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Calculated heat capacity different (lower) from experimental value?

Physics Asked by Mecury-197 on December 29, 2020

When treating gases as ideal, classical systems, the heat capacity can be found by multiplying the number of independent degrees of freedom by 1/2k, where k is Boltzmann’s Constant. However, these values are usually lower than the real values of heat capacities for gases. Why?

I was thinking that it was due to the fact that even at low/room temperature there is some population of excited vibrational levels that add energy fluctuations that are not accounted for in the classical derivation that leads to the 1/2k for each degrees of freedom. But the population of any vibrational states above the ground state is so low that I didn’t feel like that fully explained it, so I also thought that there could be some interaction of the gas molecules (since they aren’t really ideal) that is somehow adding to the heat capacity. But I couldn’t rationalize how/why this could happen.

I calculated alot of heat capacity values and compared them to the experimental heat capacities, and only noble gases had the same experimental and calculated values, and H2 was the only gas that had a higher calculated value than experimental. I feel like the noble gas trend supports my idea about vibration levels, but my explanation doesn’t seem to explain why hydrogen deviates from the general trend.

Does anyone have any insight into this?

This is part of a homework question for a stat mech class.

Clarification on calculation: I added 1/2k for each of the 3 degrees of translational freedom, and 1/2k for each of the degrees of rotational freedom. Therefore for a diatomic with 3 degrees of translational freedom, and 2 degrees of rotational the heat capacity was 5/2k.

One Answer

You don't say whether your analysis includes rotational modes. I assume it does otherwise the disagreement between experiment and the ideal gas specific heat would be profound. A linear molecule will have two rotational modes, each adding $tfrac{1}{2}R$, so the specific heat (excluding vibration) will be $C_p = tfrac{7}{2}R$.

Anyhow, hydrogen has quite a high rotational constant of $B = 60.85,text{cm}^{-1}$. The spacing between rotational energy levels is $2B$, $4B$, $6B$ etc. If we take $4B$ and convert this to an energy we get about $4.8 times 10^{-21}$ J. If we take this equal to $kT$ and divide by $k$ we get a temperature of about $350$K, so at room temperature the rotational degrees of freedom aren't fully populated. That's why $C_p$ for hydrogen is less than $tfrac{7}{2}R$, though actually it's only slightly less.

Answered by John Rennie on December 29, 2020

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