Reference: Daniel V. Schroeder, *An Introduction to Thermal Physics*, (Addison-Wesley, 2000) – Problems 2.37 – 2.38.

The entropy of an ideal gas is given by the Sackur-Tetrode formula:

where is the volume, is the energy, is the number of molecules, is the mass of a single molecule and is Planck’s constant.

We can apply this formula to the case where we begin with two *different* ideal gases and , with a total number of gas molecules divided into two volumes and , but at equal pressures and temperatures. Thus the number of type molecules can be expressed as a fraction of the total number so that , and thus . The volumes can be expressed as the same fractions of the total volume, so that and .

We now remove the partition between the gases and allow them to mix. Because they were at the same pressure and temperature before they mixed, both and remain unchanged when the gases mix, so the energy of each gas also remains unchanged. For each species of gas, the only change is the volume, which expands to the total volume .

From 1, the change in entropy in a process where only the volume changes is

The entropy changes for the two gases is therefore

Thus the total entropy change after mixing, called, appropriately, the *entropy of mixing*, is

[Note that both logarithms are negative since , so .] If so that we start out with 2 equal quantities of gases, the formula reduces to

This is the same result as equation 2.54 in Schroeder, since in that equation his is the number of molecules of *each* gas, not the total number.

It’s worth noting that this formula applies only if the two gases are different, that is, they are distinguishable. If the two gases are the same, there is essentially no change in entropy when we remove the partition. The situation is similar to Example 3 in our earlier post which dealt with two Einstein solids. Before we remove the partition, the gas in each portion of the volume is overwhelmingly likely to be at or near its most probable state. After removing the partition, the combined gas is also almost certain to be at or near the most probable macrostate for the overall system. Since the gas molecules are indistinguishable, it’s virtually impossible to tell the difference between the states before and after the partition is removed, so the entropy of the two systems are virtually identical. [I don’t understand Schroeder’s explanation following his equation 2.56, where he tries to explain the difference by doubling the amount of gas in what appears to be a fixed volume. This isn’t what happens if you start with a fixed amount of gas divided into two cells and then remove the partition.]

Another way of looking at is as follows. Suppose we start with a number of identical molecules. (This argument applies to any system in which the molecules all have similar properties and interact with each other in the same way, not just to ideal gases.) The entropy of this system is some value which may or may not be easy to calculate. Now suppose that at some point in time, we magically change of these molecules to a different species (which has similar properties to the original species as mentioned). The entropy will increase by the number of distinct ways we can choose to locate these molecules among the places available. (The entropy due to the number of possible locations and momenta of the molecules won’t change when we replace of the molecules by a different species, since that is already accounted for by . We’re interested only in the *extra* entropy generated by introducing a second species of molecule.) The number of ways of choosing locations from a total of is just so the entropy of mixing is

Using Stirling’s approximation for large , and taking as before, we get

(we’ve neglected the first term in the second line as for large it is negligible compared to the last two terms) which is the same as 5.

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JesseWhen Schroeder discusses equation 2.56, he states just beforehand that he is forgetting the partition, and simply comparing what happens if you double the gas in a fixed volume vs. adding an equal amount of a distinguishable gas. So his argument works, but you have to catch that quick line explaining that he’s moved on to a different scenario.