Free expansion of perfect gas

We now discuss an example of irreversible adiabatic change, such as the free expansion of a perfect gas, whereby a gas initially contained in one half of its container is allowed to expand into the whole container freely (see Fig. ). Since the gas is isolated from the environment, $Q = 0$. A free expansion cannot do any work and so we also have $W = 0$, which for the first law implies $\Delta E = 0$. This is obvious also from our assumption of perfect gas, because the gas particles can only exchange energies by occasional collisions among themselves and their total (kinetic) energy remains constant. Moreover, the energy of the perfect gas does not depend on volume, as since the particles of gas are not interacting their mean distance cannot matter. As a consequence, $E = E(T)$, and since $E$ is constant $T$ also remains constant. Eq.  thus implies $TdS = PdV$, and we can calculate the change of entropy by integrating

$$\Delta S = \int_{V_1}^{V_2} \frac{P}{T}dV,$$ (3.82)

where $V_1$ and $V_2$ are the volumes before and after the expansion. However, since $Q = 0$ we should have $TdS = 0$, and since $W = 0$ we should also have $PdV=0$, so what is the meaning of Eq. ? In fact, during the expansion we are in the situation discussed around Eq. , where the pressure of the gas is not well defined (and even its volume is not determined as it expands), so it is not clear what we should use in Eq. .

At this point we recall that $S$ is a function of state, and so it does not matter how the gas is expanded from $V_1$ to $V_2$: the corresponding entropies $S_1$ and $S_2$ only depend on those two extreme states. Therefore we are at liberty, for example, to use a reversible isothermal expansion to calculate the integral  (or anything else of our liking that takes us from $V_1$ to $V_2$ with constant $T$). If the gas is driven from $V_1$ to $V_2$ with a series of quasi-equilibrium states, then it follows its equation of state $PV = Nk_{\rm B}T$ and we have:

$$\Delta S = \int_{V_1}^{V_2} \frac{P}{T}dV = \int_{V_1}^{V_2} \frac{Nk_{\rm B}}{V}dV = Nk_{\rm B}\ln \frac{V_2}{V_1},$$ (3.83)

which is obviously positive. The fact that the free expansion is adiabatic is therefore a red-herring as far as the entropy change is concerned, because equilibrium thermodynamics cannot be applied. In this particular case it has been easy to identify a reversible transformation –which is all it can ever be described by equilibrium thermodynamics– to compute the actual entropy change. In other cases it may be more complicated, but the fact remains that any transformation, whether going through equilibrium states or not, always relates an initial and a final point and thermodynamic variables such as the entropy are only dependent on those two points.

Figure: Perfect gas occupying the left half of the box (left) expanding to the whole box as the partition is removed (right).