PHYSICAL CHEMISTRY IN BRIEF

CHAP. 3: FUNDAMENTALS OF THERMODYNAMICS [CONTENTS] 101

Example

Calculate the absolute molar entropy of liquid sulphur dioxide atT = 200 K and at the stan-

dard pressure 101.325 kPa. Data: T 1 = 15K, Cp(s)m = 3. 77 J mol−^1 K−^1 at T 1 , ∆S(s) =

84. 2 J mol−^1 K−^1 , Tfus= 197. 64 K,∆fusH= 7403J mol−^1 ,C(l)pm= 87. 2 J mol−^1 K−^1. Sulphur

dioxide exists in only one crystalline form.

Solution

We determine the constant in the Debye relation (3.62) from the condition

constT^3 =Cp(s)m at T=T 1.

Individual entropy contributions have the following values:

S(T 1 ) =

∫T 1

0

constT^3

T

dT=

constT 13

3

=

3. 77

3

= 1.257 Jmol−^1 K−^1.

∆fusS=

∆fusH

Tfus

=

7403

197. 64

= 37.457 Jmol−^1 K−^1.

∆S(l)=

∫T

Tfus

Cp(l)m

T

dT= 87.2 ln

200

197. 64

= 1.035 Jmol−^1 K−^1.

The absolute molar entropy of liquid sulphur dioxide isSm(T, pst) = 1.257 + 84.200 + 37.457

+ 1.035 = 123.949 J mol−^1 K−^1.

3.5.6 Helmholtz energy

3.5.6.1 Dependence on temperature and volume

The Helmholtz energy is calculated from the definition (3.12), and from the dependence of
internal energy (3.68) and entropy (3.79) onTandV

F(T, V) =F(T 1 , V 1 ) + [U(T, V)−T S(T, V)]−[U(T 1 , V 1 )−T 1 S(T 1 , V 1 )]. (3.92)