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

  • #1

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!
 

Answers and Replies

  • #2
20,844
4,543
You should have started out by substituting $$dS=\left(\frac{\partial S}{\partial T}\right)_VdT+\left(\frac{\partial S}{\partial V}\right)_TdV$$
 
  • #3
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!
 
  • #4
20,844
4,543
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.
 

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