Joule Expansion - Entropy Production

Entropy Production

It is awkward to calculate the entropy production in this process directly, because the route that the expansion takes in the time between the partition being opened and the equilibrium state being reached involves 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+
{\frac 52}nR.

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

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 as for all ideal gases.

One can also evaluate the entropy change by taking another route from the initial state to the final state, such that all the intermediary 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.

Suppose that, instead of letting the gas undergo a free expansion in which the volume is doubled, a free expansion in 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:

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

Another path that can be chosen is to let the system undergo a reversible adiabatic expansion in which the volume is doubled. The system will then perform work and the gas temperature goes down, so we'll have to supply heat to the system equal to the work performed to make sure it ends up in 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:

Heating the gas up to the initial temperature Ti increases the entropy by:

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

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 nRln 2.

Read more about this topic:  Joule Expansion

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