PHYSICAL CHEMISTRY IN BRIEF

CHAP. 4: APPLICATION OF THERMODYNAMICS [CONTENTS] 124

Solution

From the equation of state of an ideal gas we obtain

(

∂V

∂T

)

p

=

nR

p

=

V

T

By substituting into (4.39) we can see thatμJT= 0.

From the van der Waals equation we obtain (see the example in3.3.1)

(

∂V

∂T

)

p

=−n

R

Vm−b

−(VmR−Tb) 2 + 2Vam 3

and substitute into (4.39)

μJT=

RT(Vm−b)Vm^3

RT Vm^3 − 2 a(Vm−b)^2 −Vm

Cpm

=

Vm

Cpm

2 a(Vm−b)^2 −RT bVm^2

RT Vm^3 − 2 a(Vm−b)^2

4.3.6 Inversion temperature

From the definition of the Joule-Thomson coefficient (4.37) it follows that a fluid passing
through a barrier cools on expansion whenμJT>0, and heats up whenμJT<0.

p 2 < p 1 and μJT> 0 −→ T 2 < T 1 [H], (4.40)

p 2 < p 1 and μJT< 0 −→ T 2 > T 1 [H]. (4.41)

The temperature at whichμJT= 0 at a given pressure is called theinversion temperature,
Tinv. The dependence of the inversion temperature on pressure is schematically drawn in Figure
4.3.

Example

From the van der Waals equation derive the relation for the inversion temperature at zero pressure.