Difference between isobar and isochor heat capacity

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The discussion focuses on the derivation of the relationship between isobaric heat capacity (C_p) and isochoric heat capacity (C_v). It begins with the equation C_p - C_v = -T (∂V/∂p)_{T,n} (∂p/∂T)_{V,n} and explores the definitions of enthalpy and internal energy to manipulate the equations. Participants clarify the steps involved in transitioning from the combined first and second laws of thermodynamics to derive the standard relation for C_p - C_v. There is also a discussion on the significance of conducting an isobaric process and how it relates to the derivation. The conversation concludes with a correction of a typo in the equations presented.
Alexis21
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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 :)
 
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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
 
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.
 
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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