# 1.19.3: Real Gases: Liquefaction of Gases

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In the nineteenth century a major theme in science concerned the properties of gases and their liquefaction. The challenge offered by the liquefaction of gases also prompted the development of thermodynamics and the production of low temperatures. Michael Faraday is noteworthy in this context. In 1823 Faraday had liquefied chlorine, ammonia and sulfur dioxide using a combination of pressure and low temperatures [1].

Based on the observation that the densities of liquids are higher than gases the expectation was that liquefaction of a given gas would follow application of high pressure. However this turned out not to be the case. For example, Newton showed that application of 2790 atmospheres (\(\equiv 2.8 \times 10^{8} \mathrm{~N} \mathrm{~m}^{-2}\)) did not liquefy air. In 1877 Cailletet and Pictet working independently obtained a mist of oxygen by sudden expansion of gas compressed at 300 atmospheres (\(\equiv 3 \times 10^{7} \mathrm{~N} \mathrm{~m}^{-2}\)) and cooled by \(\mathrm{CO}_{2}(\mathrm{s})\). However in developing the background to this subject we turn attention to the work of Joule.

## Experiments by Joule

A gas, chemical substance \(j\), is held in a closed system. The molar thermodynamic energy \(\mathrm{U}_{j}\) is defined by equation (a) where \(\mathrm{V}_{j}\) is the molar volume and \(\mathrm{T}\), the temperature.

\[\mathrm{U}_{\mathrm{j}}=\mathrm{U}_{\mathrm{j}}\left[\mathrm{T}, \mathrm{V}_{\mathrm{j}}\right]\]

The complete differential of equation (a) describes the change in \(\mathrm{U}_{\mathrm{j}}, \mathrm{dU}_{\mathrm{j}}\), as a function of temperature and volume.

\[\mathrm{dU}_{\mathrm{j}}=\left(\frac{\partial \mathrm{U}_{\mathrm{j}}}{\partial \mathrm{T}}\right)_{\mathrm{v}} \, \mathrm{dT}+\left(\frac{\partial \mathrm{U}_{\mathrm{j}}}{\partial \mathrm{V}_{\mathrm{j}}}\right)_{\mathrm{T}} \, \mathrm{dV}_{\mathrm{j}}\]

The molar isochoric heat capacity \(\mathrm{C}_{\mathrm{Vj}}\) is defined by equation (c).

\[\mathrm{C}_{\mathrm{Vj}}=\left(\frac{\partial \mathrm{U}_{\mathrm{j}}}{\partial \mathrm{T}}\right)_{\mathrm{V}}\]

The first law of thermodynamics relates the change in \(\mathrm{U}_{j}\) to the work done on the system \(\mathrm{w}\) and heat \(\mathrm{q}\) passing from the surroundings into the system. Thus,

\[\mathrm{dU}_{\mathrm{j}}=\mathrm{q}+\mathrm{w}\]

Hence,

\[\mathrm{q}=\left(\frac{\partial \mathrm{U}_{\mathrm{j}}}{\partial \mathrm{T}}\right)_{\mathrm{V}} \, \mathrm{dT}+\left(\frac{\partial \mathrm{U}_{\mathrm{j}}}{\partial \mathrm{V}_{\mathrm{j}}}\right)_{\mathrm{T}} \, \mathrm{dV} \mathrm{V}_{\mathrm{j}}-\mathrm{w}\]

The apparatus used by Joule comprised two linked vessels having equal volumes. A tube joining the two vessels included a tap. In an experiment, one vessel was filled with gas \(j\) at a known pressure whereas the second vessel was evacuated. When the tap was opened gas flowed into the second vessel, equalizing the pressure in the two vessels. By flowing into an evacuated vessel the gas did no work because there was no confining pressure; i.e. \(\mathrm{w} =\) zero. The temperature of the gas in the containing vessel fell and that in the originally empty vessel rose by an equal amount. In other words \(\mathrm{dT}\) for the two vessel system is zero. Hence

\[\left(\frac{\partial \mathrm{U}_{j}}{\partial \mathrm{T}}\right)_{\mathrm{V}}=0\]

The clear hope was that the temperature would fall dramatically leading to liquefaction of the gas. In fact and with the benefit of hindsight the overall change in temperature \(\mathrm{dT}\) was too small to be measured. More sophisticated apparatus would show that \(\left(\frac{\partial \mathrm{U}_{\mathrm{j}}}{\partial \mathrm{T}}\right)_{\mathrm{V}} \neq 0\) because as the gas expands work is done against cohesive intermolecular interaction. It would only be zero for a perfect gas.

## Experiments by Joule and Thomson [2,3]

In a series of famous experiments carried out in an English brewery, Joule and Thomson used an apparatus in which the gas under study passed through a porous plug from high to low pressures. The plug impeded the flow of the gas such that the pressure of the gas on the high pressure side and the pressure of gas on the low pressure side remained constant. It was observed that the temperature of the gas decreased as a consequence of the work done by the gas against intermolecular cohesion.

A technological breakthrough was now made. A portion of the cooled gas was re-cycled to cool the gas on the input side. On passing through the plug the temperature of the gas fell to a lower temperature. As this process continues a stage was reached where a fraction of the gas is liquefied.

As noted above, the cooling emerges because work is done on expansion of the gases against intermolecular interaction. This is a quite general observation. When the pressure drops the mean intermolecular distance increases with the result that the temperature decreases. However there are exceptions to this generalisation. If the pressure is high the dominant intermolecular force is repulsion. Consequently when the pressure drops, work is done by the repulsive forces increasing the intermolecular distances thereby raising the temperature.

## Footnotes

[1] The account given here is based on that given by N. K. Adam, Physical Chemistry, Oxford, The Clarendon Press, 1956, chapter III.

[2] Thomson ≡ Lord Kelvin

[3] J. P. Joule and W. Thomson, Proc. Roy. Soc.,1853,**143**,3457.

[4] G.N. Lewis and M. Randall, Thermodynamics, revised by K. S. Pitzer and L Brewer, McGraw-Hill, 2nd. edn.,1961, New York, pages 47-49.