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I am self-studying Elements of Gasdynamics by Liepmann and Roshko. The authors define the specific heats of a gas at constant volume and at constant pressure as $$ c_v=\left(\frac{\partial q}{\partial T}\right)_v $$ $$ c_p=\left(\frac{\partial q}{\partial T}\right)_p $$

The author states that the specific internal energy is a state variable, so $e=e(v,T)$ and using calculus and the first law, we can write $$ de=\frac{\partial e}{\partial v}dv+\frac{\partial e}{\partial T}dT=dq-pdv $$

From this, the author states \begin{align} c_v&=\frac{\partial e}{\partial T}\\ c_p&=\frac{\partial e}{\partial T} +\left(\frac{\partial e}{\partial v}+ p\right)\left(\frac{\partial v}{\partial T}\right)_p \end{align}

How can I derive these last two expressions relating the specific internal energy and the specific heats? I am not the best with differentials.

Note: This question asks something related, but I was hoping to get some help showing how the differentials are manipulated to get from one expression to the other.

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    $\begingroup$ The authors are incorrect. Those are not the definitions of the specific heats. The definitions are $$c_{v}=\biggl (\frac{\partial u}{\partial T}\biggr)_v$$ $$c_{p}=\biggl (\frac{\partial h}{\partial T}\biggr)_P$$ $\endgroup$
    – Bob D
    Commented Aug 29, 2022 at 15:46

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I'll explain the first of the equations in some detail, perhaps enough to help you find the second answer. We simply start with the definition $$ c_v = \left( \frac{\partial q}{\partial T} \right)_v. $$ Now as you hopefully know the subscript $_v$ means that we consider the volume $v$ to be constant. We can now use the second relation $$ de = dq - p dv $$ which can be in this case treated like an equation (note that this isn't rigorous and the right correct treatment would need further considerations yet the answer would be the same), rewriting it as $$ dq = de + p dv. $$ Now again if we treat this even less rigorously we are we are considered withe the quantity $\left( \frac{\partial q}{\partial T}\right)_v$ when $v$ is constant, we can understand this quantity of telling us "how much does $q$ change if we have a small change in $T$ at constant $v$, since we know that small change of $q$ for constant $v$ is $dq = de$ we get the result $$ c_v = \frac{\partial e}{\partial T} $$ Again I was not rigorous on purpose since it seems that you don't know much about differential. I would advise to do some study on these important objects, one of the books I can recommend is Thermodynamics and an introduction to thermostatistics by Herber Callen.

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  • $\begingroup$ I read a few chapters from Herber and Callen and they were instructive. However, it has only convinced me more to study statistical thermodynamics, which may be a very deep rabbit hole indeed. $\endgroup$
    – nwsteg
    Commented Sep 19, 2022 at 0:42
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    $\begingroup$ That's how thermodynamics is. It is a very useful description, but to understand the underlying physics we always have to rely on statistical mechanics. $\endgroup$
    – Nitaa a
    Commented Sep 19, 2022 at 6:59

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