How to Derive the Joule Coefficient at Constant Internal Energy?

[JorgeMC59]

Homework Statement

Using basic thermodynamic relations, show that the Joule Coefficient is given by:

Homework Equations

$${\left(\partial T \over \partial V \right)_U}=-{T^2 \over C _v}{\partial \over \partial T} \left(P \over T \right)_V$$

The Attempt at a Solution

I started with the cyclic thermodynamic relation:
$${\left( \partial T \over \partial V \right)_U \left( \partial V \over \partial U \right)_T \left (\partial U \over \partial T \right)_V}=-1$$
Rearranging the equation:
$${\left( \partial T \over \partial V \right)_U}={-\left( \partial U \over \partial V \right)_T \left( \partial T \over \partial U \right)_V}$$
Knowing that: $C_V= \left( \partial U \over \partial T \right)_V$ → ${1 \over C_V}= \left( \partial T \over \partial U \right)_V$ I get:
$${\left( \partial T \over \partial V \right)_U}={-{1 \over C_V} \left( \partial U \over \partial V \right)_T}$$
And from there I don't know how to continue.
I hope someone here can help me, thanks in advance.

 

[Chestermiller]

Homework Statement

Using basic thermodynamic relations, show that the Joule Coefficient is given by:

Homework Equations

$${\left(\partial T \over \partial V \right)_U}=-{T^2 \over C _v}{\partial \over \partial T} \left(P \over T \right)_V$$

The Attempt at a Solution

I started with the cyclic thermodynamic relation:
$${\left( \partial T \over \partial V \right)_U \left( \partial V \over \partial U \right)_T \left (\partial U \over \partial T \right)_V}=-1$$
Rearranging the equation:
$${\left( \partial T \over \partial V \right)_U}={-\left( \partial U \over \partial V \right)_T \left( \partial T \over \partial U \right)_V}$$
Knowing that: $C_V= \left( \partial U \over \partial T \right)_V$ → ${1 \over C_V}= \left( \partial T \over \partial U \right)_V$ I get:
$${\left( \partial T \over \partial V \right)_U}={-{1 \over C_V} \left( \partial U \over \partial V \right)_T}$$
And from there I don't know how to continue.
I hope someone here can help me, thanks in advance.

You need to get (∂U/∂V)_T. Start out with

dU = TdS-PdV

so

$$dU=T\left(\left(\frac{∂S}{∂T}\right)_VdT+\left(\frac{∂S}{∂V}\right)_TdV\right)-PdV=C_vdT+\left(T\left(\frac{∂S}{∂V}\right)_T-P\right)dV$$
So,
$$\left(\frac{∂U}{∂V}\right)_T=\left(T\left(\frac{∂S}{∂V}\right)_T-P\right)dV$$

Now, you need to make use of an appropriate Maxwell relation to get the partial of S with respect to V at constant T.

Chet

 

[JorgeMC59]

You need to get (∂U/∂V)_T. Start out with

dU = TdS-PdV

so

$$dU=T\left(\left(\frac{∂S}{∂T}\right)_VdT+\left(\frac{∂S}{∂V}\right)_TdV\right)-PdV=C_vdT+\left(T\left(\frac{∂S}{∂V}\right)_T-P\right)dV$$
So,
$$\left(\frac{∂U}{∂V}\right)_T=\left(T\left(\frac{∂S}{∂V}\right)_T-P\right)dV$$

Now, you need to make use of an appropriate Maxwell relation to get the partial of S with respect to V at constant T.

Chet

Thanks for your answer Chet, I used $${\left( \partial S \over \partial V \right)_T}= {\left( \partial P \over \partial T \right)_V}$$

But now I'm stuck here: ${\left(\partial T \over \partial V \right)_U}=-{1 \over C_V}{\left[ T {\left( \partial P \over \partial T \right)_V}-P \right]dV}$

I suppose that somehow $${\left[ T {\left( \partial P \over \partial T \right)_V}-P \right]dV}={T^2{\partial \over \partial T}\left( P \over T \right)}$$ But I don't know how to get there.

 

[DrClaude]

But now I'm stuck here: ${\left(\partial T \over \partial V \right)_U}=-{1 \over C_V}{\left[ T {\left( \partial P \over \partial T \right)_V}-P \right]dV}$

Note that there is a typo in Chet's post, such that there is no $dV$ here.

I suppose that somehow $${\left[ T {\left( \partial P \over \partial T \right)_V}-P \right]dV}={T^2{\partial \over \partial T}\left( P \over T \right)}$$ But I don't know how to get there.

Correcting, you have to show that
$$\left[ T {\left( \partial P \over \partial T \right)_V}-P \right] = T^2{\partial \over \partial T}\left( P \over T \right)_V$$
Just apply the product rule on the right-hand-side.

 

[Chestermiller]

Note that there is a typo in Chet's post, such that there is no $dV$ here.


Correcting, you have to show that
$$\left[ T {\left( \partial P \over \partial T \right)_V}-P \right] = T^2{\partial \over \partial T}\left( P \over T \right)_V$$
Just apply the product rule on the right-hand-side.

Thanks for spotting that typo Dr. C.

Chet

 

[JorgeMC59]

Thanks a lot, DrClaude and Chet your answers were of great help.