Chem 420/523 Chemical Thermodynamics
Homework Assignment # 2
1. Derive explicit expression for the reversible work of isothermal expansion done in each of the following
cases: (a) dV is obtained from the equation of state pV = RT + Bp + Cp2 . (b) dV is obtained from the
Berthelot equation [Eq. (4.12) on p. 162].
Answer
(a) From the equation of state given,
RT
+ B + Cp.
p
µ
¶
RT
dVm = − 2 + C dp, at constant T .
p
Vm =
Therefore, the work done during an isothermal reversible expansion is
w=−
Z
2
1
Z
pdVm = −
Z
Z
2
2
1
µ
p −
¶
RT
+ C dp
p2
2
dp
pdp
=RT
−C
1 p
1
µ ¶
´
p2
C³ 2
−
p2 − p21 .
=RT ln
p1
2
(b) The Berthelot equation is
"
9pTc
pVm = RT 1 +
128pc T
Ã
6T 2
1 − 2c
T
!#
.
Expanding and rearranging, we get
RT
54RTc3
9RTc
−
.
+
p
128pc 128pc T 2
RT
dVm =− 2 dP, at constant T .
p
Vm =
Therefore, the work done during an isothermal reversible expansion is
w=−
Z
1
2
pdVm = RT
µ
¶
p2
.
=RT ln
p1
Z
1
2
dp
p
2. Rozen [J. Phys. Chem. (USSR) 19, 469
³ (1945);
´
³Chem.
´ Abstracts
³ 40,
´ 1712 (1946)] characterizes gases
∂p
p ∂V
p2
T
∂V
by “deviation coefficients” such as p ∂T , R ∂T , and RT ∂p . Calculate these coefficients for
V
p
T
(a) ideal gas (b)* van der Waals gas (c)* a gas that obeys the Dieterici equation of state:
pVm =
RT −a/(RT Vm )
e
.
Vm − b
Answer
1
(a) For the ideal gas,
µ
¶
·
µ
¶¸
∂p
∂ RT
R
=
=
;
∂T V
∂T Vm V
Vm
µ
¶ ·
µ
¶¸
∂V
∂ RT
R
=
= ;
∂T p ∂T
p
p
p
µ
¶ ·
µ
¶¸
∂V
∂ RT
RT
=
=− 2 .
∂p T ∂p
p
p
T
So, we get
µ
¶
T ∂p
RT
=
= 1.
p ∂T V pVm
µ
¶
p ∂V
=1.
R ∂T p
µ
¶
p2 ∂V
=−1.
RT ∂p T
(b) * For van der Waals’ gas, we have
a
RT
−
Vm − b Vm2
µ
¶
∂p
R
.
=
∂T V
Vm − b
µ
¶
RT
T ∂p
=
p ∂T V
p (Vm − b)
a
=1+ 2.
Vm
p=
Since T, p, and Vm are exact differentials, we can use Eq. (1.33) on p. 26 to get an expression for
by first calculating
³
∂T
∂V
´
p
:
³
∂V
∂T
´
p
1
a
(p + 2 )(Vm − b)
R
Vm
µ
¶
∂T
2a
1
a
=−
(Vm − b) + (p + 2 )
3
∂V p
RVm
R
Vm
3
pV − aVm + 2ab
= m
.
RVm3
T =
Therefore,
µ
∂V
∂T
¶
p
=
"µ
∂T
∂V
p
R
Also, by the same approach, we calculate
above):
³
µ
¶ #−1
=
∂V
∂T
=
p
∂V
∂p
´
T
µ
¶
RVm3
pVm3 − aVm + 2ab
pVm3
.
pVm3 − aVm + 2ab
p
by first calculating
∂p
∂V
2
¶
T
=−
³
∂p
∂V
´
T
(see the expression for p
RT
2a
2 + V3.
(Vm − b)
m
µ
¶
Vm3 (Vm − b)2
RT Vm3 − 2aVm2 + 4aVm b − 2ab2
T
µ
¶
p2 ∂V
p2 Vm3 (Vm − b)2
=
.
RT ∂p T
RT (RT Vm3 − 2aVm2 + 4aVm b − 2ab2 )
∂V
∂p
=
(c) * From the Dieterici equation of state,
RT
e−a/(RT Vm )
Vm (Vm − b)
µ
¶
∂p
R
a
R
=
×
e−a/(RT Vm ) +
e−a/(RT Vm )
∂T V
Vm (Vm − b)
RT Vm Vm (Vm − b)
p
a
= (1 +
).
T
RT Vm
µ
¶
T ∂p
a
= (1 +
).
p ∂T V
RT Vm
p=
In this case, it is not possible to write a closed
³ form
´ expression for T . Therefore, we directly differentiate
the equation given to get an expression for ∂V
:
∂T
p
³
´
"
µ
¶
R
∂Vm
RT
RT
a
=
+
+
−
p
2
(Vm − b) (Vm − b)
∂T# p (Vm − b) RT 2 Vm
µ
¶
RT
∂Vm
a
e−a/(RT Vm )
(Vm − b) RT Vm2 ∂T p
µ
¶
µ
¶
µ
¶
∂Vm
∂Vm
∂Vm
pVm
pVm
ap
ap
−
p
=
+
+
.
∂T p
T
(Vm − b) ∂T p RT 2 RT Vm ∂T p
∂Vm
∂T p
Dividing both sides by p and collecting terms containing the partial derivative, we get
µ
¶µ
¶
∂Vm
Vm
Vm
a
a
1+
=
.
−
+
(Vm − b) RT Vm
∂T p
T
RT 2
µ
¶
µ
¶µ
¶−1
p ∂Vm
pVm
Vm
ap
a
=
1+
.
+ 2 2
−
R ∂T p
RT
R T
(Vm − b) RT Vm
Proceeding in a similar manner for
³
∂V
∂p
´
T
, we get
RT
e−a/(RT Vm )
p (Vm − b)
"
µ
¶
µ
¶
µ
¶ #
∂Vm
∂Vm
∂Vm
RT
a
RT
RT
−
= − 2
+
e−a/(RT Vm )
∂p T
p (Vm − b) p (Vm − b)2 ∂p T p (Vm − b) RT Vm2
∂p T
·
µ
¶
µ
¶ ¸
∂Vm
a
∂Vm
Vm
Vm
= −
+ Vm
.
−
p
(Vm − b) ∂p T
RT Vm2
∂p T
Vm =
Collecting terms containing the partial derivative, we get
·
¸µ
¶
∂Vm
Vm
Vm
a
1+
=−
−
.
(Vm − b) RT Vm
∂p T
p
µ
¶
·
¸−1
p2 ∂Vm
a
pVm
Vm
−
=−
1+
.
RT
∂p T
RT
(Vm − b) RT Vm
3
3. Calculate w, ∆U, q, and ∆H in an isothermal
expansion of 1 mole of a gas that obeys the
³ ´ reversible
³ ´
∂p
∂U
= T ∂T
− p.]
equation of state pV = RT + Bp. [Hint: ∂V
T
V
Answer
From the equation of state, we have p(Vm − B) = RT or p =
w=−
We are given
³
∂U
∂V
´
T
=T
³
∂p
∂T
´
V
Z
2
1
Z
RT
. Therefore,
Vm − B
2
RT
dVm
1 Vm − b !
Ã
Vm,2 − B
= −RT ln
.
Vm,1 − B
pdVm = −
− p. From the equation of state,
µ
∂U
∂V
¶
T
µ
∂p
∂T
¶
=
V
R
Vm − B
RT
=
− p = 0.
Vm − B
Therefore, ∆U = 0 for an isothermal expansion process.
Since ∆U = 0, using the first law, we get
Ã
!
Vm,2 − b
.
q = −w = RT ln
Vm,1 − b
The enthalpy change is dH = d(U + pVm ) = dU + d(pVm ) = 0 + d(RT + Bp). Since the process is
isothermal, dH = Bdp. Therefore,
∆H = B(p2 − p1 ).
4. A gas obeys the equation of state pV = RT + Bp and has a heat capacity CV,m that is independent of
temperature. (a) Derive an expression relating T and V in an adiabatic reversible expansion. (b) Derive
an equation for ∆H in an adiabatic reversible expansion. (c) Derive an equation for ∆H in an adiabatic
free expansion.
Answer
(a) For an adiabatic process, dq = 0. Therefore, from the first law, we have dU = −pdVm , or
CV,m dT = −pdVm .
Substituting for p from the equation of state, this becomes
CV,m dT = −
RT
dVm .
Vm − B
This equation can be rearranged and both sides integrated as follows:
dT
dVm
=R
TÃ
Vm −!B
µ ¶
T2
Vm,2 − B
= R ln
.
CV,m ln
T1
Vm,1 − B
CV,m
4
Therefore, we get
T2
=
T1
Ã
Vm,2 − B
Vm,1 − B
!R/CV,m
.
(b) From the definitions, dH = d(U + pVm ) = CV,m dT + d(RT + Bp) = CV,m dT + RdT + Bdp. Thus,
dH = (CV,m + R)dT + Bdp.
[We cannot use the relationship that CV,m + R = Cp,m because this is not an ideal gas!] Since CV,m is
independent of tempreature (given) and R is a constant, we get
∆H = (CV,m + R) (T2 − T1 ) + B(p2 − p1 ).
(c) In a free expansion, no work is done and so, dw = 0. Also, by definition, dq = 0. Therefore, from first
law, we get
dU = dq + dw = 0!
Since dU = CV,m dT by definition, this implies that dT = 0 for this process. In other words, the
temperature remains constant. Therefore,
dH = dU + d(pVm ) = dU + RdT + Bdp = Bdp,
∆H = B(p2 − p1 ).
h
5. * Given that Cp,m = CV,m + V −
or (1.38); p. 29], show that
³
´ i³
∂H
∂p T
∂p
∂T
´
V
·
CV,m = Cp,m 1 − µJT
, and the chain rule of partial derivatives [Eq. (1.37)
µ
∂p
∂T
¶ ¸
−V
µ
¶
−V
µ
V
∂p
∂T
¶
.
V
Answer
From the given equation,
CV,m = Cp,m +
µ
∂H
∂p
¶ µ
T
∂p
∂T
V
∂p
∂T
¶
.
V
Since enthalpy s a function of T and p, i.e., H(T, p), we write
dH =
µ
∂H
∂p
¶
T
µ
∂H
dp +
∂T
¶
dT.
T
At constant enthalpy (as in the Joule-Thompson experiment), dH = 0 and we get
µ
¶
µ
¶
∂H
∂H
dp = −
dT
∂p T
∂T T
µ
¶
µ
¶ µ
¶
∂H
∂H
∂T
=−
∂p T
∂T T ∂p H
= −Cp,m µJT .
Substituting in Eq. (1), we get
·
CV,m = Cp,m 1 − µJT
µ
5
∂p
∂T
¶ ¸
V
−V
µ
∂p
∂T
¶
V
.
(1)
6. * From the hint given in Question 3 and the result derived in Question 2(a) of Homework Assignment
#1, show that Cp,m − CV,m = α2 V T /κ, where α and κ are defined in Chapter 1 [Eqs. (1.40) and (1.39),
respectively].
Answer
From Q. 3, we have
µ
∂U
∂V
From Q. 2(a) of assignment 1, we have
µ
Substituting for
³
´
∂U
∂V T
∂H
∂T
¶
µ
¶
µ
=T
T
¶
∂U
∂T
∂p
∂T
·
¶
V
µ
(2)
− p.
¶ ¸µ
¶
∂U
∂V
=
+ p+
.
∂V
∂T p
p
V
T
·
µ
¶ ¸µ
¶
∂U
∂V
.
Cp,m = CV,m + p +
∂V T
∂T p
(3)
from Eq. (2), we get
·
Cp,m = CV,m + p + T
or
Cp,m − CV,m = T
Now, by definition,
µ
∂p
∂T
¶
µ
∂p
∂T
¶ µ
µ
V
V
¸µ
∂V
∂T
−p
∂V
∂T
¶
.
p
¶
1 ∂V
V ∂T p
µ
¶
1 ∂V
κ=−
V ∂p T
α=
Using the chain rule of partial derivatives (see Eqs. (1.43) and (1.44)),
µ
∂p
∂T
¶
=
V
Substituting in Eq. (4), and recognizing that
µ
− (∂V /∂T )p
α
= .
(∂V /∂p)T
κ
∂V
∂T
¶
= αV , we get
p
Cp,m − CV,m = α2 V T /κ.
6
¶
,
p
(4)