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We're learning the relationship between Cp and Cv. We seem to be doing it a little uniquely because we're looking for equations of Cp - Cv in terms of P, T, V, and U, and in terms of P, T, V, and H

For the first one, which was also isobaric, this was the process:

H = U + PV

dH = dU + d(PV)

dH = dU + PdV + VdP 0

dH = dU + PdV

(∂H/∂T)P = (∂U/∂T)P + P(∂V/∂T)P

Cp = (∂U/∂T)P + P(∂V/∂T)P [ EQN 1 ]

dU = CvdT + (∂U/∂V)TdV

(∂U/∂T)P = Cv (∂T/∂T)P 1 + (∂U/∂V)T(∂V/∂T)P

(∂U/∂T)P = Cv + (∂U/∂V)T(∂V/∂T)P [ EQN 2 ]

Plug EQN 2 into EQN 1 to get:

Cp = Cv + (∂U/∂V)T(∂V/∂T)P + P(∂V/∂T)P

Cp - Cv = (∂U/∂V)T(∂V/∂T)P + P(∂V/∂T)P

Cp - Cv = (∂V/∂T)P[ (∂U/∂V)T + P]

Now I'm supposed to derive an equation of Cp - Cv at isochoric conditions in terms of P, T, V, and H but I'm just completely lost in a lot of parts.

Supposedly the end equation is as follows:

Cp - Cv = (∂P/∂T)v [V-(∂H/∂P)T]

I tried starting from H = U + PV and got Cv = (∂H/∂T)V - V(∂P/∂T)V before realizing that's a dead-end.

I'm sure I'll need to start off from H = U + PV but that's as far as I can imagine. Even just where to start off in will be helpful to me. I'm just super stuck.

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    $\begingroup$ Are you sure that is the equation you are supposed to end up with? $\endgroup$ Commented Nov 1 at 11:07
  • $\begingroup$ Yes. That was what our professor gave us to derive on our own. $\endgroup$ Commented Nov 2 at 5:00
  • $\begingroup$ Ah actually @ChetMiller I made a mistake on the terms. It's supposed to be in terms of P,T, V, and H in Isochoric conditions. Sorry for that. $\endgroup$ Commented Nov 2 at 5:43

1 Answer 1

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I don't know about the isochoric constraint, but in general, $$dH=C_pdT+\left(\frac{\partial H}{\partial P}\right)_TdP$$so,$$dU=dH-VdP-PdV$$$$dU=C_pdT+\left[\left(\frac{\partial H}{\partial P}\right)_T-V\right]dP-PdV$$ So, $$C_v=C_p+\left[\left(\frac{\partial H}{\partial P}\right)_T-V\right]\left(\frac{\partial P}{\partial T}\right)_V$$

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