# WORK, HEAT, AND ENERGY: THE FIRST LAW OF THERMODYNAMICS

## The Joule–Thomson Experiment

For an ideal gas U depends only on T and n, so that H depends only on T and n:

Therefore,

Joule and Thomson carried out experiments in the 1850s to determine the value of (∂H/∂P)_{T,n} for real gases. They used an apparatus equivalent to that sketched in Figure 8. This apparatus consists of two cylinders fitted with pistons and a porous

**Figure 8** The Apparatus for the Joule–Thomson Experiment (Schematic).

plug separating the two cylinders. Each side has a manometer to measure the pressure and a thermometer to measure the temperature of the gas. The entire apparatus is adiabatically insulated from the surroundings. During the experiment, one of the pistons is moved into its cylinder and the other piston is withdrawn in a way such that the pressure on each side remains constant as the gas flows irreversibly through the porous plug. If the process is slow enough a time-independent nonequilibrium state (a steady state) will be attained with a different constant temperature and constant pressure on each side. Several experiments are carried out on a given gas with different values of the pressure difference. The *Joule–Thomson coefficient* μ_{JT} is defined as the extrapolated limit

where the subscripts R and L indicate the right and left sides of the apparatus. We now show that the limit in this equation is a partial derivative at fixed enthalpy.

We choose as our system a sample of n moles of gas that flows through the porous plug after a steady state is established. We assume that irreversible processes take place only within the porous plug so that the initial and final states of our system can be treated as equilibrium states. The work done on the gas on the left side is given by

where VL_{1} is the initial volume of the left side of the apparatus and VL_{2} is the final volume of the left side of the apparatus. The work done on the gas on the right side is given by

Prior to the transfer, the pressure P_{1} of the system was equal to P_{L}, and the initial volume of the system must have been equal to the magnitude of the change in volume of the left side:

The final pressure P_{2} must be equal to PR, and the final volume V_{2} must be equal to the change in volume of the right side:

From Eqs. (20) through (23), the total work done on the system is

Any change in state

Because the apparatus is adiabatically insulated from the laboratory, no heat is transferred to or from the laboratory. Also, no heat is transferred from the system to the apparatus after the steady state is established, because the chamber on the right is then at the same temperature as the gas that exits from the plug. Therefore,

The Joule–Thomson process therefore occurs at constant enthalpy, and the Joule–Thomson coefficient is equal to a partial derivative at constant H and n:

We can use the cycle rule,

The Joule–Thomson coefficient of an ideal gas vanishes because (∂H/∂P)_{T,n} vanishes. Joule and Thomson found that the Joule–Thomson coefficient is measurably different from zero for ordinary gases at ordinary pressures. It depends on temperature and is positive at room temperature for most common gases except for hydrogen and helium. Even for these gases it is positive at some range of temperatures below room temperature. This means that for some range of temperature any gas cools on expansion through a porous plug. Expansion of a gas can be used to cool the gas enough to liquefy part of it, and the final step in the production of liquid nitrogen or liquid helium is ordinarily carried out in this way.

### Example 23

For air at 300K and 25 atm, μ_{JT} 0.173K atm^{−1}. If a Joule–Thomson expansion is carried
out from a pressure of 50.00 atm to a pressure of 1.00 atm, estimate the final temperature if
the initial temperature is equal to 300 K.

### Solution

The molecular explanation for the fact that the Joule–Thomson coefficient is positive at sufficiently low temperature is that at low temperatures the attractive intermolecular forces are more important than the repulsive intermolecular forces. When the gas expands, work must be done to overcome the attractions and the potential energy increases. If no heat is added, the kinetic energy decreases and the temperature decreases.

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