# Joule-Thomson Effect (J.T.Effect)

When a gas is allowed to expand from high to low pressure through an orifice or a porous plug under adiabatic conditions, the gas gets cooled. *The drop in temperature (dT) produced by the fall in pressure (dP) under adiabatic conditions is called Joule-Thomson Effect * and is a thermodynamic process.

The Joule Thomson effect formula is-

μ_{J.T.} = (δT/δP)_{H}

The Joule-Thomson effect is also known as the **Joule-Kelvin effect**, refers to the change which takes place in fluid’s temperature as it flows from a high pressure region to lower pressure region.

The fall in temperature is due to the decrease in Kinetic Energy of gas molecules, since a portion of it is used up in overcoming van der Waal attractive forces existing among them during expansion. Since ideal gas has no such forces, therefore, there is no expenditure of energy in overcoming these forces during expansion.

Joule-Thomson effect can be describe by means of the **Joule-Thomson coefficient**. Joule-Thomson coefficient is the partial pressure derivative with respect to temperature at constant enthalpy.

At **inversion temperature,** Joule-Thomson coefficient is equal to zero.

Joule-Thomson coefficient depends only on the temperature at which gas is allowed to expand. Above the inversion temperature, the μ_{J.T.} would be negative and gas becomes warm. Below the inversion temperature, the μ_{J.T.} would be positive and gas becomes cool. Majority of the gases falls under this category.

## Joule Thomson Experiment

An insulated tube is taken and is fitted with a porous plug in the middle and two frictionless pistons A and B on the sides. It is filled with a given volume of the gas.

Let a gas of volume V_{i} at pressure P_{i}, be forced through the plug by the slow movement of piston A. The gas in the right side of the tube expands to volume V_{f} and pressure P_{f} by moving the piston B outward. The fall in temperature is noted from the two thermometers T_{i} and T_{f}.

The work done on the piston A is P_{i}V_{i} and

work done by the gas at the piston B is P_{f}V_{f}.

Hence, the net work (W) done by the gas = P_{f}V_{f} − P_{i}V_{i}.

First law of thermodynamics under adiabatic conditons (q = 0)-

ΔE = −W

E_{f} − E_{i} = −P_{f}V_{f} − P_{i}V_{i}.

or, E_{f} − E_{i} = P_{i}V_{i} − P_{f}V_{f}.

or, E_{f} + P_{f}V_{f} = E_{i} + P_{i}V_{i}

or, H_{f} = H_{i}

or, ΔH = 0

Therefore, the the process in J.T. Experiment takes place at constant enthalpy (i.e. isenthalpic process) and not at constant internal energy.