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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?

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closed as unclear what you're asking by Martin - マーチン, Klaus-Dieter Warzecha, bon, jerepierre, Freddy May 13 '15 at 17:15

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ The second statement is vague to me. Molar mass of what? Any random species? Two nonbranched alkanes? What? $\endgroup$ – M.A.R. Apr 13 '15 at 20:32
  • $\begingroup$ Molar mass of that specific element. $\endgroup$ – Asker123 Apr 14 '15 at 1:54
  • $\begingroup$ 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... $\endgroup$ – Ndrina Limani Apr 15 '15 at 14:22
  • $\begingroup$ Could you please be more elaborate? Just a sentence isn't going to clear things up. $\endgroup$ – M.A.R. Apr 15 '15 at 15:09
  • $\begingroup$ 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. $\endgroup$ – Ndrina Limani Apr 15 '15 at 15:13
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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 $m$.

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  • $\begingroup$ 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)? $\endgroup$ – Nicolau Saker Neto May 14 '15 at 0:58
  • $\begingroup$ 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? $\endgroup$ – Asker123 May 14 '15 at 21:47

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