Expressions for constant volume and constant pressure heat capacities

My lecture handout says:


$$C\_V = \frac{\mathrm{d}q\_V}{\mathrm{d}T} = \left(\frac{\partial U}{\partial T}\right)\_{\!V}$$


I understand that $C\_V = \mathrm{d}q\_V/\mathrm{d}T$ and that $\mathrm{d}U = \mathrm{d}q\_V$, but what's the need for the partial derivative? Why can't you just substitute and get:


$$C\_V = \frac{\mathrm{d}U}{\mathrm{d}T}$$


Likewise for the constant pressure heat capacity:


$$C\_p = \frac{\mathrm{d}q\_p}{\mathrm{d}T} = \left(\frac{\partial H}{\partial T}\right)\_{\!p}$$




To complement entropid's answer, let's look at the formulation of $U$ that is slightly more useful:


$$\begin{align}
U &= q + w \\
\mathrm{d}U &= \mathrm{d}q + \mathrm{d}w \\
&= C\,\mathrm{d}T - p\,\mathrm{d}V
\end{align}$$


At constant volume, $\mathrm{d}V = 0$, so $p\,\mathrm{d}V = 0$


$$\begin{align}
\mathrm{d}U &= C\_V\,\mathrm{d}T - 0 \\
C\_V &= \frac{\mathrm{d}U}{\mathrm{d}T}
\end{align}$$


However, this notation assumes that $U$ is only a function of $T$, as in $U(T)$. It is not. In this version, $U$ is a function of $V$ and $T$: $U(V, T)$. Note that we do not need $p$ as a variable, since it is dependent only on $V$ and $T$ as well - for an ideal/perfect gas $p(V,T)=\frac{nRT}{V}$. Thus, we need partial derivative notation so that we know we are not trying to take the total derivative of $U$:


$$C\_V = \left(\frac{\partial U}{\partial T}\right)\_{\!V}$$


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The difference in **internal energy** ($\mathrm{d}U$) is a function of many different variables:


$$\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V \!$$


Therefore you must specify – via a partial derivative – with respect to which of the variables you are doing the derivative, otherwise you would need a *total derivative*, which is not what the equation requires.


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