Thermodynamics: proof about free expansion

[A_B]

Homework Statement

prove that for a free expansion the following equality holds:

$$\left( \frac{\partial T}{\partial V} \right) _{U} = - \frac{1}{C_{V}} \left[ T \left(\frac{\partial P}{\partial T} \right)_{V} - P \right]$$

Homework Equations

thermodynamic potentials, maxwell equations etc.

The Attempt at a Solution

I've tried playing around with the equations a little, but didn't get anywhere. I thought that $\left( \frac{\partial T}{\partial V} \right)_{U}$ should be zero, because if the energy remains constant, so should the temperature.

 

[Mapes]

I thought that $\left( \frac{\partial T}{\partial V} \right)_{U}$ should be zero, because if the energy remains constant, so should the temperature.

Only for an ideal gas.

A good start to this problem would be to use the identity

$$\left(\frac{\partial x}{\partial y}\right)_z \left(\frac{\partial y}{\partial z}\right)_x \left(\frac{\partial z}{\partial x}\right)_y=-1$$

and get that constant U constraint out of there.

 

[A_B]

OK, this is what I got so far:

$$\begin{align*}<br /> <br /> \left( \frac{\partial T}{\partial V} \right)_{U} &= \left[ \left( \frac{\partial T}{\partial V} \right)_{U}<br /> \left( \frac{\partial V}{\partial U} \right)_{T} \left( \frac{\partial U}{\partial T} \right)_{V} \right] / \left[ \left( \frac{\partial V}{\partial U} \right)_{T} \left( \frac{\partial U}{\partial T} \right)_{V} \right]<br /> <br /> \\ &= -1 / \left[ \left( \frac{\partial V}{\partial U} \right)_{T} C_{V} \right]<br /> <br /> \\ &= - \frac{1}{C_{V}} \left[ \frac{1}{ \left( \frac{\partial V}{\partial U} \right)_{T}} \right]<br /> <br /> \end{align*}$$

am I on the right track?

 

[A_B]

ok, i finished the problem, it continues:

$$\begin{align*}<br /> &= - \frac{1}{C_{V}} \left[ \frac{1}{ \left( \frac{\partial V}{\partial U} \right)_{T}} \right]<br /> \\ &= - \frac{1}{C_{V}} \left[ \left( \frac{\partial U}{\partial V} \right)_{T}\right]<br /> \\ &= - \frac{1}{C_{V}} \left[ \left( \frac{ \partial Q}{\partial V} \right)_{T} + \left( \frac{ \partial W}{\partial V} \right)_{T} \right]<br /> \\ &= - \frac{1}{C_{V}} \left[ T \left( \frac{ \partial S}{\partial V}\right)_{T} - P \left( \frac{ \partial V}{\partial V} \right)_{T} \right]<br /> \\ &= - \frac{1}{C_{V}} \left[ T \left( \frac{\partial S}{\partial V} \right)_{T} - P \right]<br /> \end{align*}$$

Finally, using the maxwell relation $\left( \frac{ \partial P}{\partial T} \right)_{V} = \left( \frac{ \partial S}{\partial V} \right)_{T}$ we obtain the desired relation:

$$\left( \frac{\partial T}{\partial V} \right) _{U} = - \frac{1}{C_{V}} \left[ T \left(\frac{\partial P}{\partial T} \right)_{V} - P \right]$$


The things I'm still having some trouble with is first of all the identity
$$\left(\frac{\partial x}{\partial y}\right)_z \left(\frac{\partial y}{\partial z}\right)_x \left(\frac{\partial z}{\partial x}\right)_y=-1$$
Any reference about this?

Also , it seems very plausible that
$$\frac{1}{ \left( \frac{\partial V}{\partial U} \right)_{T}} = \frac{\partial U}{\partial V} \right)_{T}$$
But I'm not 100% sure why this works.

These are both calculus questions i suppose, so they're in the wrong place here.


Thanks

Alex

 

[Mapes]

These are definitely useful identities when manipulating partial derivatives. They're called the triple product rule and the inverse function rule, respectively, if you want to search for more information.