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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_trW_int where W_tr and W_int are number of microstates due to translational and internal degrees of freedom, respectively. If this gas is non-degenerate, then

(1)

\begin{align} W_{tr}=\frac{1}{N!}\left(\frac{V}{\lambda_T^3}\right)^{N}\sim\left(\frac{e}{\lambda_T^3}\frac{V}{N}\right)^N,\;\; N \gg 1 \end{align}

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

(2)

\begin{align} S(n)=nR\ln{(V/n)}+\mbox{const} \end{align}

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

(3)

\begin{align} \bigtriangleup S=\bigtriangleup S_A+\bigtriangleup S_B = n_1 R \ln{\left(\frac{V_1+V_2}{V_1}\right)}+n_2 R\ln{\left(\frac{V_1+V_2}{V_2}\right)} \end{align}

(4)

\begin{align} = n_1 R\ln{\left(\frac{n_1+n_2}{n_1}\right)}+n_2 R\ln{\left(\frac{n_1+n_2}{n_2}\right)}=R\ln{\left(\frac{3125}{108}\right)}\approx 3.4R \end{align}

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

(5)

\begin{align} \bigtriangleup S=(n_1+n_2) R \ln{\left(\frac{V_1+V_2}{n_1+n_2} \right)}-n_1 R \ln{\left(\frac{V_1}{n_1}\right)}-n_2 R \ln{\left(\frac{V_2}{n_2}\right)}=0 \end{align}

as expected.

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