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The formula for internal pressure of a system is: $$\pi_T=T\left(\frac{\partial p}{\partial T}\right)_{\!V}-p$$

My professor said we can redefine this equation in terms of either the

  1. thermal compressibility, $\kappa$
  2. expansion coefficient, $\alpha$

How do we do this?

I was able to derive for a van der Waals gas:

  1. $\Large\kappa = \frac{V-nb}{pV-\frac{an^2}{v}+\frac{2abn^3}{v^2}}$

  2. $\Large \alpha= \frac{nR}{PV-\frac{an^2}{v}+\frac{2abn^3}{v^2}}$

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    $\begingroup$ See equations (5) through (9) of getafix's answer here: Relation between constant-pressure and constant-volume heat capacities: Cp - Cv = nR $\endgroup$ – orthocresol Feb 11 '17 at 20:02
  • $\begingroup$ @orthocresol: This is not correct for a non-ideal gas. $\endgroup$ – Chet Miller Feb 11 '17 at 21:16
  • $\begingroup$ @ChesterMiller Do you mean the answer that I linked is wrong? (Edit: The answer that I linked derives the case for a non-ideal gas. Right before equation (9) it is written that $(\partial p/\partial T)_V = \alpha/\kappa_T$, which is the same as what you have in your current answer. The title of the question does, indeed, only apply to an ideal gas.) $\endgroup$ – orthocresol Feb 11 '17 at 21:26
  • $\begingroup$ @orthocresol: The result you gave after the : is only valid for an ideal gas. But other parts of the development, particularly the result that matches the final equation I gave are correct. Still, neither the development you gave nor mine substituted back into the equation to get $\pi$. Please advise me whether you think it would be best if I removed my answer? $\endgroup$ – Chet Miller Feb 11 '17 at 21:39
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    $\begingroup$ @ChesterMiller I thought your answer was OK (and upvoted it). It was what I had in mind anyway when I linked the answer above. $\endgroup$ – orthocresol Feb 11 '17 at 21:59
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$\def\d{\mathrm{d}}$I thought I might include the slightly longer approach because the non-standard treatment of partial differentials can be confusing.


Begin with the differential of volume assuming a constant number of particles.

$$\d V= \left(\frac{\partial V}{\partial T}\right)_P\d T + \left(\frac{\partial V}{\partial P}\right)_T\d P \tag1$$

Similarly, assuming that $P = P(T,V)$ gives us

$$\d P= \left(\frac{\partial P}{\partial T}\right)_V\d T + \left(\frac{\partial P}{\partial V}\right)_T\d V.\tag2$$

Substitute $(2)$ into $(1)$ and rearrange the differentials:

$$\left[1-\left(\frac{\partial V}{\partial P}\right)_T\left(\frac{\partial P}{\partial V}\right)_T\right]\d V = \left[\left(\frac{\partial V}{\partial T}\right)_P+\left(\frac{\partial V}{\partial P}\right)_T\left(\frac{\partial P}{\partial T}\right)_V\right]\d T.$$

This equality will hold in general if and only if the corresponding coefficients of the differentials are both equal to zero. We are interested in the coefficient of $\d T$, namely

$$\left(\frac{\partial V}{\partial T}\right)_P = -\left(\frac{\partial V}{\partial P}\right)_T\left(\frac{\partial P}{\partial T}\right)_V.$$

Assuming the partial derivative of $V$ with repect to $P$ is non-zero we can solve for $\left(\frac{\partial P}{\partial T}\right)_V$.

$$\left(\frac{\partial P}{\partial T}\right)_V = -\frac{\left(\frac{\partial V}{\partial T}\right)_P}{\left(\frac{\partial V}{\partial P}\right)_T}$$

This result is identical to what Chester Miller reached.

Hint: multiply by $1$ and apply definitions of expansion coefficients to finish the derivation.

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    $\begingroup$ Great answer! I have a question though. Why do you say that "this equality will hold in general if and only if the corresponding coefficients of the differentials are both equal to zero"? Why must the coefficients be equal to 0? $\endgroup$ – Nova Feb 13 '17 at 1:15
  • $\begingroup$ @Nova It's similar to how can e.g. $ax^2 = by$ be equal for all $x$ and all $y$. We haven't really made any assumptions about the partial derivatives, hence the LHS and RHS equality "in all cases" requires both coefficients are equal to zero. Thanks for your kind words! $\endgroup$ – Linear Christmas Feb 13 '17 at 7:09
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$$dV=\left(\frac{\partial V}{\partial T}\right)_pdT+\left(\frac{\partial V}{\partial p}\right)_Tdp$$So at constant volume dV = 0 and $$\left(\frac{\partial p}{\partial T}\right)_V=-\frac{\left(\frac{\partial V}{\partial T}\right)_p}{\left(\frac{\partial V}{\partial p}\right)_T}$$

The rest is straightforward.

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