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#### Joule-Thomson Effect

Joule and Thomson (later Lord Kelvin) showed that when a compressed gas is forced through a porous plug into a region of low pressure, there is appreciable cooling.

The phenomenon of producing lowering of temperature when a gas is made to expand adiabatically from a region of high pressure into a region of low pressure, is known as Joule-Thomson Effect or Joule-Kelvin Effect.

#### Joule-Thomson Experiment

The apparatus used by Joule and Thomson to measure the temperature change on expansion of a given volume of gas is illustrated in the following figure:

An insulated tube is fitted with a porous plug in the middle and two frictionless pistons A and B on the sides.

Let a volume V1 of a gas at pressure P1 be forced through the porous plug by a slow movement of piston A. The gas in the right-hand chamberis allowed to expand to volume V2 and pressure P2 by moving the piston B outward. The change intemperature is found by taking readings on the two thermometers.

Most gases were found to undergo cooling on expansion through the porous plug. Hydrogen
and helium were exceptions as these gases showed a warming up instead of cooling.

Explanation: The work done on the gas at the piston A is P1V1 and the work done by the gas at
the piston B is P2V2.

Hence the net work (w) done by the gas is:

w = P2V2 – P1V1
ΔE = q – w (First Law)

But the process is adiabatic and, therefore, q = 0

ΔE = E2 – E1 = – w = – (P2V2 – P1V1)
E2 – E1 = – (P2V2 – P1V1)

Rearranging,

E2 + P2V2 = E1 + P1V1
H2 = H1 or ΔH = 0

Thus the process in Joule-Thomson experiment takes place at constant enthalpy.

#### Joule-Thomson Coefficient

Joule-Thomson coefficient: is the number of degrees temperature change produced per atmosphere drop in pressure under constant enthalpy conditions on passing a gas through the porous plug.

Joule-Thomson coefficient is represented by the symbol μ. Thus,

If μ is positive, the gas cools on expansion;
If μ is negative, the gas warms on expansion.

The temperature at which the sign changes is called the Inversion temperature.

Most gases have positive Joule-Thomson coefficients and hence they cool on expansion at room temperature. Thus liquefaction of gases is accomplished by a succession of Joule-Thomson expansion.

The inversion temperature for H2 is –80ºC. Above the inversion temperature, μ is negative. Thus at room temperature hydrogen warms on expansion. Hydrogen must first be cooled below –80ºC(with liquid nitrogen) so that it can be liquefied by further Joule-Thomson expansion. So is the casewith helium.

#### Explanation of Joule-Thomson Effect

We have shown above that Joule-Thomson expansion of a gas is carried at constant enthalpy.

But

H = E + PV

Since H remains constant, any increase in PV during the process must be compensated by decrease of E, the internal energy. This leads to a fall in temperature i.e., T2< T1.

For hydrogen and helium PV decreases with lowering of pressure, resulting in increase of E and T2> T1. Below the inversion temperature, PV increases with lowering of pressure and cooling is produced.

Reference: Essentials of Physical Chemistry /Arun Bahl, B.S Bahl and G.D. Tuli / multicolour edition.