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I knowv that $C_V=\frac{\frac{f}{2} Nk_B}{m}$ and $C_P=\frac{(\frac{f}{2} +1)Nk_B}{m}$. Since for solids their values are very close to each other, I would assume $\frac{f}{2} +1$ is very close to $\frac{f}{2}$. Namely, I thought I would need to have $\frac{f}{2} >> 1$. This would require a high number of degrees of freedom. However, is this the case? We are talking about a solid, of which the particles can move in less directions then, let's say, a gas molecule. Or is the very constraint that makes the number of degrees of freedom get very high? And is this the case or is there another explanation for why $C_V$ and $C_P$ have close values for solids?

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To complement the qualitative answers, here's the math. We know that $$C_P-C_V=\frac{VT\alpha^2}{\beta_T},$$ for all materials, where $C$ is the heat capacity (at constant pressure $P$ or volume $V$), $T$ is the temperature, $\alpha$ is the thermal expansion coefficient, and $\beta_T$ is the isothermal compressibility.

Solids typically have much smaller volumes and expansion coefficients than gases, although their compressibilities are also much smaller. The squared $\alpha$ term causes the heat capacities to be very similar.

A problem with oliver's answer is that it refers to the misconception that $C_P>C_V$ because the material must expand and do work against the atmosphere when pressure is held constant. (This misconception is repeated in the accepted answer here, for example.) But the equation above shows that $C_P>C_V$ even when thermal contraction occurs, i.e., when $\alpha<0$, because the term is squared. In fact, the expansion-against-the-atmosphere argument is reliable only for ideal gases (for which $\alpha=1/T$); condensed matter stores energy in internal modes when free to expand or contract, and this contributes to the finding of $C_P>C_V$.

A problem with Chet Miller's answer is that it might imply that $C_P\approx C_V$ because the compressibility $\beta_T$ is small. But $\beta_T$ is in the denominator, so decreasing it alone should make $C_P-C_V$ larger. The resolution is that low-$\beta_T$ materials are also low-$\alpha$ materials; in fact, incompressibility implies zero thermal expansion $\alpha$. As shown above, the squaring of a small $\alpha$ term causes it to dominate the expression and imply $C_P\approx C_V$.

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  • $\begingroup$ Interesting fact about thermal contraction. Never thought about that. +1 $\endgroup$
    – oliver
    Commented Apr 19, 2021 at 18:37
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Although changing the temperature of a solid under constant volume conditions can cause extremely high pressures, the corresponding volume changes under constant pressure conditions are very small, so the amount of work that a solid can do while expanding when heated, is also very small. It is this work that makes the difference between $C_P$ and $C_V$, which explains why they are not that much different.

To put it colloquially: by heating a metal block, you can probably lift a whole skyscraper by a 1mm, but you aren't able to design a lift by this principle, even if it should only carry a few people from the lobby to the first floor.

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  • $\begingroup$ Okay. So degrees of freedom do not factor into this at all? $\endgroup$
    – Agnese
    Commented Apr 18, 2021 at 15:25
  • $\begingroup$ To be honest, I don't know how degrees of freedom come into play. I have only covered the bird's eye view. Maybe someone else can answer this. $\endgroup$
    – oliver
    Commented Apr 18, 2021 at 15:28
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    $\begingroup$ @Agnese: by the way, are you sure that these formulae are valid also for solids? At first glance, these look a lot like the ones for ideal gases... For the ideal gas, internal energy only depends on temperature, but AFAIK it's more complicated for solids due to interactions between atoms/molecules... $\endgroup$
    – oliver
    Commented Apr 18, 2021 at 15:32
  • $\begingroup$ That's true. I wanted to make a quick comparison with gases, but I have just looked into it and the formulae are different for solids. It seems like the number of degrees of freedom is equal to the number of modes in a solid, which if I am not wrong is related to its lattice structure. So maybe my guess on solids having a high number of degrees of freedom could make sense, assumed the number of modes is high enough? $\endgroup$
    – Agnese
    Commented Apr 18, 2021 at 15:39
  • $\begingroup$ I mean, in a very approximate way, since I derived it in a very approximate way. $\endgroup$
    – Agnese
    Commented Apr 18, 2021 at 15:39
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It is because solids and liquids are very close to being incompressible. So it doesn't matter whether the pressure is changing or not.

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