Chemistry 5350
Advanced Physical Chemistry
Fall Semester 2013
Take Home Quiz 2
Due: September 26, 2013
First Law and State Functions
1. The internal energy of a perfect monotomic gas relative to its value at T = 0 is 23 nRT .
and ∂U
for the gas.
Calculate ∂H
∂P T
∂P T
3
U=
nRT ;
2
∂U
∂P
=0
T
by direct differentiation
5
3
nRT + nRT =
nRT
H = U + PV =
2
2
∂H
∂P
=0
T
by direct differentiation
∂CP
∂p
as a second derivative of H, and find its relation to
T
P
From this relation show that ∂C
= 0 for a perfect gas.
∂P T
2. Express
Because
Thus,
∂H
∂P T
∂Cp
∂P
T
∂
=
∂P
∂H
∂T
P
T
∂
=
∂T
∂H
∂P
∂H
∂P T
T
P
= 0 for a perfect gas, the temperature derivative also equals zero.
∂CP
∂P
=0
T
3. Write an expression for dV given that V is a function of P and T .
Deduce an expression for d ln V in terms of the expansion coefficient and isothermal compressibility
V = V (P, T )
Hence,
∂V
∂V
dV =
dP +
dT
∂P T
∂T P
1
∂V
∂V
dV
1
=
dP +
dT
d ln V =
V
V
∂P T
V
∂T P
∂V
∂V
1
1
κT = −
α=
V
∂T p
V
∂P T
d ln V = −κT dP + αdT
4. A gas obeying the following equation of state
P (V − nb) = nRT
is subjected to a Joule-Thomson expansion.
Will the temperature increase, decrease, or remain the same?
µJT =
∂T
∂P
H
1
=−
CP
∂H
∂P
T
1
=−
CP
∂V
−T
+V
∂T P
For a gas which obeys the above equation of state
µJT = −
b
CP
Since b > 0 and CP > 0, then for this gas µJT < 0 or
∂T
∂P H
< 0.
This equation indicates that when the pressure drops during the Joule-Thomson expansion,
∆P = P2 − P1 < 0
the temperature must increase
∆T
∆P
and
∆T = −
∼−
H
b
CP
b
× ∆P > 0
CP
5. At low pressure and 298 K, the experimental value of N2 for the Joule-Thomson coefficient, µJT =
(∂T /∂P )H = 0.222K atm−1 .
a. Derive the expression for the Joule-Thomson coefficient for a gas which obeys the van der Waals
equation of state
a
RT
− 2
P =
Vm − b Vm
can be shown to be
µJT =
2a
−b
RT
1
Cp,m
The van der Waals equation of state can be written as:
P V = RT −
If the very small term
ab
V2
a
ab
+ bP + 2
V
V
is neglected and the term
a
PV
is replaced by
a
,
RT
RT
a
−
+b
P
RT
a
R
∂V
= +
∂T P
P
RT 2
V =
from above
then
µJT
R
V −b
a
=
+
P
T
RT 2
∂V
2a
T
−b
−V =
∂T P
RT
∂V
2a
1
1
−T
−b
=−
+V =
CP
∂T P
CP RT
then
b. At what temperature does the Joule-Thomson coefficient for N2 vanish? For N2 , a = 1.35 ×
106 atm cm6 mol−2 , b = 38.6 cm3 mol−1 , and Cp,m = 29.125J mol−1 K−1
Set µJT = 0 to find the inversion temperature:
T =
2a
2 × 1.35 × 106 atm cm6 mol−2
= 852 K
=
Rb
(82.0578cm3 atm K−1 mol−1 ) × (36cm3 mol−1 )
µJT > 0 below the inversion temperature, and µJT < 0 above the inversion temperature.
c. Can N2 gas be cooled by a Joule-Thomson expansion if the temperature is higher than the
temperature calculated above? Explain in terms of the value given for µJT .
In an expansion of a gas, the change in pressure is negative, ∆P = P2 − P1 < 0, because P2 is
less than P1 .
Using the equation for the Joule-Thomson expansion, the change in temperature is given by
∆T = ∆P × µJT
If the temperature of the gas is less than the inversion temperature, (Tinversion < 852K, µJT >
0), the temperature of the gas decreases during the expansion.
If the temperature of the gas is greater than the inversion temperature, (Tinversion > 852K, µJT <
0), the temperature of the gas increases during the expansion.
Thus, for temperatures greater than 852 K, it is not possible to cool the N2 gas by a JouleThomson expansion
6. 2.00 moles of Kr gas at 300 K expands isothermally and reversibly from 20.0 L to 50.0 L. Calculate
q, w, ∆U , and ∆H. Assume ideal gas behavior.
∆U = 0 and ∆H = 0 for an ideal gas
q = −w
For this reversible process Pext = P , giving
Z
Z V2
P dV = −
w=−
V1
V2
V1
nRT
V2
dV = −nRT ln
V
V1
V2
= −2.00 8.314 J mol−1 K−1 (300 K) ln
w = −nRT ln
V1
50.0 L
20.0 L
= -4.57 kJ
q = −w = 4.57 kJ
7. 2.00 moles of Kr gas at 20 L and 300 K expands adiabatically and reversibly to a final volume
of 80.0 L. Calculate q, w, ∆U , and ∆H. Assume ideal gas behavior.
First determine the final temperature using CV = 32 R:
CV
CV
T2 R V2 = T1 R V1
CV
T2 R × (80.0 L) = (300 K)
3
3
T22 = 300 2 K × (0.25);
CV
R
× (20.0 L)
T2 = 119 K
q=0
∆U = w =
w = 2.00 mol ×
∆H =
Z
119K
nCV dT = nCV ∆T
300K
3
× 8.314 J mol−1 K−1 × (119 K - 300 K) = -4.51 kJ
2
1299K
nCP dT = nCP ∆T = 2 ×
300K
Z
5
× 8.314 J mol−1 K−1 × (119 K - 300 K) = -7.52 kJ
2
8. An equation of state of a gas is given by P =
nRT
.
V −nb
a. Give the total differential for the equation of state of this gas with respect to volume.
∂V
∂V
dT +
dP
dV =
∂T P
∂P T
nRT
nR
dT −
dP
dV =
P
P2
b. Solve for the change in volume as a function of state variables at constant pressure.
For constant pressure, dP = 0 and the second term in the exact differential vanishes.
Z T2
Z V2
nR
dV =
dT
P
T1
V1
(V2 − V1 ) =
nR
× (T2 − T1 )
P
c. Obtain an expression from Part b for one mole of gas under the following conditions: P = 2.00
atm, T = 273 K, b = 0.115 L/mol.
Using this data, one can obtain a value of V1
V1 =
nRT
+ b = 11.2 L
P
The equation now becomes
V2 = 11.2 L + 0.0410L K−1 × (T2 − 273 K)
d. Plot the expression from part c on a graph of volume versus temperature for this gas:
e. Add to the graph in part d the plot of similar function for P = 1.000 atm.
V2 = 22.4L + 0.0820L K−1 × (T2 − 273 K)
Label the two functions c and e.
50
Equation e
40
V
30
Equation c
20
10
0
273
373
473
T
573
9. Derive a general relation between CV and CP
∂U
∂U
dU =
dT +
dV = dq − Pext dV
∂T V
∂V T
then
dq =
∂U
∂T
dT + Pext +
V
∂U
∂V
dV
T
at constant pressure
dqP = CV dT + Pext +
∂U
∂V
dV
T
dqP
dT
= CP , we obtaine
∂V
∂U
CP − CV = P +
∂V T
∂T P
Dividing through by dT and setting
The above equation takes on a simple form for an ideal gas because
Then
CP − CV = nR
∂U
∂V T
= 0 and
∂V
∂T
P
=
nR
.
P