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What is the change of entropy that occurs when two moles of an ideal gas *A* and three moles of an ideal gas *B*, both at standard temperature and pressure are allowed to mix? What if the gases are the same, e.g., *A* and *A*?

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Answer

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As we will show, the entropy increases if the gases being mixed are not identical. According to the general principles of statistical mechanics, the entropy is *S* = *k* ln *W*, where *W* is the number of microstates that correspond to a given macrostate. For an ideal gas, we have *W* = *W _{tr}W_{int}* where

*W*and

_{tr}*W*are number of microstates due to translational and internal degrees of freedom, respectively. If this gas is non-degenerate, then

_{int}where *N* is the number of molecules, *V* is the volume, and $\lambda_T^3$ is the cube of the thermal wavelength (effectively, the "volume" occupied by a molecule at temperature *T*). The important factorial term *N*! eliminates the overcounting of states for indistinguishable particles.

The only parameters that change as a result of the mixing are *V* and the number of moles *n*. Therefore, we can write

where *R* is the universal gas constant. Note that at standard temperature and pressure *V* and *n* are directly proportional, *V* = (22.4 l)×*n*. Using this fact, the increase in entropy due to the mixing can be written as

On the other hand, if the gases were identical, e.g., A and A, then

(5)as expected.

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