PHYSICAL CHEMISTRY IN BRIEF

CHAP. 4: APPLICATION OF THERMODYNAMICS [CONTENTS] 125

c

c

c

μJT > (^0) E s

TI^0

T

C

p

XXX

XyX



1

μJT = 0

ps=ps(T)

Obr.4.3:Dependence of the inversion temperature on pressure. The Joule-Thomson coefficient is
positive inside the curve delimiting this dependence and negative outside this boundary. TI^0 is the
inversion temperature at zero pressure. Cis the critical point. The weak line ending in the critical
point is the dependence of the saturated vapour pressure on temperature. The pressure corresponding
to the pointEis the highest pressure at which the Joule-Thomson coefficient can be positive

Solution

We proceed from the formula used in the preceding example forμJTof a van der Waals gas. For

μJT= 0, the numerator in the formula is zero

RT b Vm^2 − 2 a(Vm−b)^2 = 0

and

Tinv=

2 a

Rb

(

Vm−b

Vm

) 2

.

At zero pressure isVm=∞, and hence

lim

p→ 0

Tinv= lim

Vm→∞

Tinv= lim

Vm→∞

2 a

Rb

(

Vm−b

Vm

) 2

=

2 a

Rb

, [p= 0].