1. CHEM 444 HW#7
Answer key by Jiahao Chen
due November 10, 2004
Grading policy: One point is given for each numbered equation shown in the solution, or for written
statements and/or equations to the same effect. The total score for this problem set is 25 points plus a
maximum of 5 bonus points.
Note. If you see PAQ on your script, it means “please answer the question.”
¯ ¯ ¯
1. (22-8) Determine CP − CV for a gas that obeys the equation of state P V − b = RT . [5 points]
Solution. Divide the result of Problem 3 (22-11) throughout by n to convert it into the molar equation
¯ 2
¯ ¯ ∂V ∂P
CP − CV = −T ¯ (1)
∂T P ∂V T
The required derivatives of the equation of state are
¯
∂V R
= (2)
∂T P P
∂P RT
¯ = − 2 (3)
∂V T ¯
V −b
P2
= − (4)
RT
2
¯ ¯ R P2
∴ CP − CV = −T − =R (5)
P RT
Note. You can use Equation (22.23), the way the solutions manual does. However I don’t want you to
get the idea that thermodynamics is just about looking up the right equation from the textbook.
Here we derived a result in this problem set so I’m perfectly fine with just quoting that result.
2. (part of 22-10) Use Equation 22.22 to show that
∂CV ∂2P
=T (6)
∂V T ∂T 2 V
∂CV
Show that ∂V T
= 0 for an ideal gas and a van der Waals gas. [8 points]
Solution. Equation 22.22 is
∂U ∂P
= −P + T (7)
∂V T ∂T V
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2. CHEM 444 HW#7 Solutions
which when differentiated with respect to T at constant V gives
∂ ∂U ∂P ∂P ∂2P ∂2P
=− + +T =T (8)
∂T ∂V T V
∂T V ∂T V ∂T 2 V ∂T 2 V
but recall that since U is an analytic function, the mixed derivatives are equal, and so by defini-
tion
∂ ∂U ∂2U ∂2U ∂ ∂U
= = = (9)
∂T ∂V T V
∂T ∂V ∂V ∂T ∂V ∂T V T
∂CV
≡ (10)
∂V T
which completes the proof.
The equation of state for an ideal gas is
P V = nRT (11)
and so
∂2P ∂ nR ∂CV
= =0⇒ =0 (12)
∂T 2 V ∂T V V ∂V T
and for a van der Waals gas,
RT a
P = ¯ + ¯2 (13)
V −b V
∂2P ∂ R ∂CV
⇒ = ¯ =0⇒ =0 (14)
∂T 2 V ∂T V −b ∂V T
V
Note. Quoting the origin of Maxwell’s relation is OK as long as you define new symbols that you
introduce. It is really really important that you state what are y1 , y2 , x1 and x2 , even if you just
use them once.
3. (22-11) Consider V to be a function of T and P and write out dV . Now divide through by dT at
constant V and substitute the expression for ∂P V that you obtain into Equation 22.23 to get [4
∂T
points]
2
∂V ∂P
CP − CV = −T
∂T P ∂V T
Solution. As required,
∂V ∂V
V = V (T, P ) ⇒ dV = dT + dP (15)
∂T P ∂P T
So differentiating with respect to T at constant V gives
∂V ∂V ∂P
0 = + (16)
∂T P ∂P T ∂T V
∂P ∂V ∂V ∂V ∂P
⇒ = − ÷ =− (17)
∂T V ∂T P ∂P T ∂T P ∂V T
Substituting into Equation 22-23:
∂P ∂V
CP − CV = T (18)
∂T V ∂T P
2
∂V ∂P
= −T
∂T P ∂V T
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3. CHEM 444 HW#7 Solutions
Note. Equation 17 follows immediately from the permutation rule for partial derivatives:
∂V ∂T ∂P
= −1 (19)
∂T P ∂P V ∂V T
Note. Please do not mix intensive and extensive quantities! Either put overbars over all relevant
variables, or not at all. What I mean is that
¯ ∂U
CV = = CV
∂T V
Note. When the question says to “divide through by dT at constant V ” it doesn’t mean literal division.
What it means is to change things like dP to become ∂P V . Again, please review multivariable
∂T
calculus.
4. (22-14) . Show that the enthalpy is a function of only the temperature for a gas that obeys the
¯
equation of state P V − bT = RT . [4 + 5 points]
Solution. Note that since H = H (S, P ) it is sufficient to prove that H for this gas does not vary with
pressure [1 point]. To show this, we need Equation (22.34) which we can derive as follows [total
of 5 bonus points]: begin with the differential form dH:¯
¯ ¯ ¯
dH = T dS + V dP (20)
and take the partial derivative with respect to P at constant T :
¯
∂H ¯
∂S
=T ¯
+V (21)
∂P T ∂P T
¯
∂S
We can eliminate ∂P using the Maxwell’s relation from the thermodynamic potential which
T
has natural variables P and T , i.e. the Gibbs free energy. [1 point] From the differential form
¯ ¯ ¯
dµ = dG = −SdT + V dP (22)
the Maxwell’s relation thus obtained is
¯
∂S ¯
∂V
− = (23)
∂P T ∂T P
which when substituted into Equation 21 then gives Equation (22.34) (divided throughout by n);
applying this result to the equation of state gives
¯
∂H ∂V¯
¯
= V −T (24)
∂P T ∂T P
¯ R
= V −T +b (25)
P
= 0 (26)
where the equation of state was again applied to obtain the final result.
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4. CHEM 444 HW#7 Solutions
5. (22-18) Show that
∂V
dH = V − T dP + CP dT (27)
∂T P
What does this tell you of the natural variables of H? [4 points]
Solution. Considering H = H (P, T ) gives
∂H ∂H
dH = dP + dT (28)
∂P T ∂T P
Using Equation 22-34 (none other than Equation 24 above without the overbars)
∂H ∂V
=V −T (29)
∂P T ∂T P
And that the latter derivative is the heat capacity at constant pressure
∂H ∂U
H = U + PV ⇒ = = CP (30)
∂T P ∂T P
gives the final result
∂V
dH = V − T dP + CP dT
∂T P
The result shows that the change of H with respect to P is not independent of the temperature
and therefore P and T together cannot be considered natural variables of H. [1 point]
Supplement on natural variables. The textbook gives an incredibly hand-wavy definition of nat-
ural variables. (If you wanted to be cynical, you could say that neither McQuarrie nor Simon
really understands the concept either!) Here I try to explain what exactly is meant by “the coef-
ficients... are simply thermodynamic functions.” An IUPAC Technical Report [1] defines natural
variables as “[a set of] independent variables [used] to describe the extensive state.” What this
means is that if x and y are natural variables of f , for example, then each derivative does not
depend on the other variable, i.e.
∂f ∂f ∂f ∂f
= (y) and = (x)
∂x y ∂x y ∂y x ∂y x
In this problem, we can see clearly from Equation (22.34) that ∂H T is a function of T and
∂P
hence the pair (P, T ) does not form a set of natural variables for H. Note that it is meaningless
to describe any one variable as being ’natural’. I hope this makes sense to everyone.
Note. It is dangerous to say that the derivates are complex. Yes, complex means ’not simple’, but it
also means ’not real’!
References
[1] Alberty, R. A. et. al., Use of Legendre Transformations in Chemical Thermodynamics (IU-
PAC Technical Report). Pure Appl. Chem., 73 (8), 2001, 1349-1380. Available online at
http://www.iupac.org/publications/pac/2001/pdf/7308x1349.pdf.
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