PHYSICAL CHEMISTRY IN BRIEF

CHAP. 3: FUNDAMENTALS OF THERMODYNAMICS [CONTENTS] 92

Example

Derive the relation for the calculation of theCpm(T, p)pressure dependence in a gas that obeys

the equation of state

Vm=

RT

p

+b−

a

RT^2

,

wherea, bare constants independent of temperature and pressure.

Solution

We calculate

(

∂^2 Vm

∂T^2

)

pfrom the given equation of state

(

∂^2 Vm

∂T^2

)

p

=−

6 a

RT^4

We divide equation (3.63) by the amount of substancenin order to obtain the relation between

molar quantities, and we choosep 1 = 0. Thus we obtain

Cpm(T, p) =Cp◦m(T)−

∫p

0

T

(

∂^2 Vm

∂T^2

)

p

dp ,

whereCp◦m(T) =Cpm(T,0), because a substance at zero pressure behaves as an ideal gas. We

substitute for the second derivative of volume with respect to temperature, and integrate

Cpm(T, p) =Cp◦m(T)−T

∫p

0

(

−

6 a

RT^4

)

dp=Cpm◦ (T) +

6 a

RT^3

p.

3.5.2 Internal energy

3.5.2.1 Temperature and volume dependence for a homogeneous system

By integrating the total differential (3.51) with respect to the general prescription (3.30) we
obtain

U(T, V) =U(T 1 , V 1 ) +

∫T

T 1

CV(T, V 1 ) dT+

∫V

V 1

[

T

(

∂p

∂T

)

V

−p

]

dV. (3.68)