Clarification on derivations of specific heats in fluids

[Kori Smith]

Hello! I understand what specific heats are and how to derive them. I just feel that I'm missing a little something in the methodology.

Consider the 1st law of thermodynamics and the definition of enthalpy:

1) dU = δQ -δW = δQ - PdV
2) H = Q - VP

For the derivation of C_V, dV = 0 and the relationship becomes

(∂U/∂T)_V = (∂Q/∂T)_V = C_V

For the derivation of C_P, something happens that I don't quite understand. Sources I've found say that the incremental form of the enthalpy relation is given as

3) dH = δQ - VdP

since dP = 0, it becomes

(∂U/∂T)_P = (∂Q/∂T)_P = C_P

but why do we write it like this? Wouldn't the chain rule for differentiation apply to d(VP) s.t. it becomes

d(VP) = VdP + PdV

in which case, where does the PdV component in EQ (3) go? Thanks ahead of time for the clarification!

 

[Chandra Prayaga]

The definition of enthalpy is not correct. Enthalpy is a state function, like U, T, V or p. It is defined by:
H = U + pV. Then instead of your equ. (3):

dH = dU + pdV + Vdp = δQ + Vdp, so that, for a constant pressure process, dp = 0, and
dH = δQ,
(∂H/∂T)_p = (∂Q/∂T)_p = C_p

The problem with the definition of enthalpy that you gave is that there is no state function called Q, so you cannot define any state function in terms of something called Q.

 

[Kori Smith]

The definition of enthalpy is not correct. Enthalpy is a state function, like U, T, V or p. It is defined by:
H = U + pV. Then instead of your equ. (3):

dH = dU + pdV + Vdp = δQ + Vdp, so that, for a constant pressure process, dp = 0, and
dH = δQ,
(∂H/∂T)_p = (∂Q/∂T)_p = C_p

The problem with the definition of enthalpy that you gave is that there is no state function called Q, so you cannot define any state function in terms of something called Q.

Woops! I made a typo. As usual, my issue boils down to simple errors. When I look at it again, I see that

dH = dU + pdV + Vdp = δQ - δW +pdV + Vdp = δQ + Vdp + (pdV - δW) = δQ + Vdp

since δW = pdV (pressure-volume work). Which is exactly as it should be and explains where the missing pdV went.

 

[Chandra Prayaga]

Absolutely.