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The molar heat capacity of hydrogen gas and deuterium gas are nearly the same, $\pu{28.8 J K-1 mol-1}$ and $\pu{29.2 J K-1 mol-1}$, respectively, but the absolute entropy of deuterium ($\pu{145.0 J K-1 mol-1}$) is significantly larger than that of hydrogen ($\pu{130.7 J K-1 mol-1}$).

I have seen the equation that associates the heat capacity to entropy, and I understand the equipartition theorem and different modes of motion contribute to the heat capacities. Since the heat capacities are fairly the same, what other factors can affect the entropy in these species?

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The entropy depends on the mass. The translational part is given by the Sakur-Tetrode eqn., the rotational part depends on the moment of inertia, and the vibrational part on the vibrational frequency. In each case the partition function $z$ has to be calculated then the entropy via $\displaystyle S=R\ln(z)+RT\left(\frac{\partial z}{\partial T} \right)_V -R\ln(N)+R$ (for a perfect gas). The physical reason is that for the heavier molecule there are more energy levels up to a given energy that can be occupied and so more ways of populating these levels and so more entropy.

The heat capacity is the slope of internal energy $U$ with temperature (at constant volume) $\displaystyle C_V=\left({\partial U}{\partial T}\right)_V$ and the internal energy is related to the partition function as $\displaystyle \frac{U}{T}=RT\left(\frac{\partial z}{\partial T}\right)_V$ and so also to the entropy. The classical thermodynamic relationship is $\displaystyle S=S_0+\int_0^T \frac{C_p}{T}dT$ where $C_p$ is a function of temperature.

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    $\begingroup$ I believe that the Sakur-Tetrode is applicable only for monatomic gases, if the problem would be associated with the species like Helium and Neon, then the concept of the mass would be the reasons for the deviation in entropy, as you correctly state the mass is playing a vital role in that equation. $\endgroup$ May 5 '20 at 9:15
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    $\begingroup$ Yes it was derived for an ideal monoatomic gas, but this is only because all the energy is assumed to be kinetic, i.e. $p^2/(2m)$. We can use this just to calculate the translational kinetic energy part to the entropy then add the other terms. $\endgroup$
    – porphyrin
    May 5 '20 at 10:52
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Based on the difference in mass of the two isotopes you expect a difference of $\pu{14.4 J mol-1 K-1}$ $\left(\frac{5}{2} R\ln(2)\right)$ in the entropy (computing the rotational and translational contributions - $R\ln(2)$ and $\frac{3}{2} R\ln(2)$ respectively- for a rigid rotor ideal gas using basic statistical mechanics). That's in line with what you state in the problem.

On the other hand, the heat capacity is computed by taking the temperature derivative of the logarithm of the partition functions, eliminating any mass dependence. When you do the computation this leads to the conclusion that the ideal heat capacities of the two gases should be identical - at least at this level of theory. So the similarity in heat capacities is at least suggestive that practice does not deviate much from this theory. More interesting is then the fact that they do differ.

As for the question in the title, the answer is simply that the derivative of the entropy w.r.t. temperature is related to the heat capacity as follows: $$\left(\frac{\partial S}{\partial T}\right)_V=\frac{C_V}{T}$$

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    $\begingroup$ The entropy at absolute zero is zero. If the heat capacities are only 3% different from each other over the entire temperature range, and we integrate from zero kelvin to room temperature, why do we get entropy values that are about 10% different from each other? @Lackofgoodpeople $\endgroup$ May 4 '20 at 22:36
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    $\begingroup$ @KarstenTheis The expressions I used are applicable at high T. More importantly, as you drain the systems of energy by lowering T molecules lose their translational and rotational degrees of freedom (becoming vibrational instead) during condensation and crystallization. You are left with zero point energy. $\endgroup$
    – Buck Thorn
    May 5 '20 at 4:44
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    $\begingroup$ @KarstenTheis The more appealing explanation for why mass has this effect on the entropy is what porphyrin wrote, namely that the isotopes result in different energy level gaps and so different state occupation statistics. $\endgroup$
    – Buck Thorn
    May 5 '20 at 4:47
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The relative entropies of two substances at a temperature $T'$ are not determined by their relative heat capacities at $T'$. Rather,

$$\text {Absolute entropy at } T' \equiv S(T') =\Delta S_{0 \rightarrow T'}$$ $$= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T} = \int_{0}^{T'}\frac{C_p(T)}{T} dT.$$

I.e., for each substance, you need to consider how $C_p(T)$ varies with $T$ from $T=0$ to $T=T'$. Just because the heat capacities are about the same at $T'$ doesn't mean they won't be different between $T=0$ and $T=T'$.

Intutively, the way to think about this is that the entropy is determined by how the energy levels are populated; $\frac{C_p(T)}{T}$ at any given temperature tells us the ways in which the existing populations are able to change as the temperature is increased; and the integral of $\frac{C_p(T)}{T}$ gives us the entropy by summing up these changes.

In addition, hydrogen and deuterium have different freezing and boiling points. Thus they will not always be of the same phase. This will cause further differences between their heat capacities.

To get an idea of the complexities involved in determining the relative heat capacities of these substances at lower temperatures, take a look at these two NIST publications, which give the equations of state for hydrogen and deuterium, respectively:

https://tsapps.nist.gov/publication/get_pdf.cfm?pub_id=832374

and

https://www.nist.gov/system/files/documents/2018/03/12/2fundamental_equation_of_state_for_deuterium.pdf

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