Physical Chemistry , 1st ed.

either side are fixed, it should be understood that the gas experiences a drop
in pressure as it is forced from one side to the other.
On the left side, work is done on the gas, which contributes positively to the
overall change in energy. On the right side, the gas does work, contributing
negatively to the overall change in energy. The net work wnetperformed by the
system after the first piston is completely pushed in is

wnetp 1 V 1 p 2 V 2

Since the system is adiabatic,q0, so Unetwnet, but we will write Uas
the internal energy of the gas on side 2 minus the internal energy of the gas
on side 1:
wnetU 2 U 1

Equating the two expressions for wnet:

p 1 V 1 p 2 V 2 U 2 U 1

and rearranging:

U 1

 p 1 V 1 U 2

 p 2 V 2

The combination U+ pVis the original definition ofH, the enthalpy, so for
the gas in this Joule-Thomson experiment,

H 1 H 2

or, for the gas undergoing this process, the changein His zero:

H 0

Since the enthalpy of the gas does not change, the process is called isenthalpic.
What are some consequences of this isenthalpic process?
Although the change in enthalpy is zero, the change in temperature is not.
What is the change in temperature accompanying the pressure drop for this
isenthalpic process? That is, what is ( T/ p)H? We can actually measure this
derivative experimentally, using an apparatus like the one in Figure 2.10.
The Joule-Thomson coefficientJTis defined as the change in temperature
of a gas with pressure at constant enthalpy:

JT

T

p



H

(2.34)

A useful approximation of this definition is

JT (^) 





T

p



H

2.7 Joule-Thomson Coefficients 43

Side 2:

T 2 , p 2

Side 1:

T 1 , p 1

Gas forced

through by

piston

Piston in Piston out

Adiabatic system

Porous

barrier

Figure 2.10 The isenthalpic experiment of Joule and Thomson. A description is given in
the text.