Physical Chemistry Third Edition

(C. Jardin) #1
78 2 Work, Heat, and Energy: The First Law of Thermodynamics

Thermometers

Adiabatic insulation Porous plug Adiabatic insulation

Manometers

Figure 2.8 The Apparatus for the Joule–Thomson Experiment (Schematic).

plug separating the two cylinders. Each side has a manometer to measure the pressure
and a thermometer to measure the temperature of the gas. The entire apparatus is
adiabatically insulated from the surroundings. During the experiment, one of the pistons
is moved into its cylinder and the other piston is withdrawn in a way such that the
pressure on each side remains constant as the gas flows irreversibly through the porous
plug. If the process is slow enough a time-independent nonequilibrium state (a steady
state) will be attained with a different constant temperature and constant pressure on
each side. Several experiments are carried out on a given gas with different values of the
pressure difference. TheJoule–Thomson coefficientμJTis defined as the extrapolated
limit

μJT lim
PR→PL

(

TR−TL

PR−PL

)

 lim
∆P→ 0

(

∆T

∆P

)

(definition ofμJT) (2.5-19)

where the subscripts R and L indicate the right and left sides of the apparatus. We now
show that the limit in this equation is a partial derivative at fixed enthalpy.

William Thomson, 1824–1907, later
Lord Kelvin, was a great Scottish
mathematician and physicist who was
an early supporter of Joule and became
a close personal friend. He had many
accomplishments, including the
absolute temperature scale and the
introduction of the Bell telephone into
Great Britain. However, he is reported
to have stated around 1880 that all
possible discoveries in physics had
already been made.


We choose as our system a sample ofnmoles of gas that flows through the porous
plug after a steady state is established. We assume that irreversible processes take
place only within the porous plug so that the initial and final states of our system
can be treated as equilibrium states. The work done on the gas on the left side is
given by

wL−

∫VL2

VL1

PLdVL−PL(VL2−VL1) (2.5-20)
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