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Relation between CP & CV
VPVP T
U
T
HCC
VP
VP T
U
T
PVUCC
PVUH Recall:
VPPVP T
U
T
VP
T
UCC
Can be related
Relation between CP & CV
VPPVP T
U
T
VP
T
UCC
VT
U
U is a function of T and V
dVV
UdT
T
UdU
TV
Thus;
At constant P;
PT
PV
P dVV
UdT
T
UdU
Relation between CP & CV
PT
PV
P dVV
UdT
T
UdU
Dividing by dTP ;
P
P
TVP
P
dT
dV
V
U
T
U
dT
dU
PTVP T
V
V
U
T
U
T
U
VPPVP T
U
T
VP
T
UCC
VPPVP T
U
T
VP
T
UCC
PT
VP T
VP
V
UCC
Joule Experiment
- A sample of a gas (the system) was placed in one side of the apparatus and the other side of the apparatus was evacuated.
-The initial temperature of the apparatus was measured.
-The stopcock was then opened and the gas expanded irreversibly into the vacuum.
To determine TV
U
Joule Experiment
-Because the surroundings were not affected during the expansion into a vacuum, w was equal to zero.
-The gas expanded rapidly so there was little opportunity for heat to be transferred to or from the surroundings.
- If a change in temperature of the gas occurred, heat would be transferred to or from the surroundings after the expansion was complete, and the final temperature of the surroundings would differ from the initial temperature.
Joule Experiment
UJ VT /Joule coefficient, J
To determine TV
U
UVT V
T
T
U
V
U
JVT
CV
U
Joule experiment gave and 0J 0/ TVUbecause the changes in temperature that occurred were too small to be measured by the thermometer.
Joule-Thomson experiment
Initial State
Final State
Insulated wall
To determine TP
H
Joule-Thomson experiment
rightleft www
2
1 0
2
0
1
V
V
dVPdVP
2211 VPVP
wqUUU 12
w0Adiabatic process
221112 VPVPUU Rearranging;
111222 VPUVPU
12 HH
0HIsenthalpic
process
Joule-Thomson experiment Joule-Thomson coefficient,
HJT PT /
HPT P
T
T
H
P
H
JTPT
CP
H
To determine TP
H
JT
Perfect Gases Perfect gas: one that obeys both of the
following:nRTPV
0
TV
U U is not change with V at constant T
If we change the volume of an ideal gas (at constant T),
we change the distance between the molecules.
Since intermolecular forces are zero, this distance
change will not affect the internal energy.
For perfect gas, internal energy can be expressed as a function of temperature (depends only on T):
An infinesimal change of internal energy,
Perfect Gases
TUU
dTCdU VV
V dT
UC
From
(slide 19)
Perfect Gases For perfect gas, enthalpy depends only on
T;
Thus;
An infinesimal change of enthalpy,
PVUH
nRTUH
This shows that H depends only on T for a perfect gas
THH
From
(slide19)
dTCdH PP
P dT
HC
Perfect Gases The relation of CP and CV for perfect gas;
PTVP T
VP
V
UCC
From slide 24
PP T
VP
T
VP
0
Perfect gas
nRCC VP
PV=nRT, thus PnRT
V
P
/
RCC mVmP ,,or
Perfect Gases For perfect gas,
JTPT
CP
H
JVT
CV
U
0
TV
U
0
TP
H
- Since U & H depend only on T.
Perfect gas and First LawFirst law;
For perfect gas;
dwdqdU
dwdqdU
PdVdqdTCV Perfect gas, rev. process, P-V work only
Reversible Isothermal Process in a Perfect Gas First law:
Since the process is isothermal, ∆T=0; Thus, ∆U = 0
First law becomes;
wqU
dTCdU V
0
0
wq 0
Reversible Isothermal Process in a Perfect Gas Since , thus:
Work done;
qw 0wq
2
1
PdVw
2
1
2
1
1dV
VnRTdV
V
nRT
2
1lnV
VnRT
1
2lnP
PnRTOR rev. isothermal
process, perfect gas
Since
Reversible Adiabatic Process in Perfect Gas First law,
Since the process is adiabatic,
For perfect gas, becomes,
dwdqdU
0dq
dwdU
dwdU dWdTCV dTCdU VPdV
dVV
nRT
Reversible Adiabatic Process in Perfect Gas
Since , thus
By integration;
dVV
nRTdTCV
dVV
RTdTC mV
,
2
1
2
1
, dVV
RdT
T
C mV
2
1lnV
VR
Reversible Adiabatic Process in Perfect Gas
If is constant, becomes;
OR
mVC ,
2
12
1
, lnV
VRdT
T
C mV
2
1
,1
2 lnlnV
V
C
R
T
T
mV
mVC
R
V
V ,
2
1ln
mVC
R
V
V
T
T ,
2
1
1
2
rev. adiabatic
process, perfect gas
Alternative equation can be written instead of
Consider
Reversible Adiabatic Process in Perfect Gas
mVC
R
V
V
T
T ,
2
1
1
2
222111 // TVPTVP
121122 /TTVPVP
mVC
R
V
VVPVP
,
2
11122
mVmV CRCR VPVP ,, 122
111
Since
Reversible Adiabatic Process in Perfect Gas
mVmV CRCR VPVP ,, 122
111
mVmVmV CRCCR ,,,1
RCC mVmP ,,mVmP CC ,,2211 VPVP mVmP CC ,,where
rev. adiabatic process, perfect gas, CV constant
Reversible Adiabatic Process in Perfect Gas If is constant, becomesVC wqU
wTTCV 12rev. adiabatic process, perfect gas, CV constant
Exercise A cylinder fitted with a frictionless piston
contains 3.00 mol of He gas at P = 1.00 atm and is in a large constant- temperature bath at 400 K. The pressure is reversibly increased to 5.00 atm. Find w, q and ∆ U for the process.
Solution A cylinder fitted with a frictionless piston
contains 3.00 mol of He gas at P = 1.00 atm and is in a large constant- temperature bath at 400 K. The pressure is reversibly increased to 5.00 atm. Find w, q and ∆ U for the process.
Assumption: He as a perfect gas. ∆U =0 since T is constant (U depends only on T)
1
2lnP
PnRT
0 wqU
2
1
PdVw 2
1
2
1
1dV
VnRTdV
V
nRT
2
1lnV
VnRT
Solution
1
2lnP
PnRTw
00.100.5ln400134.800.3 11 KKJmolmol
J41061.1
Jwq 41061.1
1. Reversible phase change at constant P & T. Phase change: a process in which at least one new
phase appear in a system without the occurance of a chemical reaction (eg: melting of ice to liquid water, freezing of ice from an aqueous solution.)
Calculation of First Law Quantities
PqH
wqU
VPPdVw 2
1
Calculation of First Law Quantities2. Constant pressure heating with no phase
change
VPPdVw 2
1
dTCqHT
T
PP 2
1
wqwqU P
3. Constant volume heating with no phase change
Recall:
Calculation of First Law Quantities
02
1
PdVw
2
1
dTCqU v
dT
dqC vv
qwqU
4. Perfect gas change of state:
H & U of a perfect gas depends on T only;
Calculation of First Law Quantities
dTCdH P dTCdU V
dTCH P2
1
dTCU V2
1
5. Reversible isothermal process in perfect gas H & U of a perfect gas depends on T only;
Since , thus
Calculation of First Law Quantities
dTCH P2
1
dTCU V2
1
0H 0U
2
12
1
lnV
VnRTPdVw
0 wqU wq
6. Reversible adiabatic process in perfect gas
If Cv is constant, the final state of the gas can be found from:
where
Calculation of First Law Quantities
0q Adiabatic process
dTCH P2
1
dTCU V2
1
wwqU
2211 VPVP
VP CC
Exercise CP,m of a certain substance in the temperature
range 250 to 500 K at 1 bar pressure is given by CP,m = b+kT, where b and k are certain known constants. If n moles of this substance is heated from T1 to T2 at 1 bar (where T1 & T2 are in the range of 250 to 500 K), find the expression for ΔH.
SOLUTION CP,m of a certain substance in the temperature range 250 to
500 K at 1 bar pressure is given by CP,m = b+kT, where b and k are certain known constants. If n moles of this substance is heated from T1 to T2 at 1 bar (where T1 & T2 are in the range of 250 to 500 K), find the expression for ΔH.
- P is constant throughout the heating, thus;
dTCqHT
T
PP 2
1
dTnCT
T
mP2
1
,dTkTbn
T
T 2
1
mPC ,
2
12
2 T
T
kTbTn
22
12 122
1TTkTTbn
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