❒ 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.
❒ 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)
E2+ P2V2 = E1 + P1V1
H2= H1 or ΔH = 0
Thus the process in Joule-Thomson experiment takes place at constant enthalpy.
❒ 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.
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.