
LatmLatmLLatmw
VVatmVpw ifopp
.2701815466415
15
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• Ex. 6.3
Given a balloon with a volume of 4.00x106 L. It was heated by
1.3x108 J until the volume became 4.5x106L. Assuming the
balloon is expanding against a constant external pressure of
1 atm, calculate the change in the internal energy of the
gas confined by the balloon.
4.00x106 L
1 atm1 atm
4.50x106 L
Vi
Vf
Q
popp
JJJJJE
JmPamPaLatm
LatmJLLatmJE
VVpQEVpQEwQE ifoppopp
77868
333
68668
100.81007.5103.1325.101105.0103.1
325.101.325.10110101325.1
.105.0103.1105.4105.41103.1
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HeatMolar heat capacity
Electrical work w=V.I.t → system gains this energy as heat qKinetic energy of system rises → temperature rises
Translational, rotational, vibrational energy increase
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cp,m: molar heat capacity at constant pressure heat needed to raise the temperature of 1 mole of substance by 1ºC at constant pressure.
cV,m: molar heat capacity at constant volume heat needed to raise the temperature of 1 mole of substance by 1ºC at constant volume.
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cp: heat capacity at constant pressure heat needed to raise the temperature of n mole of substance by 1ºC at constant pressure.
cV: heat capacity at constant volume heat needed to raise the temperature of n mole of substance by 1ºC at constant volume.
mpp cnc ,
mVV cnc ,
pmp
pp dT
q
nc
dT
qc
.
1,
VmV
VV dT
q
nc
dT
qc
.
1,
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Extensive and intensive properties
properties
extensive intensive
depends on amount of substance
cp, V, m, n
doesn’t depend on amount of substance
cp,m, p, T
Ratio of two extensive propertiescp,m, Mwt,
n
mMwt
n
cc p
mp ,
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dTcndTcq
dT
q
nc
dT
qc
mppp
pmp
pp
,
, .1
dTcndTcq
dT
q
nc
dT
qc
mVVV
VmV
VV
,
, .1
dVpw
dVpqdE
wqdE
opp
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dVpqdE opp
00
0.
ww
Vdvolumeconst
tindependenTcifTcnE
qEEE
qdTcnqdE
qdE
mVVm
Vif
V
T
T
VmV
E
E
V
f
i
f
i
,,
,
Vq
0w
Measure qV
→ change in
internal energy
of system DE
determined
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VmV
V
mVV
V
TV
ccnT
E
dTcnqEd
dTT
EEddVvolumeconstat
dVV
EdT
T
EEd
VTfE
,
,
0:
),(
dVV
EdTcnEd
T
mV
,
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dTcnEd
dTT
EdV
V
EdT
T
EEd
V
EgasidealFor
mV
VTV
T
,
0:
0
0
:/exp
dVV
EdT
T
EEd
dT
gasidealofncompressioansionIsothermal
TV
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HqHH
VpEH
qVpEVpE
VpVpqEE
pVpVqVVpqEE
dVpqdVpqdE
dVpqdVpqdVpqdE
pif
piiifff
iiffpif
ifpifpif
V
V
p
V
V
p
E
E
psyspopp
f
i
f
i
f
i
Enthalpy, H
First law under the conditions
- Reversible popp=psys
- constant pressure psys=const
pqHd
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p
p
qdH
VdpqdH
VdppdVpdVqdH
VdppdVdEdH
pVddEdH
pVEH
qHd
)(
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pmp
p
mpp
p
TV
ccnT
H
dTcnqHd
dTT
HHddppressureconstat
dpp
HdT
T
HHd
pTfH
,
,
0:
),(
dpp
HdTcnHd
T
mp
,
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dTcnHd
dTT
Hdp
p
HdT
T
HHd
p
HgasidealFor
mp
pTp
T
,
0:
0
0
:/exp
dpp
HdT
T
HHd
dT
gasidealofncompressioansionIsothermal
Tp
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fi zzfunctionstatezdz
dHdE
0
00
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dVpdTcn
dVpqdVV
EdT
T
EEd
ppdq
gasidealofncompressioansion
adiabaticreversible
mV
TV
sysopp
,.
0
/exp
V
dV
c
R
T
dT
dVV
RTndTcn
mV
mV
,
,.
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i
f
i
f
mV
mp
i
f
mV
mVmp
i
f
mVmp
i
f
mVi
f
V
VmV
T
T
V
V
T
T
c
c
V
V
c
cc
T
T
Rccgasidealfor
V
V
c
R
T
T
V
dV
c
R
T
dT f
i
f
i
ln1ln
lnln
:
lnln
,
,
,
,,
,,
,
,
ii
ff
i
f
i
f
i
f
i
f
i
f
i
f
i
f
i
f
Vp
Vp
RnpV
RnpV
T
T
Rn
pVTTRnpV
V
V
T
T
V
V
T
T
V
V
T
T
1
1
lnln
ln1ln
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iiff
iiifff
i
f
ii
ff
VpVp
VVpVVp
V
V
Vp
Vp
11
1
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dVpdTcn mV ,.
Adiabatic compression: work done on system → internal energy increase → T↑
00 dTdV
dVpdTcn mV ,.
Adiabatic expansion: work done by system → internal energy decrease → T↓
00 dTdV
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reversible adiabatic
work for ideal gas:
TcnnRTnRTR
cnRTnRT
cc
cw
nRTnRTcc
cnRTnRT
c
cnRTnRTw
VpVpVVpVVpw
VconstVconstVVconstV
constw
dVVconstwdVVconstw
VconstpconstpV
dVpdVpw
mVifmV
ifmVmp
mV
ifmVmp
mVif
mV
mpif
iffiifff
ifif
V
V
V
V
opp
ii
f
i
f
i
,,
,,
,
,,
,
,
,
1
1
1
1
1
1
1
1
1
1
1
11
11
1111
dVpdTcn mV ,.much easier
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Dependence of internal energy on volume changes
pT
pT
V
E
dVV
EdT
T
EEd
VT
TV
For ideal
gas: V
nR
T
p
V
nRTp
V
0
pppV
nRT
V
E
T
For VdW
gas: 2
2
V
an
bnV
RTnp
bnV
nR
T
p
V
2
2
2
2
V
an
V
an
bnV
RTn
bnV
nRT
V
E
T
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Two moles of a VdW gas were compressed at constant temperature of
25oC from 10 L to 1 L. Calculate the change in the internal energy E.
fiif
V
V
V
V
V
V
TTV
VVan
VVan
VanE
V
dVandV
V
anEd
dVV
andV
V
EdV
V
EdT
T
EEd
f
i
f
i
f
i
11111 222
22
2
2
2
2
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Vp
Vp
V
E
p
H
VdpdVpV
Edp
p
H
dTisothermal
VdppdVdVV
EdTcndp
p
HdTcn
VdppdVdVV
EdT
T
Edp
p
HdT
T
H
VdppdVdEdH
pVEH
dpp
HdT
T
HHd
TTT
TT
T
mV
T
mp
TVTV
TV
0:
,,
Dependence of enthalpy on pressure
changes
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pT
pTV
TpV
TVTVT
VT
TTT
T
VTV
p
H
T
V
p
V
T
p
p
V
V
T
T
p
Vp
V
T
pTV
p
Vpp
T
pT
p
H
pT
pT
V
E
Vp
Vp
V
E
p
H
1
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For ideal
gas: p
nR
T
V
p
nRTV
p
For
liquids
and
solids:
0
p
nRT
p
nRT
p
H
T
VTV
p
H
T
pT
VdpdTcndHVp
Hmp
T
.. ,
dTcndTT
EdV
V
EdT
T
EEd
ilitycompressiblowdV
mV
VTV
..
0
,
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Relation
between
cp and cv
pT
mVmp
TT
mVmp
T
mVmp
mpp
opp
T
mV
opp
T
mV
TV
T
Vp
V
Ecncn
dVpV
EdVpdV
V
EdTcncn
reversibledVpdVV
EdTcndTcn
dTcnqqpressureconstatprocessaFor
dVpdVV
EdTcnq
dVpqdVV
EdTcn
wqdVV
EdT
T
EEd
,,
,,
,,
,
,
,
..
..
)(..
.
.
.
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pVpV
mVmp
VT
pT
mVmp
T
V
T
pT
T
Vpp
T
pTcncn
pT
pT
V
E
T
Vp
V
Ecncn
,,
,,
..
..
p
nR
T
V
p
nRTV
p
For ideal
gas:
Rcc
p
nRp
T
Vp
V
Ecncn
mVmp
pT
mVmp
,,
,, ..
For
liquids
and
solids:
mVmp cc ,, because is rather
smallp
T
V
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Joule-Thompson Effect
p1=constant p2=constant Insulated tube →
q=0
2211
0
0
21
0
0
21
21
1
2
1
2
VpVpdVpdVpdVpdVpw
dVpdVpdwdwdw
V
V
V
V
rightleft
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00
21
111222
221112
dHH
HH
VpEVpE
VpVpEE
WWQE
TmpH
TJ
TH
mp
T
mp
TV
p
H
cnp
T
p
H
p
Tcn
dpp
HdTcn
dpp
HdT
T
HHd
,
,
,
1
0
isoenthalpic
Joule-Thompson
coefficient
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For ideal
gas:
0
Tp
H
01
,
TmpH
TJ p
H
cnp
T
Pressure drop: p1 > p2 dp
< 0dT = 0 no change in Temp
Tinincreasep
T
Tindropp
T
Tinchangenop
T
H
TJ
H
TJ
H
TJ
00
00
00
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For VdW
gas:
Can be shown at zero-
pressure limit:
bRT
a
cnp
T
mpH
TJ
21
,
J-T inversion temperature: T at which mJ-T=0
Expansion at this Temp: no change in T
bR
aT
bRT
ab
RT
a
bRT
a
cn
TJ
TJTJ
TJmpTJ
2
20
2
021
,
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TinincreasebRT
aTTif
TindropbRT
aTTif
TinchangenobRT
aTTif
TJTJ
TJTJ
TJTJ
02
02
02
If a nitrogen gas cylinder is opened at room temp:
Frost formation on valve
If a hydrogen gas cylinder is opened at room temp:
increase of valve temp
eventual explosion
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Liquefying Air
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Chemical EnergyCH4(g) +2O2(g) → CO2(g) + 2H2O(g)
C-H O=O C=O O-H
Chemical reaction:o No change in the number/nature of atomso Redistribution of Bonds (change in bonding)o Change in attraction & repulsion forces between the atoms
o Change in the potential energy Ep of molecules
Ep
R
P
Energy is conserved!
Energy difference released as heat.
Heat of reaction (Qv, Qp).
Reaction exothermic.
DHreaction= Hf – Hi = HP – HR < 0
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N2(g) + O2(g) → 2NO(g)
N≡N O=O N=O
Ep
R
P Energy is conserved!
Energy difference obtained
from
surroundings as heat.
Heat of reaction (Qv, Qp=DE,
DH).
Reaction endothermic.DHreaction= Hf – Hi = HP – HR > 0
DEreaction= Ef – Ei = EP – ER > 0
o Ep(R) > Ep(P), reaction exothermic
o Ep(R) < Ep(P), reaction endothermic
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Thermochemical equation
N2(g) + O2(g) → 2NO(g) DHo=+180.5 kJ
CH4(g) +2O2(g) → CO2(g) + 2H2O(g) DHo=-802.3 kJ
moles
1 mole of gaseous methane (CH4) reacts with two moles of
gaseous molecular oxygen producing 1 mole of gaseous carbon
dioxide, 2 moles of water vapor and 802.3 kJ of heat.
DHo: Standard heat of reaction:Standard conditions: T=25oC, p=1atm.
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Calorimetry
caloriecal
measurement
heat unit
1 cal = 4.185 J
Calorimetry = heat measurement experiments
1 Cal =1000 cal
Problem: - heat (Q) is a path function!!!
- Q differs from one way of performing
the experiment to another.
- details of the experiment must be described!!!!
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heat measurement experiments
VpQE
wQE
opp
0
tan
V
tconsV
VQE
Heat measured at constant volume:
o Equal to DEo Equal to a change in a state function!!o Details of the experiment no more important.
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However, heat measurement experiments are usually performed at constant pressure pVEH
VdppdVdEdH
VpVpQH
VpEH
VVpEEHH
VpEH
pppdpppp
VdppdVdEdH
ififif
VV
EE
HH
iffi
p
p
V
V
E
E
H
H
f
i
f
i
f
i
f
i
f
i
f
i
f
i
.
0
0
pQH Heat measured at constant pressure:
o Equal to DH
o Equal to a change in a state function!!o Details of the exp. no more important.
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csp: specific heat (specific heat capacity) heat needed to raise the temperature of 1 gram of substance by 1ºC.
C : heat capacity heat needed to raise the temperature of substance (m gram) by 1ºC.
T increase by 1ºC:
1 g csp
m gr. ? = CspcmC
Q : heat heat needed to raise the temperature of substance (m gram) by a given temperature difference, DTºC.
m gr.
CT increase by 1ºC
T increase by DTºC
? = QTcmQ
TCQ
sp
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Bomb Calorimeter
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0.5269 g of octane (C8H18) were placed in a bomb calorimeter
with a heat capacity of 11.3 kJ/ºC. The octane sample was
ignited in presence of excess oxygen. The temperature of
the calorimeter was found to increase by 2.25ºC. Calculate
DE of the combustion reaction of octane.
kJCC
kJTCQQ V 4.2525.23.11
DE defined for the reaction as written!!!!!!!!!!!
C8H18(g) +12.5O2(g) → 8CO2(g) + 9H2O(g)
DE defined for the combustion of 1 mole octane (114.2 g)!!
0.5269 g QV
114.2 g ? = DE
DE=-(114.2 g x 25.4 kJ)/0.5269g=-5505 kJ
n
Q
Mwtm
Q
m
QMwtE VVV
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When 1.5 g of methane (CH4) was ignited in a bomb
calorimeter with 11.3 kJ/ºC heat capacity, the temperature
rised by 7.3ºC. When 1.15 g hydrogen (H2) was ignited in
the same calorimeter, the temperature rised by 14.3ºC.
Which one of the two substances has a higher specific heat
of combustion (i.e. heat evolved upon the combustion of 1
g of substance)? kJCC
kJTCCHQV 833.73.114
kJCC
kJTCHQV 1623.143.112
1.5 g QV=83 kJ
1 g ? =55 kJ/g
1.15 g QV=162 kJ
1 g ? =141 kJ/g
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Coffee-Cup Calorimeter
50 mL of 1.0 M HCl at 25ºC were
added to 50 mL of 1.0 M NaOH at
25ºC in a coffee-cup calorimeter.
The tempe-rature was found to rise
to 31.9ºC. Calculate the heat of the
neutraliza-tion reaction!
Was caused the temperature to increase?
Exothermic Reaction
HCl(aq) +NaOH(aq) → NaCl(aq) + H2O(l)
H+(aq) +OH-
(aq) → H2O(l)
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heat evolved = heat gained + heat
gained by reaction by solution by
calorimeter rcalorimetesolutionsprct TCTcmQ
Assumptions:
Ccal=0 (very small mass)
Solution ≈ water (csp)solution=(csp)water=4.18 Jg-1ºC-1
(density)solution=(density)water=1 g/mL
gmLmL
gVdm 1001001
JCCg
JgQrct 2.28849.618.4100
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HCl(aq) +NaOH(aq) → NaCl(aq) + H2O(l)
nHCl=MHClxVHCl = 1 mol/L x 0.050 L = 0.050 mol
nNaOH=MNaOHxVNaOH = 1 mol/L x 0.050 L = 0.050 mol
0.050 mol 0.050 mol
0.050 mol 0.050 mol
0.050 mol H2O 2884.2 J mol
1 mol H2O ? Qp=57,684 J/molH2O
DH= -57,684 J/molH2O
DH= -57.7 kJ/molH2O
n
QH p
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Hess’s Law
N2(g) + 2O2(g)
initial
2NO2(g)
final
2NO(g)
O2(g) O2(g)
path 1
path 2
)(2)(2)(
)()(2)(2
)(2)(2)(2
21
22
2
22
ggg
ggg
ggg
if
NOONOforH
NOONforH
NOONforH
HH
HHH
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N2(g) + O2(g) → 2NO(g) DH2a
2NO(g) + O2(g) → 2NO2(g) DH2b
N2(g) + 2O2(g) → 2NO2(g) DH1
DH1=DH2a+DH2b
The enthalpy of a given chemical reaction is constant,
regardless of the reaction happening in one step or many
steps.
If a chemical equation can be written as the sum of
several other chemical equations (steps), the enthalpy
change of the first chemical equation equals the sum of
the enthalpy changes of the other chemical equations
(steps).
Hess’s Law:
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Rules for manipulating thermochemical equations
- If equation is multiplied by a factor, multiply DH by this factor.
N2(g)+3H2(g) → 2NH3(g) DH=-92 kJ
2x (N2(g)+3H2(g) → 2NH3(g) DH=-92 kJ)
2N2(g)+6H2(g) → 4NH3(g) DH=-184 kJ
1/2x (N2(g)+3H2(g) → 2NH3(g) DH=-92 kJ)
1/2N2(g)+3/2H2(g) → NH3(g) DH=-46 kJ
- If equation is reversed, change the sign of DH
2NH3(g) → N2(g) + 3H2(g) DH=+92 kJ
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The enthalpy of combustion of graphite is -394 kJ/mol.
The enthalpy of combustion of diamond is -396 kJ/mol.
Calculate DH for the reaction:
Cgraphite → Cdiamond
Solving Strategy• Write the given data in form of thermochemical equations:
CG + O2(g) → CO2(g) DH=-394 kJ
CD + O2(g) → CO2(g) DH=-396 kJ• Construct the equation of interest from the given data:
1 mole cgraphite is needed as reactant. Take the equation in the given data that contains cgraphite. Check the number of moles and whether it is on the reactant side. Manipulate if necessary.
CG + O2(g) → CO2(g) DH=-394 kJ1 mole cdiamond is needed as product. Take the equation in the given data
that contains cdiamond. Check the number of moles and whether it is on the product side. Manipulate if necessary.
CO2(g) → CD + O2(g) DH=+396 kJ
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Sum the resulting equations and their DH values:
CG + O2(g) → CO2(g) DH=-394 kJ
CO2(g) → CD + O2(g) DH=+396 kJ
Cgraphite → Cdiamond DH=+2 kJ
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Given:
2 B(s)+3/2 O2(g) → B2O3(s) DH=-1273 kJ
B2H6(g)+3 O2(g) → B2O3(s) + 3 H2O(g) DH=-2035 kJ
H2(g)+1/2 O2(g) → H2O(l) DH=-286 kJ
H2O(l) → H2O(g) DH=+44 kJ
Calculate DH for
2 B(s) + 3 H2(g) → B2H6(g)
2 B(s)+3/2 O2(g) → B2O3(s) DH=-1273 kJ
B2O3(s) + 3 H2O(g) → B2H6(g)+3 O2(g) DH=+2035 kJ
3H2(g)+3/2 O2(g) → 3 H2O(l) DH=3x(-286) kJ
2 B(s) + 3 H2O(g) + 3 H2(g) → B2H6(g)+3 H2O(l) DH=-96 kJ
3 H2O(l) → 3 H2O(g) DH=3x(+44) kJ
2 B(s) + 3 H2(g) → B2H6(g DH=+36 kJ
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Heat of Formation
Formation reaction:
reaction of forming 1 mole of product from the
elements in their stable form at 25ºC and 1
atm.Heat of formation = DH of formation reaction = DFH
Standard heat of formation = DHº of formation reaction = DFHº
DFHº(NO(g)): ½ N2(g)+½ O2(g) → NO(g) DHº
DFHº(CO(g)): Cgraphite(s)+½ O2(g) → CO(g) DHº
DFHº(O(g)): ½ O2(g) → O(g) DHº
DFHº(Cdiamond(s)): Cgraphite(s) → Cdiamond(s) DHº
DFHº(O2(g)): O2(g) → O2(g) DHº=0
DFHº(Cgraphite(s)): Cgraphite(s) → Cgraphite(s) DHº=0
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01,25
atmC
elementsstableH o
oF
tsreac
oiFiproducts
oiFi
orct HnHnH
tan
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CH4(g) +2O2(g) → CO2(g) + 2H2O(g)DH=?
CG(s)+ O2(g) → CO2(g) DFH(CO2)
2x (H2(g)+1/2 O2(g) → H2O(g) ) 2xDFH(H2O)
CH4(g) → CG(s) + 2 H2(g) -DFH(CH4)
O2(g) → O2(g) -DFH(O2)=0
CH4(g) +2O2(g) → CO2(g) + 2H2O(g)
DH=DFH(CO2)+ 2xDFH(H2O) - DFH(CH4) - DFH(O2)
DH=DFH(CO2)+ 2xDFH(H2O) – [DFH(CH4) +DFH(O2)]
tsreac
oiFiproducts
oiFi
orct HnHnH
tan
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4 NH3(g) +7 O2(g) → 4 NO2(g) + 6 H2O(l)DH=?
DH= 4xDFH(NO2)+ 6xDFH(H2O) – 4xDFH(NH3)
2 Al(s) +Fe2O3(s) → Al2O3(s) + 2 Fe(s)DH=?
DH= DFH(Al2O3)+ 2xDFH(Fe) – [DFH(Fe2O3)+ 2xDFH(Al)]
DH= DFH(Al2O3) – DFH(Fe2O3)
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2 CH3OH(l) +3 O2(g) → 2 CO2(g) + 4 H2O(l)DH=?
DH= 2xDFH(CO2)+ 4xDFH(H2O) – 2xDFH(CH3OH)
DH= 2x(-394 kJ)+ 4x(-286 kJ) – 2x(-239 kJ)=-1454 kJ
2 mol CH3OH -1454 kJ
2x32 g
?
-1454 kJ
1 g = -22.7 kJ/g
Calculate the heat of combustion of methanol
(CH3OH(l)) in kJ/g and compare its value with that of
octane (C8H18(l)).
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C8H18(l) +12.5 O2(g) → 8 CO2(g) + 9 H2O(l)
DH= 8xDFH(CO2)+ 9xDFH(H2O) – DFH(C8H18)
DH= 8x(-394 kJ)+ 9x(-286 kJ) – (-276 kJ)=-5450 kJ
1 mol C8H18 -5450 kJ
114 g
?
-5450 kJ
1 g = -47.8 kJ/g