PHYSICAL CHEMISTRY IN BRIEF

CHAP. 3: FUNDAMENTALS OF THERMODYNAMICS [CONTENTS] 99

Solution

We use equation (3.80) and rewrite it to the relation between molar quantities. Thus we get

Sm(T, Vm) =Sm◦(T 1 , Vmst) +

∫T

T 1

CV◦m(T)

T

dT+Rln

Vm

Vmst

+

∫V

∞

[(

∂p

∂T

)

Vm

−

R

Vm

]

dVm.

From the van der Waals equation we have

(

∂p

∂T

)

Vm

=

R

Vm−b

We substitute for the derivative and forCV◦m, and integrate:

Sm(T, Vm) =S◦m(T 1 , Vmst) +

∫T

T 1

A+BT

T

dT+Rln

Vm

Vmst

+R

∫Vm

∞

(

1

Vm−b

−

1

Vm

)

dVm

= S◦m(T 1 , Vmst) +Aln

T

T 1

+B(T−T 1 ) +Rln

Vm

Vmst

+Rln

Vm−b

Vm

Finally, the dependence on the amount of substance is, according to3.2.5,

S(T, Vm, n) =nSm(T, Vm).

3.5.5 Absolute entropy

This term is understood as the value of entropy in a given state (T, p) [see3.2.8]. To calculate
it we proceed from the state (T = 0,pst= 101.325 kPa), and by way of a sequence of thermo-
dynamic processes we arrive at the state (T,p) while summing the entropy changes during the
individual processes

S(T, p) =S(T 1 , pst) + ∆S(s)+ ∆fusS+ ∆S(l)+ ∆vapS+ ∆S(g), (3.91)