0
$\begingroup$

I've been investigating various properties of the Gruneisen parameter and in my calculations the temperature derivative of the isentropic bulk modulus keeps coming up, i.e.

$$\left( \frac{\partial K_S}{\partial T}\right)_V.$$

Is there an intuitive way to think about this derivative or a way to write it in a different form? So far this is my thinking: $$ \left( \frac{\partial K_S}{\partial T}\right)_V = \frac{\partial}{\partial T}\left( -V\left(\frac{\partial P}{\partial V}\right)_S\right)_V \\ = V\frac{\partial}{\partial V}\left( \left(\frac{\partial P}{\partial T}\right)_V\right)_S \\ = -V\frac{\partial}{\partial V}\left( \alpha K_T\right)_S \\ = -V\alpha \left( \frac{\partial K_T}{\partial V}\right)_S + VK_T \left( \frac{\partial \alpha}{\partial V}\right)_S $$ where $\alpha$ is the thermal expansion coefficient and $K_T$ is the isothermal bulk modulus. But I think $$ \left( \frac{\partial \alpha}{\partial V}\right)_S = \frac{\partial}{\partial V}\left( \left(\frac{\partial V}{\partial T} \right)_P \right)_S = \frac{\partial}{\partial T}\left( \left(\frac{\partial V}{\partial V} \right)_S \right)_P = 0 $$ therefore $$ \left( \frac{\partial K_S}{\partial T}\right)_V = -V\alpha \left( \frac{\partial K_T}{\partial V}\right)_S. $$

But I'm not sure this helps me. If the derivative was of the isentropic bulk modulus at constant entropy or the isothermal bulk modulus at constant temperature, I feel like this would be easier to interpret as the third derivative of either the internal energy or Helmholtz free energy respectively, but I'm not sure how to deal with transforming the derivative properly without the cyclic rule.

EDIT - I was missing a factor of V. Thanks to Chester Miller for pointing it out.

Any ideas are appreciated :-)

$\endgroup$
  • $\begingroup$ I thought the bulk modulus is defined as $V\left(\frac{\partial P}{\partial V}\right)_T$ $\endgroup$ – Chet Miller Mar 12 '18 at 21:38
  • $\begingroup$ @ChesterMiller, that's the isothermal bulk modulus. The isentropic bulk modulus is $V\left(\frac{\partial P}{\partial V}\right)_S$. They're related by $K_S = K_T \gamma$ where $\gamma = \frac{C_P}{C_V}$. $\endgroup$ – JeffP Mar 12 '18 at 22:06
  • $\begingroup$ Yes, but you seem to have omitted the factor of V in your first equation $\endgroup$ – Chet Miller Mar 12 '18 at 22:10
  • $\begingroup$ @ChesterMiller, ah yes, thank you! I believe since that initial temperature derivative is at constant V though, the factor of V slides through and simply multiples everything from there out. I've corrected the question. $\endgroup$ – JeffP Mar 13 '18 at 15:43
  • $\begingroup$ Did you not like my answer? $\endgroup$ – Chet Miller Mar 13 '18 at 17:03
1
$\begingroup$

This is not an answer per se, but it's too long for a comment.

Your maneuver in your first step $$ \frac{\partial}{\partial T}\left( \left(\frac{\partial P}{\partial V}\right)_S\right)_V = \frac{\partial}{\partial V}\left( \left(\frac{\partial P}{\partial T}\right)_V\right)_S $$ is not necessarily valid. Partial derivatives only necessarily commute if you're holding variable #1 constant when you're taking the derivative with respect to variable #2, and vice versa; so it is true that $$ \frac{\partial}{\partial S}\left( \left(\frac{\partial P}{\partial V}\right)_S\right)_V = \frac{\partial}{\partial V}\left( \left(\frac{\partial P}{\partial S}\right)_V\right)_S $$ or the same thing with $T$ replaced with $S$ everywhere. But the "mixed" version is not necessarily true.

To show that this can in fact fail to be true, let $f = x^2 + x + y/x$, and let $z = xy$. We wish to take the derivatives $$ \frac{\partial}{\partial x}\left( \left(\frac{\partial f}{\partial y}\right)_z\right)_y \stackrel{?}{=} \frac{\partial}{\partial y}\left( \left(\frac{\partial f}{\partial x}\right)_y\right)_z $$ (Your first step is precisely this maneuver with $f = P$, $x = T$, $y = V$, and $z = S$.) For the left-hand side, we have $$ \left(\frac{\partial f}{\partial y}\right)_z = \frac{\partial}{\partial y} \left( \frac{z^2}{y^2} + \frac{z}{y} + \frac{y^2}{z} \right)_z \\ = -2\frac{z^2}{y^3} - \frac{z}{y^2} + \frac{2y}{z} \\ = -2 \frac{x^2}{y} - \frac{x}{y} + \frac{2}{x} $$ and so $$ \mathrm{LHS} = \frac{\partial}{\partial x} \left( -2 \frac{x^2}{y} - \frac{x}{y} + \frac{2}{x} \right)_y = - \frac{4x}{y} - \frac{1}{y} - \frac{2}{x^2}. $$ Meanwhile, for the right-hand side, we have $$ \left(\frac{\partial f}{\partial x}\right)_y = 2x + 1 - \frac{y}{x^2} \\ = \frac{2z}{y} + 1 - \frac{y^3}{z^2} $$ and so the right-hand side is $$ \mathrm{RHS} = \frac{\partial}{\partial y} \left( \frac{2z}{y} + 1 - 2 \frac{y^3}{z^2} \right)_z = - \frac{z}{y^2} + 0 - 3\frac{y^2}{z^2} = - \frac{2x}{y} + 0 - \frac{3}{x^2}. $$ Obviously, we have $\mathrm{LHS} \neq \mathrm{RHS}$; in fact, none of the three terms in this expression agree.


EDIT: To fix this, we can use the chain rule which holds so long as we're holding the same things constant in all the derivatives: $$ \frac{\partial}{\partial T}\left( \left(\frac{\partial P}{\partial V}\right)_S\right)_V = \left(\frac{\partial S}{\partial T} \right)_V \frac{\partial}{\partial S} \left( \left(\frac{\partial P}{\partial V}\right)_S\right)_V \\ = \frac{C_V}{T} \frac{\partial}{\partial V} \left( \left(\frac{\partial P}{\partial S}\right)_V\right)_S \\ = \frac{C_V}{T} \frac{\partial}{\partial V} \left( \left( \frac{\partial T}{\partial S}\right)_V \left(\frac{\partial P}{\partial T}\right)_V \right)_S \\ = \frac{C_V}{T} \frac{\partial}{\partial V} \left( \frac{T}{C_V} \left(\frac{\partial P}{\partial T}\right)_V \right)_S \\ = \frac{\partial}{\partial V} \left( \left(\frac{\partial P}{\partial T}\right)_V \right)_S + \left(\frac{\partial P}{\partial T}\right)_V \frac{\partial}{\partial V} \left( \ln \left( \frac{T}{C_V} \right) \right)_S. $$

$\endgroup$
  • $\begingroup$ Thank you for pointing out the error! This might be a problem elsewhere in my analysis. But would this error then be solved by multiplying by $C_V$ a la the chain rule? In other words is this true: $$\frac{\partial}{\partial T}\left(\left(\frac{\partial P}{\partial V}\right)_S\right)_V = \left(\frac{\partial S}{\partial T}\right)_V \frac{\partial}{\partial S}\left(\left(\frac{\partial P}{\partial V}\right)_S\right)_V = C_V \frac{\partial}{\partial S}\left(\left(\frac{\partial P}{\partial V}\right)_S\right)_V? $$ $\endgroup$ – JeffP Mar 13 '18 at 16:01
  • $\begingroup$ @JeffP: Pretty sure that $(\partial S/\partial T)_V = C_V/T$, not $C_V$. But yes, using the chain rule appropriately should allow you to perform these steps; you'd just end up with extra factors in front of each term. $\endgroup$ – Michael Seifert Mar 13 '18 at 16:07
  • $\begingroup$ Yes you're right. I was just about to edit my comment when I realized that I couldn't anymore. I'll go through the math and see where it leads me. $\endgroup$ – JeffP Mar 13 '18 at 16:08
  • $\begingroup$ After going through the math a bit more, the extra terms add quite a bit of complexity but also some needed clarity. Since you pointed out my my most egregious error, I think I'll say this answered my question. Thanks for the input! $\endgroup$ – JeffP Mar 13 '18 at 16:44
1
$\begingroup$

You have $$C_p=C_v+T\left(\frac{\partial P}{\partial T}\right)_V\left(\frac{\partial V}{\partial T}\right)_P$$ Also, $$\left(\frac{\partial V}{\partial T}\right)_P=-\frac{\left(\frac{\partial P}{\partial T}\right)_V}{\left(\frac{\partial P}{\partial V}\right)_T}$$ So, $$C_p=C_v-T\frac{\left[\left(\frac{\partial P}{\partial T}\right)_V\right]^2}{\left(\frac{\partial P}{\partial V}\right)_T}$$where P and $C_v$ are now regarded as functions of V and T. So, $$K_S=\frac{C_p}{C_v}K_T=K_T+V\frac{T}{C_v}\left[\left(\frac{\partial P}{\partial T}\right)_V\right]^2$$ The partial derivative of this with respect to T at constant V can readily be determined.

$\endgroup$
  • $\begingroup$ When I read your answer last it didn't have the expansion of $K_S$ so I felt Michael Seifert got at a core issue in my analysis. I'll need to work through the math from your formulation to see if this helps me understand the derivative better. $\endgroup$ – JeffP Mar 13 '18 at 19:17
  • $\begingroup$ At first thought, a felt like providing the relationship between Cp and Cv would be a big enough hint. Then I had second thoughts, and fleshed out the answer some more. $\endgroup$ – Chet Miller Mar 14 '18 at 1:37
  • $\begingroup$ I went through the math a bit but I think this just shifts the derivatives to other quantities and I'm not sure whether it helps. I don't have my notes with me right now but the resulting expression involves $\frac{\partial}{\partial V}\left(\alpha K_T\right)$ and $\frac{\partial}{\partial T}\left(\alpha K_T\right)$. Is there something I'm missing about those derivatives? EDIT - I can't remember what variables those derivatives are held constant with respect to but they're probably V and T. I'll check tomorrow. $\endgroup$ – JeffP Mar 15 '18 at 2:54
  • $\begingroup$ Would you like me to complete the analysis by taking the partial derivative of this equation with respect to temperature at constant volume? $\endgroup$ – Chet Miller Mar 15 '18 at 3:03
  • $\begingroup$ What is the final relationship you are trying to achieve? With all these variables present, it is difficult to guess the final relationship. Tell me that, and I'll provide the derivation of how to get there. $\endgroup$ – Chet Miller Mar 15 '18 at 12:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.