What is the proof for (∂u/∂T)_P=c_P – Pβv?

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SUMMARY

The proof for the equation (∂u/∂T)_P = c_P – Pβv is established using the first law of thermodynamics and properties of ideal gases. The derivation begins with the expression d'Q = dh - vdP, followed by expanding d'Q and du in terms of dP and dT. For ideal gases, the relationship (∂u/∂v)_T = (∂h/∂P)_T = 0 simplifies the proof, confirming that internal energy U is solely a function of temperature T.

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Homework Statement



To prove that (∂u/∂T)_P=c_P – Pβv where _P =>P constant;β=>co-eff. of vol exp.

Homework Equations


The Attempt at a Solution



I proved it for ideal gases.
Write d'Q=dh-vdP
Now expand d'Q with 1st law and du(in 1st law) in terms of dP and dT.Since du is a total differential.
Expand dh as total differential in dP and dT.

Now I used the property of ideal gases:(∂u/∂v)_T=(∂h/∂P)_T=0
The rest is a bit manipulation.

Can anyone say how to prove this in general?
The book does not mention specifically that "use ideal behaviour" or so.
There must be some way to get it.
Please help.
 
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Use the first law.
Since [tex]dU = \delta Q - PdV[/tex], take the partial derivative wrt to T at constant pressure, and you get the answer.

For ideal gases, it's even easier, since U is only a function of T
 
Thank you...
Earlier I did not get this...
 

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