Why is it that increasing the molar mass generally increases the Entropy? [closed]

I've just found read that entropy increases through these reasons:

1. Changing the phase of the elements to a gas.
2. Increasing the molar mass of that specific element increases the entropy.

I understand the first one, but I can't grasp the second one.
Because naturally you would think that $\ce{H2}$ would have a higher entropy than $\ce{Ne}$ because one mole of it is generally lighter.

But however it is said that Ne has a higher entropy because of its lower molar mass.

Why?

• The second statement is vague to me. Molar mass of what? Any random species? Two nonbranched alkanes? What? – M.A.R. Apr 13 '15 at 20:32
• Molar mass of that specific element. – Asker123 Apr 14 '15 at 1:54
• If increasing the molar mass means bonding that specific element with other elements (form a compound), then the more the atoms are, the more is the possibility to have disorder.Although that second statement is quite blury... – Ndrina Limani Apr 15 '15 at 14:22
• Could you please be more elaborate? Just a sentence isn't going to clear things up. – M.A.R. Apr 15 '15 at 15:09
• You don't need to write an essey in order to be clear. I tried to help and that was my thought because, i repeat : the second statement is hard to understand. GOOD LUCK. – Ndrina Limani Apr 15 '15 at 15:13

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&=\left(\frac{2\pi mkT}{h^2}\right)^{3/2}XYZ\\ &=\left(\frac{2\pi mkT}{h^2}\right)^{3/2}V \end{align}

For a system of $N$ particles

\begin{align}Z&=\frac{z^N}{N!}\\&=\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 &=\frac{U-U_0}{T}+k\ln Z\\ &=\frac{U-U_0}{T}+k\ln \frac{z^N}{N!}\\ &=\frac{U-U_0}{T}+kN\ln z-k\ln N!\\ &\approx\frac{U-U_0}{T}+kN\ln z-kN\ln N+kN\\ &=\frac{U-U_0}{T}+nR\ln z-nR\ln nN_\text{A}+nR\\ &=\frac{U-U_0}{T}+nR\ln\left[z\cdot\frac{1}{nN_\text{A}}\cdot \operatorname{e}\right] \\ &=\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 &=\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]\\ &=\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]\\ &=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\}\\ &=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 $m$.

• Interesting answer. Does the relationship between entropy and molar mass really depend on the Schrödinger equation solution for a particle in a box, or could it be obtained without having to delve into quantum mechanics (perhaps at the expense of not deducing the proportionality constant)? – Nicolau Saker Neto May 14 '15 at 0:58
• But to be fair, is there anyway you can compress all that and express it in English or some sort of concepts that go together to get this final answer? – Asker123 May 14 '15 at 21:47