Homework 2David Sirajuddin

Physics 715 - Statistical MechanicsProfessor Michael Ramsey-Musolf

February 22, 2010

1. (Huang 1.5)

A substance has the following properties:

At a constant temperature T0 the work done by it on expansion from V0 to V is

W = RT0 lnV

V0

The entropy is given by

S = RV

V0

(T

T0

)a

where V0, T0, and a are fixed constants.

(a) Calculate the Helmholtz free energy, F

The Helmholtz free energy F is defined as

F = E − TS

where E, T , and S are the internal energy, temperature, and entropy of the systemrespectively. Differentiating this equation:

dF = dE − (TdS + SdT )

And, recalling that the internal energy dE = TdS − PdV , for pressure P andvolume V , admits

dF = ���TdS − PdV −���TdS − SdT→ dF = −PdV − SdT (1)

Such that Huang’s Eqn. (1.19) is retrieved [1]. It is noted that the variables Vand T have incidentally been chosen to be independent. The remaining parameter(pressure P ) may be obtained from the equation of state (see part (b)). The aboveequation furnishes the following definitions:

1

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

(∂F

∂V

)T

= −P and

(∂F

∂T

)V

= −S (2)

Which may formally be obtained by independently letting V be constant (dV →0), and T be constant (dT → 0) in Eqn. (1) followed by solving for the re-maining parameter. Thus, the subscripts corresponding to the partial differen-tiation explicitly indicate which quantity is held constant during the operation(e.g. (∂F/∂V )T instructs to differentiate F with respect to V while holding thetemperature T constant). The partial differential equations (2) are solved in orderto find the state function F . Preliminarily, a partial integration is performed onthe left-hand equation over the interval V0 ≤ V ′ ≤ V ,

(∂F

∂V

)T

= −P∫ F (V,T )

F (V0,T )

∂F ′ = −∫ V

V0

P∂V ′︸ ︷︷ ︸=−W

F (V, T )− F (V0, T ) = −WF (V, T ) = −W + F (V0, T )

F (V, T ) = −RT0 ln

(V

V0

)+ F (V0, T ) (3)

(Primed dummy variables are used for clarity). In the last step, the expressionfor work given in the problem statement has been inserted. Due to the partialintegration, an arbitrary function F (V0, T ) must be appended given that the onlyconstraint on the above expression for F (V, T ) follows from Eqn. (2), i.e. thatthe F (V0, T ) term vanishes under partial differentiation with respect to V so thatEqn. (2) is recovered. This restriction allows for the integration ”constant” tobe, in general, a function of the temperature T .

In order to solve for the function F (V0, T ), a partial derivative with respect to Vmay be performed on Eqn. (3) so as to establish an equivalence with the righthand equation of Eqn. (2). Beginning with Eqn. (3):

(∂F

∂T

)V

=������������(

∂

∂T

)V

RT0 ln

(V

V0

)︸ ︷︷ ︸

=0, not a function of T

+

(∂

∂TF (V0, T )

)V(

∂F

∂T

)V

=

(∂

∂TF (V0, T )

)V

(4)

Noting that the problem statement informs S = R VV0

(TT0

)a

, the right-hand equa-

tion in Eqn. (3) may be written as2

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

(∂F

∂T

)V

= −S = −RVV0

(T

T0

)a

(5)

Equating the right-hand sides of Eqns. (4) and (5) and integrating over theinterval [T0, T ],

(∂

∂TF (V0, T )

)V

= −RVV0

(T ′

T0

)a

∫ F (V0,T )

F (V0,T0)

∂F ′ = −RVV0

∫ T

T0

(T ′

T0

)a

∂T ′

F (V0, T0)− F (V0, T ) = . . .

F (V0, T ) = −RVV0

∫ T

T0

(T ′

T0

)a

∂T ′ + F (V0, T0) (6)

The integral over T is solved by the substitution,

θ =T ′

T0

⇒ T0dθ = dT ′

so that ∫ T

T0

(. . .)dT ′ ⇒ T0

∫ T/T0

1

(. . .)dθ

where the partial integration has been replaced by ordinary integration for con-venience, with the constant volume condition taken to be understood.

The integral is then,

∫ T

T0

(T ′

T0

)a

dT ′ = T0

∫ T/T0

1

θadθ

=T0

a+ 1

[θa+1

]T/T0

1∫ T

T0

(T ′

T0

)a

dT ′ =T0

a+ 1

[(T

T0

)a+1

− 1

]

Inserting this result into Eqn. (6) gives

F (V0, T ) = −RVV0

T0

a+ 1

[(T

T0

)a+1

− 1

]+ F (V0, T0)

Finally, substituting this result into Eqn. (3) yields the final result,

3

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

F (V, T ) = −RT0 ln

(V

V0

)+ F (V0, T )

F (V, T ) = −RT0 ln

(V

V0

)− V

V0

RT0

a+ 1

[(T

T0

)a+1

− 1

]+ F (V0, T0)

F (V, T )− F (V0, T0) = −RT0 ln

(V

V0

)− V

V0

RT0

a+ 1

[(T

T0

)a+1

− 1

]

Letting ∆F (V, T ) ≡ F (V, T )−F (V0, T0), the Helmholtz free energy is determinedto be

∆F (V, T ) = −RT0

{ln

(V

V0

)+V

V0

1

a+ 1

[(T

T0

)a+1

− 1

]}(7)

For some constant F (V0, T0) that may be specifed by boundary conditions.

(b) Find the equation of state

Using result (7) with the left-hand Eqn. (2), the equation of state is computed:

P = −(∂ (∆F )

∂V

)T

= −(∂F (V, T )

∂V

)T

−���������(∂F (V0, T0)

∂V

)T︸ ︷︷ ︸

=0, constant

= −(∂F (V, T )

∂V

)T

= −(∂

∂V

)T

{ln

(V

V0

)+V

V0

1

a+ 1

[(T

T0

)a+1

− 1

]}

= +RT0

{1/��V0

V/��V0

+1

V0

1

a+ 1

[(T

T0

a+1

− 1

)]}

P = RT0

{1

V+

1

V0

1

a+ 1

[(T

T0

a+1

− 1

)]}(8)

Despite the universal gas constant R appearing in these equations, the equationof state (8) reveals that the substance in question is not an ideal gas, whereP = NRT/V .

4

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

(c) Find the work done at an arbitrary constant temperature T .

The equation of state P = P (V, T ), above, may be used to find the work W doneat arbitrary temperature T in the expansion from volume V0 to V .

W =

∫ V

V0

PdV

=

∫ V

V0

RT0

{1

V+

1

V0

1

a+ 1

[(T

T0

a+1

− 1

)]}dV

= RT0

{lnV +

V

V0

1

a+ 1

[(T

T0

a+1

− 1

)]}V

V0

= RT0

{lnV − lnV0 +

V − V0

V0

1

a+ 1

[(T

T0

a+1

− 1

)]}it then follows that

W = RT0

{lnV

V0

+V − V0

V0

1

a+ 1

[(T

T0

a+1

− 1

)]}(9)

2. Prove the following two relations:

TdS = CPdT − T (∂V/∂T )PdP (10)

(∂CP/∂P )T = −T (∂2V/∂T 2)P (11)

Eqn. (10) proof

Beginning with the dQ Eqn. (1.6) from Huang [1]:

dQ =

[(∂U

∂T

)P

+ P

(∂V

∂T

)P

]dT +

[(∂U

∂P

)T

+ P

(∂V

∂P

)T

]dP

where the notation of Huang is adopted, U is the internal energy, Q represents heat, andall other quantities are as previously defined. Noting that the enthalpy H = U + PV ,the specific heat at constant pressure CP can be computed as follows,

5

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

CP =

(∂H

∂T

)P

=

(∂

∂T

)P

(U + PV )

=

(∂U

∂T

)P

+∂

∂T(PdV )

=

(∂U

∂T

)P

+�

������(

∂P

∂TdV

)P︸ ︷︷ ︸

=0,P=constant

+P

(∂V

∂T

)P

Cp =

(∂U

∂T

)P

+ P

(∂V

∂T

)P

(12)

Inputting this result into the dQ equation, and letting dQ = TdS implies

dQ =

[(∂U

∂T

)P

+ P

(∂V

∂T

)P

]︸ ︷︷ ︸

=CP

dT +

[(∂U

∂P

)T

+ P

(∂V

∂P

)T

]dP

TdS = CpdT +

[(∂U

∂P

)T

+ P

(∂V

∂P

)T

]︸ ︷︷ ︸

¬

dP (13)

The solution strategy proceeds by proving term ¬ is identical to −T (∂V/∂T )P inEqn. (10). Motivated by this end, the desired form of term ¬ may be obtained bymanipulating the above equation. Specifically, it is noted that dS itself is an exactdifferential,

dS =Cp

TdT +

1

T

[(∂U

∂P

)T

+ P

(∂V

∂P

)T

]dP

which necessarily constrains the terms on the right-hand side of the equation to satisfy:

∂

∂P

[CP

T

]T

­

=∂

∂T

[1

T

(∂U

∂P

)T

+P

T

(∂V

∂P

)T

]P

®

(14)

Evaluating both sides of the above equation, beginning with the left-hand side (term­), the specific heat CP is replaced with expression (12):

6

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

­ =∂

∂P

(CP

T

)T

=1

T

(∂CP

∂P

)T

, T is held constant

=1

T

∂

∂P

[(∂U

∂T

)P

+ P

(∂V

∂T

)P

]T

=1

T

(∂2U

∂P∂T

)P,T

+1

TP

(∂2V

∂P∂T

)P,T

+1

T�

��

��(∂P

∂P

)T

=1

(∂V

∂T

)P

­ =1

T

(∂2U

∂P∂T

)P,T

+1

TP

(∂2V

∂P∂T

)P,T

+1

T

(∂V

∂T

)P

And, the right-hand side term ® of Eqn. (14) admits

® =∂

∂T

[1

T

(∂U

∂P

)T

+P

T

(∂V

∂P

)T

]P

=

(∂

∂T

)P

[1

T

(∂U

∂P

)T

]+

(∂

∂T

)P

[P

T

(∂V

∂P

)T

]=

∂

∂T

(1

T

)P︸ ︷︷ ︸

=−1/T 2

(∂U

∂P

)T

+1

T

(∂2U

∂T∂P

)T,P

+∂

∂T

(P

T

)P︸ ︷︷ ︸

=−P/T 2

(∂V

∂P

)T

+P

T

(∂2V

∂T∂P

)T,P

® = − 1

T 2

(∂U

∂P

)T

+1

T

(∂2U

∂T∂P

)T,P

− P

T 2

(∂V

∂P

)T

+P

T

(∂2V

∂T∂P

)T,P

Equating terms ­ and ® above, Eqn. (14) yields,

∂

∂P

[CP

T

]T

­

=∂

∂T

[1

T

(∂U

∂P

)T

+P

T

(∂V

∂P

)T

]P

®

���������1

T

(∂2U

∂P∂T

)P,T

¯

+���������P

T

(∂2V

∂P∂T

)P,T

°

+1

T

(∂V

∂T

)P

= − 1

T 2

(∂U

∂P

)T

+���������1

T

(∂2U

∂T∂P

)T,P

¯

− P

T 2

(∂V

∂P

)T

+���������P

T

(∂2V

∂T∂P

)T,P

°

where correspsonding terms are cancelled in pairs as per the labels ¯ and °. Thecancellation is substantiated on the enforcement that the functions U and V are uni-formly convergent such that the order of partial differentiation may be interchanged(i.e. ∂P∂TU = ∂T∂PU and ∂P∂TV = ∂T∂PV ). Thus,

7

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

1

T

(∂V

∂T

)P

= − 1

T 2

(∂U

∂P

)T

− P

T 2

(∂V

∂P

)T

−T(∂V

∂T

)P

=

(∂U

∂P

)T

+ P

(∂V

∂P

)T

= ¬

So that desired form of term ¬ of the TdS equation (14) is retrieved. Inputting thisresult into Eqn. (14) provides the final result,

TdS = CpdT +

[(∂U

∂P

)T

+ P

(∂V

∂P

)T

]︸ ︷︷ ︸

¬

dP

TdS = CPdT − T (∂V/∂T )PdP 3

Eqn. (11) proof

The second relation (Eqn. (11)),(∂CP

∂P

)T

= −T(∂2V

∂T 2

)P

can be proved by using result (10) above:

TdS = CPdT − T (∂V/∂T )PdP

Equations are developed by setting P and T constant independently, and solving forthe remaining parameter. Letting P be constant implies dP → 0 in the above equation,such that it can be written

(TdS)P = (CPdT )P

→(∂S

∂T

)P

=CP

T

→(

∂2S

∂P∂T

)P,T

=1

T

(∂CP

∂P

)T

(15)

where a subsequent differentiation with respect to P has been performed to obtainan equation for (∂CP/∂P )T as per Eqn. (11). The additional relation is obtained byallowing T to be constant so that dT → 0, implying

8

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

(��TdS)T = −[��T

(∂V

∂T

)P

dP

]T

→(∂S

∂P

)T

= −(∂V

∂T

)P

→(

∂2S

∂T∂P

)T,P

= −(∂2V

∂T 2

)P

(16)

The left-hand sides of Eqns. (15) and (16) are recognized to be identical provided thatthe differential operators commute. Thus, upon equating the right-hand terms of bothequations,

1

T

(∂CP

∂P

)T

= −(∂2V

∂T 2

)P

which gives, (∂CP

∂P

)T

= −T(∂2V

∂T 2

)P

3

And, the proposition is proved.

3. (a) Derive equations 14(a-e) from Lecture V [2]

The equations to be proven are as follows,

(∂E

∂P

)T

= −T(∂V

∂T

)P

− P(∂V

∂P

)T

(17)(∂H

∂V

)T

= T

(∂P

∂T

)V

+ V

(∂P

∂V

)T

(18)(∂H

∂P

)= V − T

(∂V

∂T

)P

(19)(∂E

∂T

)P

= CP − P(∂V

∂T

)P

(20)(∂H

∂T

)V

= CV + V

(∂P

∂T

)V

(21)

where the notation of Ramsey-Musolf is used [2] (E is the internal energy, andall other quantities are as previously defined), and the equation numbers havebeen shifted in order to avoid confusion with previous equation labels used in thisassignment. Obtaining these results is done with aid of the following diagram [2]:

9

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

Figure 1: Diagram summarizing thermodynamic parameter relationships [2]

Which visually summarizes the relationships of thermodynamic quantities andpotentials. Partial differentiation with respect to one of the flanking parametersof a given off-diagonal element (while the other flanked quantity is held constant)admits the corresponding thermodynamic quantity traced out by the diagonalarrow followed. Following the arrow implies that one performs this derivativewith respect to the parameter it connects with, while the other quantity thatflanks the potential of interest is held constant. Going against an arrow suffers achange by a minus sign.

Eqn. (17) proof

The result is found by differentiating the first law with respect to the pressure Pat constant temperature T :

dE = TdS − PdV (22)(∂E

∂P

)T

=

(∂

∂P

)T

(TdS)−(∂

∂P

)T

(PdV )

=�

��

��(∂T

∂P

)T

=0 , T =constant

dS + T

(∂S

∂P

)T

−

�

������(

∂P

∂PdV

)T

=0

+ P

(∂V

∂T

)T

T(

∂E

∂P

)T

= T

(∂S

∂P

)T

±

− P(∂V

∂P

)T

(23)

From Figure 1, the relation S = −(∂G/∂T )P , is used to evaluate term ± above:

10

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

± = T

(∂S

∂P

)T

= −(∂

∂P

)T

(∂G

∂T

)P

= −(∂

∂T

)P

(∂G

∂P

)T︸ ︷︷ ︸

=V

± = −(∂

∂T

)P

V = −(∂V

∂T

)P

Substituting this result into term ± into Eqn. (23) completes the derivation

(∂E

∂P

)T

=

±

T

(∂S

∂P

)T

− P(∂V

∂P

)T(

∂E

∂P

)T

= −(∂V

∂T

)P

− P(∂V

∂P

)T

3

Eqn. (18) proof

Definition (18): (∂H

∂V

)T

= T

(∂P

∂T

)V

+ V

(∂P

∂V

)T

is furnished by the definition of enthalphy dH = TdS+V dP . Differentiating thisequation,

(∂H

∂V

)T

=

(∂

∂V

)T

(TdS) +

(∂

∂V

)T

(V dP )

=�

��

��(∂T

∂V

)T

=0 , T =constant

dS + T

(∂S

∂V

)T

+���

����(∂V

∂VdP

)T

=0

+ V

(∂P

∂V

)T(

∂H

∂V

)T

= T

(∂S

∂V

)T

²

+ V

(∂P

∂V

)T

From Figure 1, the relation S = −(∂F/∂T )V , is used to evaluate term ² above:

11

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

² = T

(∂S

∂V

)T

= −(∂

∂V

)T

(∂F

∂T

)V

= −(∂

∂T

)V

(∂F

∂V

)T︸ ︷︷ ︸

=−P

² = −(∂

∂T

)V

(−P ) = +

(∂P

∂V

)T

Inputting this result into term ² into Eqn. (24) completes the proof

(∂H

∂V

)T

=

²

T

(∂S

∂V

)T

+ V

(∂P

∂V

)T(

∂H

∂V

)T

= T

(∂P

∂V

)V

+ V

(∂P

∂V

)T

3

Eqn. (19) proof

The same procedure as in the derivation of Eqn. (18) above is used to demonstratethe validity of Eqn. (19), which reads(

∂H

∂P

)T

= V − T(∂V

∂T

)P

This time, the enthalphy dH = TdS + V dP is differentiated with respect to P atconstant T ,

(∂H

∂P

)T

=

(∂

∂P

)T

(TdS) +

(∂

∂P

)T

(V dP )

=�

��

��(∂T

∂P

)T

=0 , T =constant

dS + T

(∂S

∂P

)T

+���

����(∂V

∂PdP

)T

=0

+ V�

��

��(∂P

∂P

)T

=1(∂H

∂P

)T

= T

(∂S

∂P

)T

³

+ V

From Figure 1, it is seen that S = −(∂G/∂T )P , is used to evaluate term ³ above:12

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

³ = T

(∂S

∂P

)T

= −(∂

∂P

)T

(∂G

∂T

)P

= −(∂

∂T

)P

(∂G

∂P

)T︸ ︷︷ ︸

=V

³ = −(∂

∂T

)P

(V ) = −(∂V

∂P

)T

Combining this result with Eqn. (25) shows that,

(∂H

∂P

)T

=

³

T

(∂S

∂P

)T

+ V(∂H

∂P

)T

= −T(∂V

∂T

)P

+ V 3

Eqn. (20) proof

Verifying Eqn. (20), (∂E

∂T

)P

= CP − P(∂V

∂T

)P

is accomplished by differentiating the First Law, dE = TdS − PdV , with respectto T at constant P .

dE = TdS − PdV (24)(∂E

∂T

)P

=

(∂

∂T

)P

(TdS)−(∂

∂T

)P

(PdV )

=�

������(

∂T

∂T

)P

dS

=0

+ T

(∂S

∂P

)T

−

���

����(∂P

∂PdV

)T

=0

+ P

(∂V

∂T

)T

T(

∂E

∂P

)T

= T

(∂S

∂P

)T

´

− P(∂V

∂P

)T

(25)

Term ´ can be shown to be identical to the specific heat at constant pressure Pby invoking the definition of enthalphy: dH = TdS + V dP . Holding P constantimplies dP → 0, such that

13

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

(dH)P = (TdS)P

⇒(∂

∂T

)P

(H) =

(∂

∂T

)P

(TdS)(∂H

∂T

)P

=�

������(

∂T

∂T

)P

dS︸ ︷︷ ︸=0, T=constant

+T

(∂S

∂T

)P(

∂H

∂T

)P

= T

(∂S

∂T

)P

≡ CP = ´

Where term ´ has been identified as the specific heat CP . Inserting this resultinto Eqn. (25) admits,(

∂E

∂T

)P

= CP − P(∂V

∂T

)P

3

Thus, proving statement (20).

Eqn. (21) proof

Equation (21), (∂H

∂T

)V

= CV + V

(∂P

∂T

)V

is proven in a similar manner to Eqn. (20). The definition of enthalphy, dH =TdS + V dP , is first differentiated with respect to T at constant V .

(∂H

∂T

)V

=

(∂

∂T

)V

(TdS) +

(∂

∂T

)V

(V dP )

=∂

∂T(TdS)V +

��

��

(∂V

∂T

)︸ ︷︷ ︸

=0, V =constant

dP + V

(∂P

∂T

)V(

∂H

∂T

)V

=��

�����(

∂T

∂T

)V

dS

=0

+ T

(∂S

∂P

)T

+ V

(∂P

∂T

)V(

∂H

∂T

)V

= T

(∂S

∂P

)T

µ

+ V

(∂P

∂T

)V

(26)

Term µ is shown to be identical to the specific heat CV at constant volume byemploying the First Law (dE = TdS −PdV ). Allowing V to be constant impliesdV → 0, so that

14

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

(dE)V = (TdS)V

⇒(∂

∂T

)V

(E) =

(∂

∂T

)V

(TdS)(∂E

∂T

)V

=��

�����(

∂T

∂T

)V

dS︸ ︷︷ ︸=0

+T

(∂S

∂T

)V(

∂E

∂T

)V

= T

(∂S

∂T

)V

≡ CV = µ

Where term µ has been identified as the specific heat CV . Substituting this resultinto Eqn. (26) validates that,(

∂H

∂T

)V

= CV + V

(∂P

∂T

)V

3

(b) Use the results from part (a) to determine the change in internal energy ∆E whenan ideal gas goes from pressure P1 to pressure P2 and the change in enthalpy ∆Hwhen the system’s temperature changes from T1 to T2.∆E calculation

Equation (17) of part (a) may be used to directly find the change in internalenergy ∆E = E(P2, T )− E(P1, T ). Equation (17) reads,(

∂E

∂P

)T

= −T(∂V

∂T

)P

− P(∂V

∂P

)T

For an ideal gas law, the equation of state V = NkT/P , may be used in the termsabove. Computing the necessary derivatives,(

∂V

∂T

)P

=Nk

Pand

(∂V

∂P

)T

= −NkTP 2

Inputting these expressions into the above equation,

(∂E

∂P

)T

= −T(Nk

P

)−��P

(NkT

P�2

)= −NkT

P︸ ︷︷ ︸−V

+NkT

P︸ ︷︷ ︸V

= −V + V(∂E

∂P

)T

= 0

Integrating over the pressure for P1 ≤ P ′ ≤ P2 implies15

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

∆E ≡ E(P2, T )− E(P1, T ) = 0 3

∆H calculation

Equation (21), (∂H

∂T

)V

= CV + V

(∂P

∂T

)V

can be used in tandem with the ideal gas law (P = NkT/V ) to find the change inenthalpy, ∆H = H(P, T2)−H(P, T1), when an ideal gas changes from a temper-ature T1 to T2. Computing the second term on the right-hand side of the aboveequation,

V

(∂P

∂T

)V

= ��V

(Nk

��V

)= Nk

So that the above equation gives,

(∂H

∂T

)V

= CV +Nk

= ��CV + (CP −�

�CV )(∂H

∂T

)V

= CP

Where the identification Nk = CP −CV follows from the discussion of Huang (pg.9). This substitution has tacitly assumed that the specific heat CV is independentof temperature. Should the specific heat CP also be independent of temperaturefor the range [T1, T2], the change in enthalpy may be directly obtained,

∆H ≡ H(P, T2)−H(P, T1) = Cp(T2 − T1) 3

4. Show that

CP − CV = −T [(∂P/∂T )V ]2

(∂P/∂V )T

Equating the two TdS equations (1.13) and (1.14) of Huang, and choosing P and V tobe independent variables (where the temperature T may be obtained by the equationof state), the following equation is arrived at [1]:

[(CP − CV )

(∂T

∂V

)P

− T(∂P

∂T

)V

]dV = −

[(CP − CV )

(∂T

∂P

)V

− T(∂V

∂T

)P

]dP

16

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

Since the parameters V and P have been chosen as independent, the coefficients (squarebracketed terms) of their differential elements must vanish. Equating the right-handcoefficients to zero admits the result quoted in the problem statement.

(CP − CV )

(∂T

∂P

)V

− T(∂V

∂T

)P

= 0

CP − CV = T(∂V/∂T )P

(∂T/∂P )V

The lemmas provided by Huang are used to recast this equation into the desired formgiven in the problem statement [1, pg. 20]. Lemma (b) implies,

CP − CV = = T(∂V/∂T )P

(∂T/∂P )V

(27)

= T

(∂V

∂T

)P

1

(∂T/∂P )V︸ ︷︷ ︸=(∂P/∂T )V

CP − CV = T

(∂V

∂T

)P

(∂P

∂T

)V

(28)

The term (∂V/∂T )P is replaced by use of the cyclic permutation lemma (c),(∂V

∂T

)P

(∂T

∂P

)V

(∂P

∂V

)T

= −1 (29)

An expression for (∂V/∂T )P is obtained from this equation by aid of lemma (b), whichallows the following identies,

(∂T

∂P

)V

=1

(∂P/∂T )V

⇒(∂T

∂P

)V

(∂P

∂T

)V

= 1(∂P

∂V

)T

=1

(∂V/∂P )T

⇒(∂P

∂V

)T

(∂V

∂P

)T

= 1

Thus, multiplying Eqn. (29) by the indicated derivatives above such that these prod-ucts are equal to unity isolates an equation for (∂V/∂T )P (which may be substitutedback into the original equation [Eqn. (28)] to recover the result). Performing thismultiplication on Eqn. (29) admits

(∂V

∂T

)P �����������[(

∂T

∂P

)V

(∂P

∂T

)V

]=1

�����������[(∂P

∂V

)T

(∂V

∂P

)T

]=1

= −(∂P

∂T

)V

(∂V

∂P

)T

17

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

giving (∂V

∂T

)P

= −(∂P

∂T

)V

(∂V

∂P

)T

Using this result in Eqn. (28),

CP − CV = T

[−(∂P

∂T

)V

(∂V

∂P

)T

](∂P

∂T

)V

= −T(∂P

∂T

)2

V

(∂V

∂P

)T

Finally, using lemma (b) once more implies (∂V/∂P )T = 1/(∂V/∂P )T so that theabove equation can be written in its final form

CP − CV = −T [(∂P/∂T )V ]2

(∂P/∂V )T

3

which verifies the equation.

5. Show that (∂T

∂P

)S

=

(T

CP

)(∂V

∂T

)P

(30)

From the TdS Eqn. (1.14) in Huang [1]:

TdS = CPdT − T(∂V

∂T

)P

dP

Letting the entropy S be constant implies dS → 0,

0 = CPdT − T(∂V

∂T

)P

dP

CPdT = T

(∂V

∂T

)P

dP

Solving for (∂T/∂P )S yields the proposed equation,(∂T

∂P

)S

=T

CP

(∂V

∂T

)P

3

18

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

6. Show that

limT→0

(∂P

∂T

)V

→ 0

The proof follows from a direct application of the Third Law. That is to say, thatthe entropy at absolute zero approaches a constant, which may be taken to be zero.Equivalently, the result follows from pure mathematical considerations after an initialphysical definition is introduced. This definition is given as Eqn. (1.40) in Huangas the entropy S(T ) at a state corresponding to a temperature T calculated from aninitial state at absolute zero [1, pg. 26]:

S(A) =

∫ T

0

CV (T )dT

T

where, in the notation of Huang, the term CR is taken to be CV . The claim is madethat (

∂P

∂T

)V

=

(∂S

∂V

)T

and will be proven a posteriori. Proceeding with the definition of the entropy S(T ), itcan be written

(∂S

∂V

)T

=

(∂

∂V

)T

∫ T

0

CVdT

T

=

∫ T

0

(∂CV

∂V

)T

µ

dT

T(31)

Term µ may be recasted by aid of the TdS equation (1.13) [1].

TdS = CPdT + T

(∂P

∂T

)V

dV

dS =CP

TdT +

(∂P

∂T

)V

dV

where the parameters T and V are independent. Letting each of these quantities beconstant independently, and solving for the remaining parameter allows the followingtwo relations to be written,

19

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

(∂S

∂T

)V

=CV

T

⇒(

∂2S

∂V ∂T

)V,T

=1

T

(∂CV

∂V

)T

(32)

and

(∂S

∂V

)T

=

(∂P

∂T

)V

⇒(

∂2S

∂T∂V

)T,V

=

(∂2P

∂T 2

)V

(33)

where a subsequent differentiation has been performed on each equation. Enforcingthat the partial differential operators commute, the left-hand sides are identical andthe right-hand terms may be equated:(

∂CV

∂V

)T

= T

(∂2P

∂T

)V

(34)

It is noted that the the left-hand term is exactly that of term µ of Eqn. (31). Insertingthis result implies

(∂P

∂T

)V

=

∫ T

0

(∂2P

∂T 2

)V

=

[(∂P

∂T

)V

]T

0(∂P

∂T

)V

=

[(∂P

∂T

)V

]T

−[(

∂P

∂T

)V

]T=0

Where, for want of a clearer notation, the subscripts labeling the square brackets referto the temperature at which the quantity is evaluated at (rather than indicate whichparameter to hold constant under partial differentiation). It is seen that the originalclaim has been proven (i.e. (∂P/∂T )V = (∂S/∂V )T Evaluating this expression in thelimit T → 0, directly proves the posit of the problem statement,

limT→0

(∂P

∂T

)V

→ 0 3

20

Sirajuddin, David Homework 2 – Physics 715, Spring 2010

References

[1] Huang, Kerson Statistical Mechanics 1987. John Wiley & Sons, Inc. Canada.

[2] Ramsey-Musolf, Michael Physics 715 - Statistical Mechanics Class Notes, University ofWisconsin - Madison. Spring 2010.

21