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Chemistry 5350Advanced Physical Chemistry
Fall Semester 2013
First Law and State Functions
Take Home Quiz 2
Due: September 26, 2013
1. The internal energy of a perfect monotomic gas relative to its value at T = 0 is 3
2nRT .
Calculate(
∂H∂P
)
Tand
(
∂U∂P
)
Tfor the gas.
U =
(
3
2
)
nRT ;
(
∂U
∂P
)
T
= 0
by direct differentiation
H = U + PV =
(
3
2
)
nRT + nRT =
(
5
2
)
nRT
(
∂H
∂P
)
T
= 0
by direct differentiation
2. Express(
∂CP
∂p
)
Tas a second derivative of H, and find its relation to
(
∂H∂P
)
T
From this relation show that(
∂CP
∂P
)
T= 0 for a perfect gas.
(
∂Cp
∂P
)
T
=
[
∂
∂P
(
∂H
∂T
)
P
]
T
=
[
∂
∂T
(
∂H
∂P
)
T
]
P
Because(
∂H∂P
)
T= 0 for a perfect gas, the temperature derivative also equals zero.
Thus,(
∂CP
∂P
)
T
= 0
3. Write an expression for dV given that V is a function of P and T .
Deduce an expression for d lnV in terms of the expansion coefficient and isothermal compressibility
V = V (P, T )
Hence,
dV =
(
∂V
∂P
)
T
dP +
(
∂V
∂T
)
P
dT
d lnV =dV
V=
(
1
V
)(
∂V
∂P
)
T
dP +
(
1
V
)(
∂V
∂T
)
P
dT
α =
(
1
V
)(
∂V
∂T
)
p
κT = −
(
1
V
)(
∂V
∂P
)
T
d lnV = −κTdP + 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
{
−T
(
∂V
∂T
)
P
+ V
}
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
)
H
∼ −
b
CP
and
∆T = −
b
CP
×∆P > 0
5. At low pressure and 298 K, the experimental value of N2for 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 Waalsequation of state
P =RT
Vm − b−
a
V 2m
can be shown to be
µJT =
(
2a
RT− b
)
1
Cp,m
The van der Waals equation of state can be written as:
PV = RT −
a
V+ bP +
ab
V 2
If the very small term abV 2 is neglected and the term a
PVis replaced by a
RT, then
V =RT
P−
a
RT+ b
(
∂V
∂T
)
P
=R
P+
a
RT 2
from aboveR
P=
V − b
T+
a
RT 2
then
T
(
∂V
∂T
)
P
− V =2a
RT− b
µJT = −
1
CP
{
−T
(
∂V
∂T
)
P
+ V
}
=1
CP
(
2a
RT− b
)
b. At what temperature does the Joule-Thomson coefficient for N2vanish? For N
2, 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
Rb=
2× 1.35× 106atm cm6 mol−2
(82.0578cm3 atm K−1 mol−1)× (36cm3mol−1)= 852 K
µJT > 0 below the inversion temperature, and µJT < 0 above the inversion tempera-
ture.
c. Can N2gas 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 isless 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 N2gas by a Joule-
Thomson expansion
6. 2.00 moles of Kr gas at 300 K expands isothermally and reversibly from 20.0 L to 50.0 L. Calculateq, 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
w = −
∫ V2
V1
PdV = −
∫ V2
V1
nRT
VdV = −nRT ln
V2
V1
w = −nRT lnV2
V1
= −2.00 8.314 J mol−1 K−1 (300 K) ln
(
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 volumeof 80.0 L. Calculate q, w, ∆U , and ∆H. Assume ideal gas behavior.
First determine the final temperature using CV = 3
2R:
TCV
R
2V2 = T
CV
R
1V1
TCV
R
2× (80.0 L) = (300 K)
CV
R × (20.0 L)
T3
2
2= 300
3
2K× (0.25); T2 = 119 K
q = 0
∆U = w =
∫
119K
300K
nCV dT = nCV∆T
w = 2.00 mol×3
2× 8.314 J mol−1 K−1
× (119 K - 300 K) = -4.51 kJ
∆H =
∫
1299K
300K
nCPdT = nCP∆T = 2×5
2× 8.314 J mol−1 K−1
× (119 K - 300 K) = -7.52 kJ
8. An equation of state of a gas is given by P = nRTV−nb
.
a. Give the total differential for the equation of state of this gas with respect to volume.
dV =
(
∂V
∂T
)
P
dT +
(
∂V
∂P
)
T
dP
dV =
(
nR
P
)
dT −
(
nRT
P 2
)
dP
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.
∫ V2
V1
dV =
∫ T2
T1
nR
PdT
(V2 − V1) =nR
P× (T2 − T1)
c. Obtain an expression from Part b for one mole of gas under the following conditions: P = 2.00atm, T = 273 K, b = 0.115 L/mol.Using this data, one can obtain a value of V1
V1 =nRT
P+ b = 11.2 L
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.
0
10
20
30
40
50
273 373 473 573
T
V Equation c
Equation e
9. Derive a general relation between CV and CP
dU =
(
∂U
∂T
)
V
dT +
(
∂U
∂V
)
T
dV = dq − PextdV
then
dq =
(
∂U
∂T
)
V
dT +
[
Pext +
(
∂U
∂V
)
T
]
dV
at constant pressure
dqP = CV dT +
[
Pext +
(
∂U
∂V
)
T
]
dV
Dividing through by dT and setting dqPdT
= CP , we obtaine
CP − CV =
[
P +
(
∂U
∂V
)
T
](
∂V
∂T
)
P
The above equation takes on a simple form for an ideal gas because(
∂U∂V
)
T= 0 and
(
∂V∂T
)
P= nR
P.
ThenCP − CV = nR
