We start with $\delta q = 0$ and $dU = C_{V}(T)dT = \delta w$. Why can we take the heat capacity at constant volume, when this process is an expansion so volume increases?


This is valid for ideal gas whose molar number is constant $n$. Why?

When a fluid changes volume, the equation $$ dU =dQ - pd V $$ is obeyed. Formally dividing by $dT$ we obtain

$$ \frac{dU}{d T} = \frac{d Q}{dT} - p\frac{d V}{d T}. $$

If we now consider only processes where $V$ remains constant, the relation

$$ \frac{dU}{d T}\bigg|_{V=const.} = \frac{d Q}{d T}\bigg|_{V=const.} $$ is obeyed.

Because of the right-hand side, we call this quantity heat capacity at constant volume and denote it $C_V(T,V)$. Generally it is a function of both $T$ and $V$ and is sufficient to express the change in the internal energy in the above way only when $V=const.$

However, for ideal gas internal energy is a function of $T$ only - let me denote it $U^{(id)}(T)$. This means that the condition on the left-hand side is superfluous - its presence does not matter. The value of the derivative can be calculated without any condition on the process: $$ \frac{dU(T,V)}{d T}\bigg|_{V=const.} = \frac{dU^{(id)}(T)}{d T}. $$ Thus the heat capacity at constant volume can be calculated as ratio of delta $U$ to delta $T$, process being immaterial. Because of the right-hand side, we know the result is a function of $T$ only - let me denote this function $C_V^{(id)}(T)$.

Now any change in the internal energy of ideal gas, irrespective of whether any other quantity is constant or not, can be written as

$$ dU^{(id)}(T) = \frac{dU^{(id)}}{dT}(T)d T. $$ From the previous equation it follows that for any process, $$ dU^{(id)}(T) = C_V^{(id)}(T)dT. $$

  • $\begingroup$ This looks like it's in contradiction with the wikipedia page on heat capacity. For the ideal gas we have $C_V-C_P = R$. $\endgroup$ – Steven Mathey Oct 13 '14 at 18:47
  • $\begingroup$ Where do you see contradiction? $\endgroup$ – Ján Lalinský Oct 13 '14 at 18:54
  • $\begingroup$ You can use $U^{(id)}(T)$ to compute both heat capacities. $C_V = dU^{(id)}/dT(T) = C_P$ since the pressure does not appear in $U^{(id)}(T)$ as well. $\endgroup$ – Steven Mathey Oct 13 '14 at 18:57
  • $\begingroup$ No, heat capacity is not defined via internal energy as $dU/dT$, but via heat as $dQ/dT$. The latter depends on whether the process has constant $V$ or constant $P$. $\endgroup$ – Ján Lalinský Oct 13 '14 at 18:59

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