Joule’s experiment based on free adiabatic expansion of a gas failed to detect the existence of intermolecular attractions. Therefore in 1952, in collaboration with Lord Kelvin(Thomson), he devised a very sensitive technique known as Porous Plug Experiment.
The principle of Joule Thomson experiment consists in forcing a gas at a constant pressure, through a porous plug thermally insulated from outside.
The porous plug consists of a compressed porous material like cotton, or silk fibers having a number of fine holes.
The gas on passing through the plug under a large pressure, it’s molecule get further apart from one another, so that the gas has to do some internal work against intermolecular attractions, if these exists.
This work will come from Internal energy of the gas. Therefore the gas passing through the plug suffer a fall in temperature, since a cooling of gas was demonstrated by Joule and Thomson indicates the existence of intermolecular attractions, & hence it is called Joule-Thomson effect.
The apparatus used in Porous plug experiment is shown in
|Fig. 1. Apparatus used in Porous plug experiment|
The gas under test is compressed to a high pressure with the help of position P and is passed through a long spiral tube S immersed in water bath, maintained at constant temperature.
The compressed gas enters the tube B fitted with a Porous plug. A vessel L fitted with cotton surrounds the tube B to provide a good thermal insulation.
When the gas flows through the plug, the temperature of gas on the two sides of plug is measured by Platinum resistance thermometer T1 & T2 and it was found that all gases show a change in temperature in passing through the porous plug. This temperature change is known as Joule Thomson effect.
Theory of Joule Thomson Expansion: Law of conservation of Enthalpy
Let us consider a simple arrangement as shown in fig.2 representing the porous plug experiment.
|Fig 2. Initial State|
CC’ is a cylinder with non-conducting walls having a porous plug G and fitted with two non conducting pistons 1 and 2 . A fixed mass of gas having a volume V1 is filled between the piston 1 and porous plug. Let this gas be highly compressed to a pressure P1 and pressure on other side of plug be P2, the piston 2 being kept behind the plug preventing any gas from passing through it.
Due to this large difference of pressure, the gas flows through the porous plug & get throttled or ‘wire drawn’i.e., its molecules get further apart.
Let now position 1 moved slowly inward in such a way so as to keep a constant pressure P1 on left hand side of plug as the gas escapes. The position 2 will move slowly outward so as to keep lower constant pressure P2 on the right hand side of the plug. After all the gas has passed through the plug the position 1 and 2 will be in position shown in fig.3 .
The gas will now occupy a greater volume, say V2.
If X1 & X2 are the positions through which piston 1 and 2 move and A is the area of cross section of the pistons, then,
External work done on the gas by piston 1 = P1 × A × X1
External work done by the escaping gas on piston 2
= P2 × A × X2
Therefore, Net external work done by the gas in passing through
the plug = P2V2 – P1V1
|Fig 3. Final State|
As the cylinder is isothermally insulated, Therefore external work done by the gas should be at cost of its own internal energy. Hence internal energy decreases.
Now according to first law of thermodynamics,
This is law of conservation of enthalpy
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