Joule expansion

File:JouleExpansion.svg

The Joule expansion, in which a volume V_i = V₀ is expanded to a volume V_f = 2V₀ in a thermally isolated chamber.

The Joule expansion is an irreversible process in thermodynamics in which a volume of gas is kept in one side of a thermally isolated container (via a small partition), with the other side of the container being evacuated. The partition between the two parts of the container is then opened, and the gas fills the whole container.

The Joule expansion is also called free expansion. The process is a useful exercise in classical thermodynamics, as it is easy to work out the resulting increase in entropy in an ideal gas, the so-called entropy production. If the gas is not ideal, the process is more complex and is called the Joule–Thomson effect.

This type of expansion is named after James Prescott Joule who used this expansion, in 1845, in his study for the mechanical equivalent of heat, but this expansion was known long before Joule e.g. by John Leslie, in the beginning of the 19th century, and studied by Joseph-Louis Gay-Lussac in 1807 with similar results as obtained by Joule.

Description

We consider an ideal gas under some pressure, P_i at temperature T_i, confined to one half of a thermally isolated container (see drawing). The gas occupies a volume V_i = V₀, mechanically separated from the other half of the container, which has an equal volume V₀, and is under lower or zero pressure. The tap (solid line) between the two halves of the container is then suddenly opened, and the gas expands to fill the entire container, which has a total volume of V_f = 2V₀. Both the previous and new temperature and pressure (T_f, P_f) follow the Ideal Gas Law, so that initially we have P_iV_i = nRT_i and then, when the tap is opened, we have P_fV_f = nRT_f (where R is the molar ideal gas constant).

Because this ideal system is thermally isolated, it cannot exchange heat with its surroundings. Also, since the system's volume is kept constant, the system cannot perform work on its surroundings. As a result, the change in internal energy ΔU = 0, and because U is strictly a function of temperature for the ideal gas, we know that T_i = T_f (the temperature of the gas does not change). This implies that P_iV₀ = P_f(2V₀), and thus the pressure halves; i.e. P_f = ½P_i.

Entropy production

It is awkward to calculate the entropy production in this process directly, because between the time the partition is opened and the time equilibrium is reached, the system passes through states that are far from thermal equilibrium. However, entropy is a function of state, and therefore the entropy change can be computed directly from the knowledge of the final and initial equilibrium states. For an ideal monatomic gas, the entropy as a function of the internal energy U, volume V, and number of moles n is given by the Sackur–Tetrode equation:

$S = nR \ln \left[\left(\frac VN\right) \left(\frac{4\pi m}{3h^2}\frac UN \right)^{\frac 32}\right]+ {\frac 52}nR.$

In this expression m the particle mass and h Planck's constant. For a monatomic ideal gas U = (3/2)nRT = nC_VT, with C_V the molar heat capacity at constant volume. In terms of classical thermodynamics the entropy of an ideal gas is given by

$S(V,T)=S_0+nR\ln\left(\frac{V}{V_0}\right)+nC_V\ln\left(\frac{T}{T_0}\right)$

where S₀ is the, arbitrary chosen, value of the entropy at volume V₀ and temperature T₀. It is seen that a doubling of the volume at constant U or T leads to an entropy increase of ΔS = nR ln(2). This result is also valid if the gas is not monatomic, as the volume dependence of the entropy is the same for all ideal gases.

A second way to evaluate the entropy change is to choose a route from the initial state to the final state where all the intermediate states are in equilibrium. Such a route can only be realized in the limit where the changes happen infinitely slowly. Such routes are also referred to as quasistatic routes. In some books one demands that a quasistatic route has to be reversible, here we don't add this extra condition. The net entropy change from the initial state to the final state is independent of the particular choice of the quasistatic route, as the entropy is a function of state.

Here is how we can effect the quasistatic route. Instead of letting the gas undergo a free expansion in which the volume is doubled, a free expansion is allowed in which the volume expands by a very small amount δV. After thermal equilibrium is reached, we then let the gas undergo another free expansion by δV and wait until thermal equilibrium is reached. We repeat this until the volume has been doubled. In the limit δV to zero, this becomes an ideal quasistatic process, albeit an irreversible one. Now, according to the fundamental thermodynamic relation, we have:

$\mathrm{d}U = T \mathrm{d}S - P \mathrm{d}V.$

As this equation relates changes in thermodynamic state variables, it is valid for any quasistatic change, regardless of whether it is irreversible or reversible. For the above defined path we have that dU = 0 and thus TdS=PdV, and hence the increase in entropy for the Joule expansion is

$\Delta S=\int_i^f\mathrm{d}S=\int_{V_{0}}^{2V_{0}} \frac{P\,\mathrm{d}V}{T}=\int_{V_0}^{2V_0} \frac{n R\,\mathrm{d}V}{V}=n R\ln 2.$

A third way to compute the entropy change involves a route consisting of reversible adiabatic expansion followed by heating. We first let the system undergo a reversible adiabatic expansion in which the volume is doubled. During the expansion, the system performs work and the gas temperature goes down, so we have to supply heat to the system equal to the work performed to bring it to the same final state as in case of Joule expansion.

During the reversible adiabatic expansion, we have dS = 0. From the classical expression for the entropy it can be derived that the temperature after the doubling of the volume at constant entropy is given as:

$T = T_i2^{-R/C_V} = T_i2^{-2/3}$

for the monatomic ideal gas. Heating the gas up to the initial temperature T_i increases the entropy by the amount

$\Delta S = n\int_{T}^{T_i}C_{V}\frac{\mathrm{d}T'}{T'}= nR \ln 2.$

We might ask what the work would be if, once the Joule expansion has occurred, the gas is put back into the left-hand side by compressing it. The best method (i.e. the method involving the least work) is that of a reversible isothermal compression, which would take work W given by

$W=-\int_{2V_0}^{V_0} P\,\mathrm{d}V =- \int_{2V_0}^{V_0} \frac{nRT}{V} \mathrm{d}V=nRT\ln 2=T\Delta S_{gas}.$

During the Joule expansion the surroundings do not change, so the entropy of the surroundings is constant. So the entropy change of the so-called "universe" is equal to the entropy change of the gas which is nR ln 2.

Real-gas effect

Joule performed his experiment with air at room temperature which was expanded from a pressure of about 22 bar. Air, under these conditions, is almost an ideal gas, but not quite. As a result the real temperature change will not be exactly zero. With our present knowledge of the thermodynamic properties of air we can calculate that the temperature of the air should drop by about 3 degrees Celsius when the volume is doubled under adiabatic conditions. However, due to the low heat capacity of the air and the high heat capacity of the strong copper containers and the water of the calorimeter, the observed temperature drop is much smaller, so Joule found that the temperature change was zero within his measuring accuracy.

References

The majority of good undergraduate textbooks deal with this expansion in great depth; see e.g. Concepts in Thermal Physics, Blundell & Blundell, OUP ISBN 0-19-856770-7