Physical Chemistry Third Edition

2.5 Enthalpy 79

whereVL1is the initial volume of the left side of the apparatus andVL2is the final

volume of the left side of the apparatus. The work done on the gas on the right side is

given by

wR−

∫VR2

VR1

−PR(VR2−VR1) (2.5-21)

Prior to the transfer, the pressureP 1 of the system was equal toPL, and the initial

volume of the system must have been equal to the magnitude of the change in volume

of the left side:

V 1 VL1−VL2 (2.5-22)

The final pressureP 2 must be equal toPR, and the final volumeV 2 must be equal to

the change in volume of the right side:

V 2 VR2−VR1 (2.5-23)

From Eqs. (2.5-20) through (2.5-23), the total work done on the system is

wwL+wRP 1 V 1 −P 2 V 2 −∆(PV) (2.5-24)

Exercise 2.25

a.Show that for any change in state

∆(PV)P 1 ∆V+V 1 ∆P+(∆P∆V) (2.5-25)

b.When can∆(PV) equalP∆V? When can it equalV∆P? When can it equalP∆V+V∆P?

Because the apparatus is adiabatically insulated from the laboratory, no heat is

transferred to or from the laboratory. Also, no heat is transferred from the system

to the apparatus after the steady state is established, because the chamber on

the right is then at the same temperature as the gas that exits from the plug.

Therefore,

q 0 (2.5-26)

∆Uq+ww−∆(PV) (2.5-27)

∆H∆U+∆(PV) 0 (2.5-28)

The Joule–Thomson process therefore occurs at constant enthalpy, and the

Joule–Thomson coefficient is equal to a partial derivative at constantHandn:

μJT

(

∂T

∂P

)

H,n

(2.5-29)