Joule-Thomson Effect

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.

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