# VdP term in definition of constant volume specific heat

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1. May 3, 2016

### rastafik

1. The problem statement, all variables and given/known data
I know that for constant volume ∂q=du and so du=Cv.dT
However i dont understand how did we get to ∂q=du by neglecting the vdP term of enthalpy
What im trying to say is, is enthalpy this ∆U+P∂V+V∂P or this ∆U+P∂V? I dont understand since the definition of enthalpy is derived out of a constant pressure volume change
And why snt specific heat at constant volume Cv=∆U+V∂P/dT instead of Cv=∆U/dT?

2. Relevant equations

3. The attempt at a solution
From what I see:
∂Qnet,in=∆U+∂Wnet,out
∂Qnet,in=∆U+P∂V+V∂P

C=∂Qnet,in/dT

For constant pressure:
Cp=∆U+P∂V/dT assuming ∆U+P∂V is enthalpy then ∆H=Cp.∆P

For constant volume:
Cv=∆U+V∂P/dT but all the books say it actually is just ∆U/dT
where did the VdP term go? if we add heat to a fixed volume won't its pressure increase and so VdP would be relevant?

2. May 3, 2016

### Staff: Mentor

dW = PdV, not d(PV).