# Clarification on derivations of specific heats in fluids

1. Nov 22, 2015

### 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 CV, dV = 0 and the relationship becomes

(∂U/∂T)V = (∂Q/∂T)V = CV

For the derivation of CP, 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 = CP

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!

2. Nov 22, 2015

### 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 = Cp

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.

3. Nov 22, 2015

### Kori Smith

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.

4. Nov 22, 2015

Absolutely.