physical chemistry - Relation between constant-pressure and constant-volume heat capacities: Cp - Cv = nR

For an ideal gas, we have

$$C_p - C_V = nR$$

where
$C_p$ is heat capacity at constant pressure,
$C_V$ is heat capacity at constant volume,
$n$ is amount of substance, and
$R=N_\mathrm A\cdot k_\mathrm B=8.314\,462\,618\,153\,24\ \mathrm{J\ mol^{-1}\ K^{-1}}$^[source] is the molar gas constant.

How can I prove this?

Answer

The heat capacities are defined as

$$C_p = \left(\frac{\partial H}{\partial T}\right)_{\!p} \qquad \qquad C_V = \left(\frac{\partial U}{\partial T}\right)_{\!V} \tag{1}$$

and since $H = U + pV$, we have

$$\begin{align} C_p - C_V &= \left(\frac{\partial H}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} \tag{2} \\ &= \left(\frac{\partial U}{\partial T}\right)_{\!p} + \left(\frac{\partial (pV)}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} \tag{3} \end{align}$$

Ordinarily, we have $\mathrm{d}(pV) = p\,\mathrm{d}V + V\,\mathrm{d}p$, but under conditions of constant pressure $\mathrm{d}p = 0$ and so we can write

$$C_p - C_V = \left(\frac{\partial U}{\partial T}\right)_{\!p} + p\left(\frac{\partial V}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} \tag{4}$$

For an ideal gas, $V = nRT/p$ ($n$ being constant throughout this whole discussion - or else $nR$ in the original equation makes no sense!) and so

$$\left(\frac{\partial V}{\partial T}\right)_{\!p} = \frac{nR}{p} \tag{5}$$

Substituting $(5)$ into $(4)$ gives

$$C_p - C_V = \left(\frac{\partial U}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} + nR \tag{6}$$

At this stage, you have two choices.

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The easy way

From the equipartition theorem, the internal energy of an ideal gas is given by

$$U = \left(\frac{\text{Degrees of freedom}}{2}\right)nRT \tag{7}$$

The number of degrees of freedom doesn't matter, because it is a constant for any given gas. The important point is that $U = U(T)$, i.e. internal energy is a function of only temperature. Therefore,

$$\left(\frac{\partial U}{\partial T}\right)_{\!p} = \left(\frac{\partial U}{\partial T}\right)_{\!V} = \frac{\mathrm{d}U}{\mathrm{d}T} \tag{8}$$

and the desired result, $C_p - C_V = nR$, immediately follows.

The "easy way" is described in a large number of textbooks, or web pages. I was even told to use it in one of my tutorials, at Oxford no less. It is not incorrect. However, I don't like the easy way, because it invokes the equipartition theorem, which (in my opinion) makes it only half a proof, unless one proves the equipartition theorem along with it.

In fact, you don't need to use equipartition to derive this result. The only thing that we need to use is $pV = nRT$. If you are not convinced: read on!

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The hard way

Without invoking the equipartition theorem, we don't know that $U = U(T)$. In general, we would expect $U$ to be a function of all three variables: $U = U(V, T, p)$. However, for an ideal gas, $p$ itself is a function of $(V,T)$: $p(V,T) = nRT/V$; so, we can eliminate the $p$-dependence of $U$, since the $p$-dependence is adequately described by the $(V,T)$ dependence. The total differential for $U = U(V,T)$ is

$$\mathrm{d}U = \left(\frac{\partial U}{\partial V}\right)_{\!T}\,\mathrm{d}V + \left(\frac{\partial U}{\partial T}\right)_{\!V}\,\mathrm{d}T \tag{9}$$

Using a similar argument, we could also treat $U$ as a function of $p$ and $T$ instead. The total differential for $U = U(p,T)$ is

$$\mathrm{d}U = \left(\frac{\partial U}{\partial p}\right)_{\!T}\,\mathrm{d}p + \left(\frac{\partial U}{\partial T}\right)_{\!p}\,\mathrm{d}T \tag{10}$$

And the third total differential we need is that of $V = V(p,T)$:

$$\mathrm{d}V = \left(\frac{\partial V}{\partial p}\right)_{\!T}\,\mathrm{d}p + \left(\frac{\partial V}{\partial T}\right)_{\!p}\,\mathrm{d}T \tag{11}$$

Substituting $(11)$ into $(9)$, we have

$$\begin{align} \mathrm{d}U &= \left(\frac{\partial U}{\partial V}\right)_{\!T} \left[\left(\frac{\partial V}{\partial p}\right)_{\!T}\,\mathrm{d}p + \left(\frac{\partial V}{\partial T}\right)_{\!p}\,\mathrm{d}T \right] + \left(\frac{\partial U}{\partial T}\right)_{\!V}\,\mathrm{d}T \tag{12} \\ &= \left[\left(\frac{\partial U}{\partial V}\right)_{\!T}\left(\frac{\partial V}{\partial p}\right)_{\!T}\right]\mathrm{d}p + \left[\left(\frac{\partial U}{\partial V}\right)_{\!T}\left(\frac{\partial V}{\partial T}\right)_{\!p} + \left(\frac{\partial U}{\partial T}\right)_{\!V}\right]\mathrm{d}T \tag{13} \end{align}$$

Equations $(10)$ and $(13)$ have very similar forms: they both look like $\mathrm{d}U = a\,\mathrm{d}p + b\,\mathrm{d}T$. We can equate the coefficients of $\mathrm{d}T$ in both equations to get

$$\left(\frac{\partial U}{\partial T}\right)_{\!p} = \left(\frac{\partial U}{\partial V}\right)_{\!T}\left(\frac{\partial V}{\partial T}\right)_{\!p} + \left(\frac{\partial U}{\partial T}\right)_{\!V} \tag{14}$$

and therefore, looking back to equation $(6)$ and substituting equation $(14)$ in, we see that

$$\begin{align} C_p - C_V &= \left(\frac{\partial U}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} + nR \tag{6} \\ &= \left(\frac{\partial U}{\partial V}\right)_{\!T}\left(\frac{\partial V}{\partial T}\right)_{\!p} + nR \tag{15} \end{align}$$

The final step is that, for an ideal gas,

$$\left(\frac{\partial U}{\partial V}\right)_{\!T} = 0 \tag{16}$$

and the proof of this can be found in this question: Internal pressure of ideal gas. This yields

$$C_p - C_V = nR \tag{17}$$

as desired.