Show that (du/dv)t=T(dp/dT)v-p - please explain

[ConstantinL]

Homework Statement

Show that (du/dv)_T = T(dp/dt)_v - p

Homework Equations

Using Tds = du + pdv and a Maxwell relation

The Attempt at a Solution

I've solved the problem, but I'm not entirely sure my method is correct.

Tds = du + pdv ---> du = Tds - Pdv

- Using dF=(dF/dx)_ydx +(dF/dy)_xdy
du=(du/dT)_v+(du/dv)_Tdv

- Therefore Tds - Pdv = (du/dT)_v+(du/dv)_Tdv

- Divide by dv:
(du/dT)_vdT/dv + (du/dv)_T = T(ds/dv)_T - p

Now, to get the right answer, this term:

(du/dT)_vdT/dv

must equal zero, but I'm not sure why - please can somebody explain?

Then you simply insert Maxwell relation -(ds/dv)_T = -(dp/dT)_v
and rearrange to get the correct answer.

Many thanks for any help!

 

[Chestermiller]

You should have started out by substituting $$dS=\left(\frac{\partial S}{\partial T}\right)_VdT+\left(\frac{\partial S}{\partial V}\right)_TdV$$

 

[ConstantinL]

You should have started out by substituting $$dS=\left(\frac{\partial S}{\partial T}\right)_VdT+\left(\frac{\partial S}{\partial V}\right)_TdV$$

Thanks for your help. I still get to a similar problem unfortunately. I get to here:

(ds/dT)_vdT/dv + (ds/dv)_T = (du/dv)_T1/T + p/T

How do I get ride of the (ds/dT)_vdT/dv term?

Many thanks!

 

[Chestermiller]

Thanks for your help. I still get to a similar problem unfortunately. I get to here:

(ds/dT)_vdT/dv + (ds/dv)_T = (du/dv)_T1/T + p/T

How do I get ride of the (ds/dT)_vdT/dv term?

Many thanks!

T(ds/dT)_vdT + T(ds/dv)_TdV = (du/dv)_TdV+(du/dT)_VdT + pdV
Collect factors of dV and dT.