Physical Chemistry Third Edition

4.2 Fundamental Relations for Closed Simple Systems 165

b.

∆S

∫P 2

P 1

(

∂S

∂P

)

T,n

dP−

∫P 2

P 1

(

∂V

∂T

)

P,n

dP−

∫P 2

P 1

V αdP

−

∫P 2

P 1

V(T 1 ,P 1 )[1+κT(P−P 1 )]αdP

−V(T 1 ,P 1 )

∫P 2

P 1

[1−κT(P−P 1 )]αdP

−V(T 1 ,P 1 )α(P 2 −P 1 )−V(T 1 ,P 1 )κTα

1

2

(P−P 1 )^2

c. ∆S−(18.^0 ×^10 −^6 m^3 )(2.^07 ×^10 −^4 K−^1 )(99.00 atm)(101325 Pa atm−^1 )

+(18. 0 × 10 −^6 m^3 )(4. 57 × 10 −^10 Pa−^1 )(2. 07 × 10 −^4 K−^1 )

×

1

2

(99.00 atm)(101325 Pa atm−^1 )^2

− 0 .0374 J K−^1 + 0 .000086 J K−^1 − 0 .0373 J K−^1

The smallness of this result indicates that assuming a constant volume for a liquid is

numerically a good approximation.

EXAMPLE 4.5

The pressure virial equation of state was shown in Eq. (1.3-4), and it was shown in an example

thatA 2 , the second pressure virial coefficient, is equal toB 2 , the second virial coefficient.

Find an expression for (∂S/∂P)T,nfor a gas obeying the pressure virial equation of state

truncated at theA 2 term.

Solution

(

∂S

∂P

)

T,n

−

(

∂V

∂T

)

P,n

−

(

∂

∂T

(

nRT

p

+nA 2

))

P,n

−

nR

p

−n

dA 2

dT

Exercise 4.3

Evaluate (∂S/∂P)T,nfor 1.000 mol of argon at 1.000 atm and 298.15 K. For argon at this

temperature,B 2 is equal to− 15 .8cm^3 mol−^1 anddB 2 /dTis approximately equal to 0.25 cm^3

mol−^1 K−^1.

EXAMPLE 4.6

Derive the expression for the entropy change for an isothermal pressure change of a gas

described by this truncated pressure virial equation of state.

Solution

∆S

∫P 2

P 1

(

∂S

∂P

)

T,n

dP−

∫P 2

P 1

(

nR

P

+n

dA 2

dT

)

dP

−nRln

(

P 2

P 1

)

−n

dA 2

dT

(P 2 −P 1 )