Concise Physical Chemistry

c03 JWBS043-Rogers September 13, 2010 11:24 Printer Name: Yet to Come

46 THE THERMODYNAMICS OF SIMPLE SYSTEMS

3.9 THE JOULE AND JOULE–THOMSON EXPERIMENTS

The Joule experiment is important because it showed that there arenothermal effects

arising from expansion of a gas. The Joule–Thomson experiment is important because

it showed that therearethermal effects arising from expansion of a gas. What?

Wait! There’s an explanation....

The Joule series of experiments was the earlier of the two. It was carried out

with simple apparatus and relatively insensitive thermometers. The Joule apparatus

consisted of two chambers, one filled with a gas at pressurepand the other evacuated,

with the chambers being connected by a short tube with a stopcock. The apparatus

was placed in a bath and allowed to come to thermal equilibrium. The stopcock

was opened, allowingexpansion of the gasinto the evacuated half of the apparatus.

The temperature change was measured and a null result was recorded—hence the

conclusion that expansion of a gas produces no thermal effect. The Joule experiment

is analogous to Boyle’s law in that it is almost correct for most gases under mild

conditions (pressure change not too great). It is an ideal gas law. Joule happened to

be rich (he never had to be distracted by earning a living) and he was also smart. He

did not really believe the results of his experiment.

The Joule–Thomson experiment is a refinement of the Joule experiment intended

to find the very thing that the Joule experiment failed to find: the thermal effect of

expanding a gas. By means of a piston, a gas was driven through a porous plug from

one chamber at high pressure into a second chamber at low pressure, thus expanding

the gas (for more detail, see Klotz and Rosenberg, 2008). He then measured the

temperature difference between the two chambers and found that it was not null. The

result was more in accord with modern experience with highly pressurized gases. The

expanding gas cools. The cooling factor is called the Joule–Thomson coefficientμJT

μJT≡

(

∂T

∂p

)

H

The partial on the right-hand side is subscriptedHbecause the process, though it in-

volves a change inT, involves no change inH.Itisisenthalpic. The Joule–Thomson

coefficient is usually positive (for an expansion,dp<0 andμJT>0, so the gas

cools). There are a few exceptions, which warm up on expansion starting from some-

where around room temperature. For these few gases,μJT<0 at room temperature;

but if they are expanded at low temperatures,μJTchanges to a positive value. As

specific examples, nitrogen hasμJT∼= 0 .6 at 200 K; hence N 2 can be cooled by

expansion starting at 200 K and brought to a temperature so low that it is liquefied.

Helium hasμJT∼=− 0 .06 and cannot be liquefied by expansion starting at 200 K.

Helium must be precooled far below 200 K to whereμJTbecomes positive for it to be

liquefied by expansion. If expansion is carried out at a sufficiently low temperature,

all gases can be liquefied.^3 The point at whichμJTchanges from+to – is called the

inversion temperature.

(^3) Some special effects arise with helium isotopes.