Difference between isobar and isochor heat capacity

  • Thread starter Alexis21
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  • #1
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Hello,

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

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

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

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

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

Answers and Replies

  • #2
Chandra Prayaga
Science Advisor
650
149
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 CV dT, so:

T dS = [(∂U/∂T)V + p ] dV + CV dT

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

CP - CV = [(∂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)

CP - CV = 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):

CP - CV = 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
 
  • #3
6
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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.
 
  • #4
Chandra Prayaga
Science Advisor
650
149
So let us start with the equation before that line:


T dS = [(∂U/∂T)V + p] dV + CV 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 + CV

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

The left hand side is precisely Cp. Now take the CV to the left and you get

Cp - CV = [(∂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
 

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