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I have the following in my notes and can't quite work out why it's the case.

We have $dQ = TdS$ for constant $T$.

We also know that $\frac{dQ}{dT} = C_v$ if $V, N$ are kept constant.

There is also $\frac{dQ}{dT} = C_v = T\left(\frac{\partial S}{\partial T}\right)_{V, N}$.

I understand that in the first equation, $T$ is a constant and one can take the derivative with respect to $T$ and only get one term i.e. $T\left(\frac{\partial S}{\partial T}\right)_{V, N}$ but the physical sense of doing this is lost to me. Does it mean that if temperature is not constant, I can't use the first equation and what does it even mean to then take the derivative there with respect to $T$?

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    $\begingroup$ Does T have to be constant in the first equation though? $\endgroup$ – Aaron Stevens May 14 '18 at 15:13
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You're not taking a derivative, you're dividing by a differential.

$dQ = TdS$ is the definition of $dS$ and does not require constant $T$. Rather, it requires that the process defining $dQ$ be reversible. We are always allowed to divide through by $dT$ on both sides of this equation (here we are strictly imagining that these quantities are small but not yet infinitesimal, we will take the limit in a few lines.)

This gives $\frac{dQ}{dT} = T \frac{dS}{dT}$. Passing to the limit that $dT$ becomes infinitesimal we get $C_v = T \left(\frac{\partial S}{\partial T}\right)_v$.

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