Why is it that increasing the molar mass generally increases the Entropy?
The allowed energy levels for a particle of mass $m$ in a box of length $X$ are given by
$$E_i=\frac{h^2}{8mX^2}i^2\quad i=1,2,\ldots$$

(Already this equation gives a hint that an ideal gas of high molar mass has more thermally accessible translational states than a gas of low molar mass. Therefore, the molar entropy of a gas of high molar mass is greater.)

In this case, the partition function for translational motion in the $x$-direction is approximately
$$z=\left(\frac{2\pi mkT}{h^2}\right)^{1/2}X$$

The partition function for translational motion in three dimensions is
$$\begin{align}
z&amp;=\left(\frac{2\pi mkT}{h^2}\right)^{3/2}XYZ\\
&amp;=\left(\frac{2\pi mkT}{h^2}\right)^{3/2}V
\end{align}$$

For a system of $N$ particles

$$\begin{align}Z&amp;=\frac{z^N}{N!}\\&amp;=\frac{1}{N!}\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}V\right]^N\end{align}$$

The statistical entropy is
$$\begin{align}
S &amp;=\frac{U-U_0}{T}+k\ln Z\\
&amp;=\frac{U-U_0}{T}+k\ln \frac{z^N}{N!}\\
&amp;=\frac{U-U_0}{T}+kN\ln z-k\ln N!\\
&amp;\approx\frac{U-U_0}{T}+kN\ln z-kN\ln N+kN\\
&amp;=\frac{U-U_0}{T}+nR\ln z-nR\ln nN_\text{A}+nR\\
&amp;=\frac{U-U_0}{T}+nR\ln\left[z\cdot\frac{1}{nN_\text{A}}\cdot \operatorname{e}\right] \\
&amp;=\frac{U-U_0}{T}+n R \ln \left[\left(\frac{2\pi m k T}{h^2}\right)^{3/2} \cdot \frac{\operatorname{e}V}{nN_\text{A}}\right] \\
\end{align}$$

The internal energy of a monoatomic ideal gas is

$$U=U_0+\tfrac{3}{2}nRT$$

which leads to the Sackur–Tetrode equation:

$$\begin{align}
S &amp;=\frac{U-U_0}{T}+nR\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\cdot\frac{\operatorname{e}V}{nN_\text{A}}\right]\\
&amp;=\tfrac{3}{2}nR+nR\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\cdot\frac{\operatorname{e}V}{nN_\text{A}}\right]\\
&amp;=nR\left\{\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\cdot\frac{\operatorname{e}V}{nN_\text{A}}\right]+\tfrac{3}{2}\right\}\\
&amp;=nR\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2} \cdot \frac{\operatorname{e}^{5/2}V}{nN_\text{A}}\right]
\end{align}$$

This equation implies that the entropy $S$ of a monoatomic ideal gas depends on the mass $m$.