Difference between isobar and isochor heat capacity

[Alexis21]

Hello,

I want to show:
C_p - C_v = -T \big( \frac {\partial V}{\partial p} \big)_{T,n} \big( \frac {\partial p}{\partial T} \big)_{V,n}^2

I started by doing this:
C_p - C_v= \big( \frac {\partial H}{\partial T} \big)_{p,N} - \big( \frac {\partial U}{\partial T} \big)_{V,n}

Applying the definitions of enthalpy and energy:
dH = TdS + V dp + \mu dn
and
dU = TdS - p dV + \mu dn

I can rewrite the equation like this:
= V \big(\frac {\partial p}{\partial T} \big) + p \big( \frac {\partial V}{\partial T} \big)
(while TdS and µdn terms cancel out each other)

Now I do not know how to continue. Can anyone help :)

 

[Chandra Prayaga]

Start with:

T dS = dU + p dV (Combined 1st and 2nd laws) (1)

Write U as a function of T and V:

T dS = [(∂U/∂T)_V dT + (∂U/∂V)_T dV] + p dV

= [(∂U/∂T)_V + p ] dV + (∂U/∂T)_V dT

The last term is C_V dT, so:

T dS = [(∂U/∂T)_V + p ] dV + C_V dT

Now do an isobaric process, and divide by dT to get:

C_P - C_V = [(∂U/∂T)_V + p ] dV (∂V/∂T)_P (2)


which is a standard relation derived, for example, in Sears

Now, starting again with the combined 1st and 2nd law (equ (1)):

dU + p dV = T dS

Divide by dV keeping T constant :

( ∂U/∂V)_T + p = T (∂S/∂V)_T (3)

Substituting this for the term in square brackets in equ (2)

C_P - C_V = T (∂S/∂V)_T (∂V/∂T)_P (4)

Use one of Maxwell's equations (Ref: Sears) to substitute for the first partial derivative on the right of equ (4):

C_P - C_V = T (∂P/∂T)_V (∂V/∂T)_P (5)

Now use the standard relation for partial derivatives (also ref: Sears)

(∂V/∂T)_P (∂T/∂P)_V (∂P/∂V)_T = -1

to substitute for the second partial derivative on the right of equ (5) and get what you need.

Please let me know if you need clarification

 

[Alexis21]

Thank you for your answer :)

I have got a question on that:
What does it exactly mean when you say 'Now do an isobaric process'. I don't see how the C_p drops in.

 

[Chandra Prayaga]

So let us start with the equation before that line:


T dS = [(∂U/∂T)_V + p] dV + C_V dT

In an isobaric process, each of the changes dS, dV and dT will have certain values. So when we divide by that value of dT, we get:

T (∂S/∂T)_p = [(∂U/∂T)_V + p] (∂V/∂T)_p + C_V

The constant p on the partial derivatives signifying the isobaric process.

The left hand side is precisely C_p. Now take the C_V to the left and you get

C_p - C_V = [(∂U/∂T)_V + p] (∂V/∂T)_p

This is equation (2) in what I wrote earlier. Incidentally, there was a typo in the earlier equation (2). There is a dV extra which should be erased.

Let me know if you need any more help