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:

$$C_p = \frac{\mathrm{d}q_p}{\mathrm{d}T} = \left(\frac{\partial H}{\partial T}\right)_{\!p}$$


2 Answers 2


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 $U$:

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


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.

  • $\begingroup$ But since $dU=dq_V$ why can't you just substitute this in? $\endgroup$
    – RobChem
    Commented Jan 5, 2015 at 20:52
  • 1
    $\begingroup$ $dU=dq_v$ only at constant volume. You need the partial derivative notation to specify that you are at constant volume. Likewise for constant pressure. $\endgroup$
    – Ben Norris
    Commented Jan 5, 2015 at 23:27

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