I'm trying to prove the following relation

$$c_p -c_v = \Biggl[p+\biggl(\frac{\partial E}{\partial V}\biggr)_T\Biggr]\biggl(\frac{\partial V}{\partial T}\biggr)_p$$

Using the first law of thermodynamics as a start, I have gotten to the stage of

$$c_p\,\mathrm{d}T = c_v\,\mathrm{d}T + \Biggl[\biggr(\frac{\partial E}{\partial V}\biggr)_T + p\Biggr]\mathrm{d}V$$

Can I simply say in my next line "Dividing by $\mathrm{d}T$ and keeping pressure constant yields"?

Then I would get the original expression but I'm not sure this is mathematically sound.

  • $\begingroup$ mathematically no problem to do that $\endgroup$ – Nosrati Oct 17 '17 at 19:20
  • $\begingroup$ Because it's equivalent to differentiating the equation with respect to $T$. $\endgroup$ – Chemomechanics Oct 17 '17 at 21:01

When there is only one independent variable, procedures such as dividing throughout by infinitesimals like $dT$ are unambiguous (perhaps even rigorous; only someone acquainted with infinitesimal calculus can tell). However in multivariable calculus, in particular thermodynamics which always contains at least two independent variables, such a procedure is potentially confusing. For example, reasoning as "Dividing by $dT$ and keeping (say) enthalpy constant yields..." would result in incorrect conclusion. So why are we permitted to hold only pressure constant while transforming from the infinitesimal version (containing $dT,dV$) to partial-derivative version (containing $\frac{\partial V}{\partial T}|_p$)?

To see the reason let us go through the derivation. First law states: \begin{align} T~dS=dE+p~dV \end{align} which is true of arbitrary infinitesimal process. If we take entropy $S$ as a function of $T,p,$ and internal energy $E$ as a function of $T,V,$ then: \begin{align} dS&=\frac{\partial S}{\partial T}\Biggr|_pdT+\frac{\partial S}{\partial p}\Biggr|_pdp\\ dE&=\frac{\partial E}{\partial T}\Biggr|_VdT+\frac{\partial E}{\partial V}\Biggr|_TdV \end{align} Substituting into first law, and using definitions $T\frac{\partial S}{\partial T}|_p=c_p,~\frac{\partial E}{\partial T}|_V=c_V$, gives: \begin{align} c_pdT+T\frac{\partial S}{\partial p}\Biggr|_pdp=c_VdT+\left(p+\frac{\partial E}{\partial V}\Biggr|_T \right)dV \end{align} This is again valid for arbitrary infinitesimal process.

Now we apply the equation above to a constant pressure process, to obtain: \begin{align} c_pdT=c_VdT+\left(p+\frac{\partial E}{\partial V}\Biggr|_T \right)dV \end{align} Above equation is valid only for constant pressure process. That is why when you divide by $dT$ throughout any partial derivatives that result must be taken to be at constant pressure: \begin{align} c_p=c_V+\left(p+\frac{\partial E}{\partial V}\Biggr|_T \right)\frac{\partial V}{\partial T}\Biggr|_p \end{align}

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  • $\begingroup$ I can't +1 your comment, I don't have enough rep. But thanks very much for the concise response. $\endgroup$ – Patrick Moloney Oct 21 '17 at 12:19

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