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PHY303 Statistical Thermodynamics Fall 2012 Prof. D. Garanin
Assignment 2, with solutions
1 Entropy change in the isobaric-isochoric cycle of an ideal gas
Show that the entropy change in the cyclic process of an ideal gas, that is represented by a rectangle in the (P, V ) diagram, is zero.
Figure 1: Isobar-isochore cycle.
Solution: In the isobatic process of an ideal gas, the in nitesimal amount of heat is given by δQ = dU + P dV = CV dT + P dV.
From the equation of state of the ideal gas P V = νRT follows T =
P V
νR ,^ dT^ =^
P dV νR Substituting this into
dS = δQ T
one obtains dS = CV P dV /(νR) + P dV P V /(νR) = (CV + νR) dV V
= CP
dV V
In the isochoric process of the ideal gas, δQ is given by
δQ = CV dT = CV V dP νR
thus dS = δQ T = CV^ V dP P V = CV^ dP P
In our cyclic process,
∆SAB =
ˆ P 2
P 1
CV
dP P = CV ln
P 2
P 1
∆SCD = CV ln
P 1
P 2 =^ −∆SAB
∆SBC =
ˆ V 2
V 1
CP
dV V = CP ln
V 2
V 1
∆SDA = CP ln V^1 V 2
= −∆SBC.
The total entropy change ∆S = ∆SAB + ∆SBC + ∆SCD + ∆SDA = 0, as it should be.
2 Entropy change in the isobaric-isochoric-isothermic cycle of an ideal gas
Show that the entropy change in the cyclic process of an ideal gas that include an isobar, an isochor, and an isotherm is zero.
Figure 2: Isobar-isochor-isotherm cycle.
Solution: Using the results of the solution of the previous problem, one nds
∆SBC = CP ln
V 2
V 1
0 , ∆SCD = CV ln
P 1
P 2
In the isothermal process of an ideal gas dU = 0 thus δQ = P dV and
dS = δQ T
P dV T = νR dV V
This yields
∆SDB = νR ln
V 1
V 2
Using the equation of state of the ideal gas, on the ends of the isotherm one has
P 1 P 2
V 1
V 2
⇒ ∆SCD = CV ln
V 1
V 2
The total entropy change over the cycle is
∆S = ∆SBC + ∆SCD + ∆SDB = (CP − CV − νR) ln
V 2
V 1
as it should be.
3 Entropy of a perfect gas
Calculate the entropy of a perfect gas as a function of (V, T ) by integration using S =
δQ/T. Solution: De ne S(V 0 , T 0 ) = S 0 as a reference point and calculate the entropy S(V, T ) via the integral of δQ/T over a path (V 0 , T 0 ) ⇒ (V, T ), that is,
S(V, T ) = S 0 +
(V,Tˆ )
(V 0 ,T 0 )
δQ T
As the the entropy is a state function, its value does not depend on the path. Thus one can choose the most convenient path, for instance, (V 0 , T 0 ) ⇒ (V 0 , T ) ⇒ (V, T ). At the rst stage only the temperature is changing while the work is zero, thus
δQ = dU =
∂U
∂T
V
dT = CV dT.
Integration for the perfect gas (CV = const) proceeds as follows:
S(V 0 , T ) = S 0 +
ˆ^ T
T 0
CV dT T =^ S^0 +^ CV^ ln^
T
T 0.
Using Eq. (2) (^) ( ∂U ∂V
S
νRT V
P V
V
= −P,
as it should be. Thus we have obtained P = νR V γ^ exp
S
CV
in the V, S variables. This formula also could be obtained from Eq. (1) and the equation of state, similarly to Eq. (2). Now, to check the Maxwell identity, Using Eq. (2) one calculates ( ∂T ∂V
S
νR CV V γ^ exp
S
CV
On the other hand, from Eq. (4) one obtains
−
∂P
∂S
V
νR CV V γ^ exp
S
CV
∂T
∂V
S
as expected.
5 Thermodynamic potentials F and G of the perfect gas
Express thermodynamic potentials F and G of the perfect gas in terms of their natural variables and check relations similar to those in the preceding problem. Solution: Using the de nition of F and the formulas for U and S of a perfect gas, one obtains
F = U − T S = CV T − T CV ln T V γ−^1 = −CV T ln
T V γ−^1 /e
Since dF = −SdT − P dV,
one can identify
−S =
∂F
∂T
V
, −P =
∂F
∂V
T
The entropy follows from Eq. (5) as
S = −
∂F
∂T
V
= −CV + CV + CV ln T V γ−^1 = CV ln T V γ−^1
that is a known result. The pressure is
P = −
∂F
∂V
T
CV T ln V γ−^1
∂V
T
= CV T (γ − 1)
∂ (ln V ) ∂V
T
= CV^ T^ (γ^ −^ 1) V = νRT V
also a known result. The Maxwell relation (^) ( ∂S ∂V
T
∂P
∂T
V is now checked as follows (^) ( ∂S ∂V
T
∂V
CV ln T V γ−^1 = CV (γ − 1) ∂ ln V ∂V
νR V
On the other hand, (^) ( ∂P ∂T
V
νR V
∂S
∂V
T
as expected. For the Gibbs thermodynamic potential G all calculations are parallel to those for F , only one has to express all the formulas via P instead of V , using the equation of state of the ideal gas.
6 Thermodynamics from F
The Helmholtz free energy of a certain gas has the form
F = − ν^2 a V −^ νRT^ ln (V^ −^ νb) +^ J(T^ ). Find the equation of state of this gas, as well as its internal energy, entropy, heat capacities CP and CV and, in particular, their dierence CP − CV. Solution: To nd the eqiation of state, one has to nd P that will ve a function of the native variables V, T :
P = −
∂F
∂V
T
ν^2 a V 2 +^
νRT V − νb.^ (6)
Rearranging this formula, one obtains (^) (
P + ν^2 a V 2
(V − νb) = νRT, (7)
the van der Waals equation of a non-ideal gas. Next, the entropy is given by S = −
∂F
∂T
V
= νR ln (V − νb) − J′(T ).
Now the internal energy becomes
U = F + T S = − ν
(^2) a V
+ J(T ) − T J′(T ).
The heat capacity CV can be found as
CV =
∂U
∂T
V
= −T J”(T )
or as CV = T
∂S
∂T
V
= −T J”(T ).
Finding
CP = T
∂S
∂T
P requires more work. An explicit way to do this is to express V in the form V = V (P, T ) everywhere with the help of Eq. (7). However, this V is a solution of a cubic equation that is better to avoid. Also this method is inconvenient to study CP − CV because both heat capacities have to be functions of the same variables. Thus it is better to use the implicit method considering S = S(V, T ) but with V = V (P, T ). Then one obtains
CP = T
∂S
∂T
V
+ T
∂S
∂V
T
∂V
∂T
P
= CV + T
∂S
∂V
T
∂T
∂V
P
In this formula (^) ( ∂S ∂V
T
= νR V − νb
whereas (^) ( ∂T ∂V
P
νR
∂V
P +
ν^2 a V 2
(V − νb) =
νR
[
2 ν^2 a V 3 (V − νb) +
P +
ν^2 a V 2
)]
Here one can eliminate P using Eq. (6) that yields ( ∂T ∂V
P
νR
[
2 ν^2 a V 3 (V − νb) + νRT V − νb
]
Gathering the terms, one obtains
CP − CV = (νR)
2 T
V − νb
[
− 2 ν
(^2) a V 3 (V − νb) + νRT V − νb
]
further CP − CV = νR νRT νRT − (2ν^2 a/V 3 ) (V − νb)^2 and, nally,
CP − CV = νR 1 − 2 ν^2 a(V^ −νb) 2 νRT V 3
νR.
One can see that at high temperatures and large volumes the additional term in the denominator becomes small and the Meyer's relation for the ideal gas arises.
