Solution to Homework 2
MAE 216: Statistical Thermodynamics
University of California, Davis, Winter Quarter 2014
1
Ideal Gas and Maxwell’s Relations (10 points)
1.a
Starting from dU = T dS − P dV , show that the equation of state P V = mRT in
fact implies that U can only depend on T .
In general, U (T, V, P ). However, due to the equation of state P V = mRT , only two of the
three variables T , V and P are independent. Without loss of generality, we choose U (T, V ) =
). All other thermodynamic functions thus also depend on T and V alone, e.g., S(T, V ).
U (T, V, mRT
V
We now calculate the total derivative of S(T, V )
∂S ∂S dS =
dT +
dV
(1)
∂T V
∂V T
which we then use to calculate the total derivative of U (T, V )
∂S ∂S ∂S ∂S dT + T
dV − P dV = T
dT + T
− P dV
dU = T dS − P dV = T
∂T V
∂V T
∂T V
∂V T
. (2)
By equating the cross derivatives of the Helmholtz free energy F (T, V ) = U (T, V ) − T S(T, V ), we
obtain one of Maxwell’s relations (see lecture notes)
∂S ∂P =
(3)
∂T V
∂V T
which can may be used in equation (2)
∂P ∂S dU = T
dT + T
− P dV
∂T V
∂T V
.
(4)
From the equation of state, we have
mRT
P =
V
∂P mR
T
=T
=P
∂T V
V
⇒
(5)
which gives
∂S (
(
()(dV
(P
dU = T
dT + (
(P(−
∂T V
Hence, U can only depend on T .
1
⇒
∂S dU = T
dT
∂T V
.
(6)
1.b
Now consider a Carnot Cycle using an ideal gas as the working substance
with internal energy U = 32 N kB T (where k is a constant). Using the fact that
dU = δQ − P dV , show that the adiabatic portions of the cycle must obey the
relation P V γ = constant and solve for the value of γ.
By definition of an adiabatic process, δQ = 0 and thus dU = −P dV . Using the equation of state
P V = mRT = N kB T , the internal energy becomes U = 32 P V . Considering U (P, V ) as a function
of P and V alone, its total derivative gives
∂ 3
∂ 3
3
dU =
P V dV +
P V dP = P dV + V dP
.
(7)
∂V 2
∂P 2
2
P
V
Equating to dU = −P dV , we obtain
3
P dV + V dP = −P dV
2
dP
5 dV
=−
P
3 V
⇒
.
(8)
Integrating both sides from state (P0 , V0 ) to state (P1 , V1 ), we obtain
Z
P1
P0
dP
5
=−
P
3
Z
V1
V0
dV
V
⇒
5
ln P1 − ln P0 = (ln V0 − ln V1 )
3
⇒
P1
= ln
ln
P0
V0
V1
5/3
. (9)
Exponentiating both sides and rearanging the terms gives
5/3
P0 V0
5/3
= P1 V1
,
(10)
which implies that P V γ with γ = 5/3 is constant during the adiabatic process.
2
Hard core gas (5 points)
A first correction for the ideal gas behaviour is to account for the fact that each gas molecule has
some volume, which we denote here by b. To correct for the excluded volume introduced by each
particle, we arrive at the modified equation of state, P (V − N b) = mRT .
Using a similar approach to problem 1 above, show that the heat capacity at constant
volume, CV , can only be a function of T .
As in problem 1.a, we can express U as a function of T and V alone (this time U (T, V ) =
U (T, V, VmRT
)) and equations (1)–(4) can be obtained in the same way. Equation 5 becomes
−N b
∂P mR
mRT
⇒
T
=T
=P
(11)
P =
V − Nb
∂T V
V − Nb
which leaves equation (6) unaffected. Hence, once again, U is a function of T alone. As a consequence,
∂U dU
CV =
=
(12)
∂T V
dT
is also a function of T alone.
2