JouleThomson: Background

Joule-Thomson Expansion | Liquefaction | Free Expansion | Adiabatic Expansion

Free Expansion

Free Expansion of a Gas

Imagine a gas confined within an insulated container as shown in the figure below. The gas is initially confined to a volume V1 at pressure P1 and temperature T1. The gas then is allowed to expand into another insulated chamber with volume V2 that is initially evacuated. What happens? Letís apply the first law.

We know from the first law for a closed system that the change in internal energy of the gas will be equal to the heat transferred plus the amount of work the gas does, or . Since the gas expands freely (the volume change of the system is zero), we know that no work will be done, so W=0. Since both chambers are insulated, we also know that Q=0. Thus, the internal energy of the gas does not change during this process.

We would like to know what happens to the temperature of the gas during such an expansion. To proceed, we recall that U is a state function. This means that we only need to know the inital and final states of U to obtain the change in U. The difference in U is independent of path. Thus we can use the powerful thermodynamic concept of constructing an imaginary reversible path that connects the initial and final states of the gas. Recall that Q and W by themselves are not state functions! Recall also that the actual free expansion is not a reversible process, and we canít apply thermodynamics to the gas during the expansion. However, once the system has settled down and reached equilibrium after the expansion, we can apply thermodynamics to the final state.

We know that the internal energy depends upon both temperature and volume, so we write

†††††††††††††††††††††††††††††††††††††† (1)

where we have kept the number of molecules in the gas (N) constant. The first term on the right side in equation (1) simply captures how U changes with T at constant V, and the second term relates how U changes with V and constant T. We can simplify this using Eulerís reciprocity relation, (where x,y,z are U,V,T):

††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††† (2)

to obtain an expression for the change in gas temperature

†††††††††††††††††††††††††††††††††††††††††††† (3)

The term is a property of the gas, and is called the differential Joule coefficient. This name is in honor of James Prescott Joule, who performed experiments on the expansion of gases in the mid-nineteenth century. If we can either measure or compute the differential Joule coefficient, we can then say how temperature changes (dT) with changes in volume (dV). Letís see how we might compute the Joule coefficient from an equation of state. The simplest possible equation of state is the ideal gas, where PV=nRT. The easiest way to find the Joule coefficient is to compute and . Note that we have left off the subscript ďNĒ for brevity, but we still require that the number of molecules in our system is constant.

We can use the following identity

†††††††††††††††††††††† ††††††††††††††††††††††††† † (4)

to show that so that .

If the gas is described by the van der Waals equation of state

††††††††††††††††††††††††††††††††††††††††††††††††††††††††††† (5)

you can show that the term in the numerator of equation (3) is given by


Think about what equation (6) is telling us. Recall that the parameter ďaĒ in the van der Waals equation of state accounts for attractive interactions between molecules. Equation (6) therefore states that the internal energy of a system expanded at constant temperature will change, and this change is due to attractive interactions between molecules. Since the ideal gas equation of state neglects these interactions, it predicts no change in the internal energy upon expansion at constant temperature, but the van der Waals equation of state does account for this. The term in the denominator of equation (3) is nothing more than the constant volume heat capacity . It can be shown that CV is never negative and only depends upon temperature for the van der Waals equation of state. Since the parameter a is also never negative, equations (3) and (6) tell us that the temperature of a real gas will always decrease upon undergoing a free expansion.How much the temperature decreases depends upon the state point and the parameter a. Molecules having strong attractive interactions (a large a) should show the largest temperature decrease upon expansion. We can understand this behavior in a qualitative sense by imagining what happens to the molecules in the system when the expansion occurs. On average, the distance between any two molecules will increase as the volume increases. If the intermolecular forces are attractive, then we expect that the potential energy of the system will increase during the expansion. This potential energy increase will come at the expense of the kinetic or thermal energy of the molecules. Therefore the raising of the potential energy through expansion causes the temperature of the gas to decrease.

We can compute how much the temperature is expected to decrease during a free expansion using the van der Waals equation of state. If one performs this calculation for the expansion of oxygen from 10 bar at 300 K into a vacuum, the temperature is found to be reduced by roughly 4.4 K. Therefore, free expansion is not a very efficient means of achieving low temperatures. To liquefy gases, we need to use a different process.

Joule-Thomson Expansion | Liquefaction | Free Expansion | Adiabatic Expansion