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EXAM – 02 (CHEM 3301) Spring 2011
Marking Scheme
- (45 points) Thermodynamic Relationships - “Maxwell’s Relations”
Using the first and second laws, or any other way you choose. For FULL Credit
YOU MUST JUSTIFY each step
a) (5 pts) Show that the fundamental equation for dU is given by
dU = TdS - PdV
dU = δq + δw (first law)
dU = δq – P ext
dV (definition of work)
If the process is assumed to be reversible, P ext
=P
gas
=P and dq rev
= TdS (from
second law of thermodynamics)
dU=TdS-PdV
b) (5 pts) Treat U as a function of V and S. Write out its total derivative, dU, in
terms of dV, dS and partial derivatives of H.
U (V, S)
dU = (∂U/∂V) S
dV + (∂U/∂S) V
dS
c) (5pts) Show that P
V
U
S
and that T
S
U
V
From a) dU = – PdV + TdS
From b) dU = (∂U/∂V) S
dV + (∂U/∂S) V
dS
As both equations for dU are equal, then the coefficients for dS and dV
must be equal.
- P = (∂U/∂V)
S
and T = (∂U/∂S) V
OR From a)From a) dU =dU = – – PdV + TPdV + T dSdS
(dU/dS)= (dU/dS)= - - P(dV/dS) + TP(dV/dS) + T
When the volume is constant dV=0 When the volume is constant dV=0 T = (T = (∂∂U/U/∂∂S)S) VV
From a)
From a)
dU =
dU =
PdV + TdS
PdV + TdS
(dU/dV)= (dU/dV)= - - P(dV/dV) + T(dS/dV)P(dV/dV) + T(dS/dV)
When the entropy is constant dS=0 When the entropy is constant dS=0 - - P = (P = (∂∂U/U/∂∂V)V)
S
S
d) (10 pts) Show that
V S
V
T
S
P
U is a state function. Therefore dU is an exact differential
That means that [(∂/∂V (∂U/∂S)
V
]
S
= [(∂/∂S (∂U/∂V)
S
]
V
From c) T = (∂U/∂S) V
and - P = (∂U/∂V) S
S V
S
P
V
T
e) (5 pts) Show that
T
Cv
T
S
V
, what is this relation useful for?
From a) dU = TdS – PdV
Dividing the whole expression by dS and considering the volume is constant
(∂U/∂S)
V
= T - P(∂V/∂S)
V
(at constant V, dV = 0)
(∂U/∂T)
V
(∂T/∂S)
V
= T
Since (∂U/∂T)
V
=C
v
, C
v
(∂T/∂S)
V
= T
T
Cv
T
S
V
The value of
V
T
S
can be estimated through C v
and T, which can be
MEASURABLE in the lab.
h) (5 pts) Calculate the change in entropy when the pressure of one mole of an
ideal gas is increased from 1 to 2.3 atmospheres at 300 K. (NB 2.3 ~e, and ln
e=1. Assume R=10 J.K
.mol
V
nR
V
S
T
2
1
2
1
S
S
V
V
dV
V
nR
dS
1
2
ln
V
V
! S = nR , since, at constant T,
1 1 2 2
PV = PV
2
1
ln
P
P
! S = nR
=1 mol x 10 J. mol
. K
ΔS= - 10 J. K
OR
i
f
V
V
S nR ln can be used together with
i i f f
PV = PP (at constant T) to derive
f
i
P
P
S nR ln
OR Use
f
i
P
P
S nR ln directly.
(30 points) Thermodynamics of mixing
a) (5 points) The chemical potential of an ideal gas is given by
A
o
+ RT ln (p
A
/p
o
) Define each of the quantities in the formula.
μ A
- the chemical potential of substance A
μ
o
- the chemical potential of A at 1.0 bar (standard Chemical potential of A)
P
A
- the partial pressure of A
P
0
- the standard pressure at 1.0 bar (i.e. 1 bar)
T- temperature
R- 8.3 J. mol
. K
(Gas constant)
b) (10 points) Consider two gases n
A
moles of A and and n
B
moles of B, each at
pressure p. The gases are mixed at constant temperature and total pressure.
Calculate the initial Gibbs energy, the final Gibbs energy and
show that ΔG
mix
= nRT (x
A
ln x
A
+ x
B
ln x
B
) Justify all steps.
G
initial
= G
A, initial
+ G
B, initial
= n A
μ A
o
RT ln (p/p
o
) + n B
μ B
o
RT ln (p/p
o
G
final
= G
A, final
+ G
B, final
= n A
μ A
o
RT ln (p A
/p
o
) + n B
μ B
o
RT ln (p B
/p
o
ΔG
mix
= G
final
- G
initial
= n
A
RT ln (p
A
/p) + n
B
RT ln (p
B
/p)
p A
o
= p B
o
= p
p final
= p
Dalton’s law p A
/p = x A
, p B
/p = x B
ln (p A
/p A
o
) = ln (p A
/p final
) = ln x A
ln (p
B
/p
B
o
) = ln (p
B
/p
final
) = ln x
B
ΔG
mix
= G
final
- G
initial
= n A
RT ln x A
RT ln x B
n A
= nx A ,
n B
= nx B
(n = n A
ΔG
mix
= nx
A
RT ln x
A
B
RT ln x
B
ΔG
mix
= nRT (x A
ln x A
ln x B
e) (10 points) These results for ΔG
mix
, ΔS
mix
and ΔH
mix
also apply real
gases/liquids. Why? Discuss this from a molecular perspective while mentioning
differences and similarities of ideal gases and real gases/liquids.
ΔG
mix
= ΔH
mix
mix
In ideal gases, the interactions between species are neglected. The enthalpy of mixing is
zero. The mixing is driven by the entropy alone. Mixing is always spontaneous for ideal
systems.
Then G mix
= TΔS
mix
for ideal gases
In real gases/liquids we do have interactions. The enthalpy of mixing is NOT zero.
If ΔH mix
<0, then mixing will ALWAYS occur.
If ΔH mix
0, then mixing may not always occur.
In fact, if ΔH mix
> TΔS
mix
, then ΔG mix
- The gases/liquids will NOT mix. They are
immiscible.
3. Non ideal solutions (Circle the CORRECT answer)
(20 points)
i) The vapor pressure of pure acetone is
a) 12 kPa
b) 18 kPa
c) 35 kPa
d) 46 kPa
ii) The total vapor pressure of a 50:50 mixture of acetone and chloroform is
a) 12 kPa
b) 18 kPa
c) 30 kPa
d) 35 kPa
iii) The region where acetone obeys Raoult’s Law (p A
= x A
p A
) is indicated by
the arrow labeled 1 2 3 4 5 6 or 7
iv) The region where acetone obeys Henry’s Law (p A
= x A
K
A
) is indicated by
the arrow labeled 1 2 3 4 5 6 or 7
v) The Henry’s Law constant for acetone is
a) 12 kPa
b) 24 kPa
c) 35 kPa
d) 36 kPa
1
2
3
4
7
