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In book "Element of Gasdynamics" (authors Liepmann and Roshko) at paragraph "1.7 The First Law Apllied to Reversible Process. specific Heats." There is written (specifying that I'm going to report only some formulas and not the entire test between each formula and other) :

$$\mathrm{d}e=\mathrm{d}q-p\cdot \mathrm{d}v \quad(1.15) $$

$$c=dq/dt \quad(1.16)$$

$$c_v=(dq/dT)_v \quad(1.17a)$$

$$c_p=(dq/dT)_p \quad(1.17b)$$

$$\mathrm{d}e=\frac{\partial(e)}{\partial(v)}\cdot dv+\frac{\partial(e)}{\partial(T)}\cdot dT=dq-p\cdot dv$$

$$c_v=\frac{\partial(e)}{\partial(T)} \quad(1.18)$$

$$c_p=\frac{\partial(e)}{\partial(T)}+(\frac{\partial(e)}{\partial(v)}+p)\cdot(\frac{\partial(v)}{\partial(T)})_p \quad(1.19)$$

Why in equation (1.18) is not indicated that $\frac{\partial(e)}{\partial(T)}$ is keeping "$v$" constant? Why in equation (1.19) is not indicated that $\frac{\partial(e)}{\partial(T)}$ is keeping "$p$" constant?

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  • $\begingroup$ In order to write my question I tried to use Mathjax indication but the output is not helping so much. $\endgroup$ – d.pensopositivo Jan 25 '17 at 20:21
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    $\begingroup$ Essentially you just need to add $$ as in $\frac{1}{2}$ renders as $\frac{1}{2}$. You can edit now (when/if my edit is improved) to fix the type setting as you wish. $\endgroup$ – snulty Jan 25 '17 at 20:29
  • $\begingroup$ They should have indicated what is being held constant. Bad notation. $\endgroup$ – Deep Jan 26 '17 at 6:22
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This notation is really bad but seems mathematically correct.

$e$ is defined as only dependent of $v$ and $T$. As such, eq (1.15) gives the total derivative of $e$ regarding these variables.

Eq (1.19) takes for granted that $v$ does not vary and as such equates the PARTIAL derivative $\frac{\partial e}{\partial T}$ with $c_v$. A better notation would be $\left.\frac{\mathrm{d}e}{\mathrm{d}T}\right|_v=c_v$ precising that your equating the total variation of energy relative to any variation of temperature without volume change with $c_v$.

The partial derivative notation just says that for any transformation, you're equating a not measurable part of the change in energy (the part related to temperature change) to $c_v$. Both are correct but one is obviously less understandable at first sight.

In the same spirit you don't have to precise that $\frac{\partial e}{\partial T}$ is taken with $p$ constant because $\mathrm{d}e$ does not depend on $p$.

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About first question: "Why in equation (1.18) is not indicated that ∂(e)/∂(T) is keeping "v" constant?" I can answer that the author writes an important note before the formula de=∂(e)/∂(v)⋅dv+∂(e)/∂(T)⋅dT=dq−p⋅dv. in the note it's written: "It is customary in thermodynamics to indicate the variables which are kept constant in forming a partial derivative by a subscript, e.g., $$(∂e/∂v)_T $$ means that T is kept constant. We shall use this notation only where there could be any doubt about the variable that is considered constant. " So it's considered implicit in equation (1.18) that $$(∂e/∂v)$$ is keeping v constant. About second question: "Why in equation (1.19) is not indicated that ∂(e)/∂(T) is keeping "p" constant?" the objection (my objection !) is wrong because the addend should be $$(\frac{\partial(e)}{\partial(T)})_v\cdot(\frac{\partial(T)}{\partial(T)})_p $$, so the book considers, without any doubt about the variable, that $$(\frac{\partial(e)}{\partial(T)})_v$$ can be written $$\frac{\partial(e)}{\partial(T)}$$ and $$(\frac{\partial(T)}{\partial(T)})_p$$ is equal to 1 in any case.

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  • $\begingroup$ Please consider adding an explanation instead of claiming that the author has stated the reason. It would be useful for those who don't have the book. $\endgroup$ – Yashas Jan 26 '17 at 9:41
  • $\begingroup$ "the author writes an important note before the formula" I'm sorry, but... doesn't this note pretty much answer your question? What are you asking if the answer is written in the book? $\endgroup$ – Steeven Jan 26 '17 at 10:19
  • $\begingroup$ Steeven, you are right, I thought and finally, I'm answering, I understood that I'm wrong and I'm publishing the answers to the objections. Could be it useful? $\endgroup$ – d.pensopositivo Jan 26 '17 at 10:40
  • $\begingroup$ I put my proper answer to my same question. Everyone is welcome to advice me, if in his/her opinion I wrote something not right, not complete or not clear. Also, the other answers are very welcome, too. $\endgroup$ – d.pensopositivo Jan 26 '17 at 10:55

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