Question

How do you derive this relation for enthalpy using this equation for internal pressure?

Answer 1

Well, we can derive it, but the final result we are "shooting for" is incorrect. It ought to give zero for ideal gases, but right now it won't.

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We apparently have to begin with:

`((delU)/(delV))_T = T((delP)/(delT))_V - P`

Split the left-hand side:

`((delU)/(delP))_T((delP)/(delV))_T = T((delP)/(delT))_V - P`

Dividing the pressure derivative over, we obtain:

`((delU)/(delP))_T = [T((delP)/(delT))_V - P] ((delV)/(delP))_T`

Use the definition that `H = U + PV` to get:

`((delH)/(delP))_T - ((del(PV))/(delP))_T = [T((delP)/(delT))_V - P] ((delV)/(delP))_T`

The left-hand side's second derivative uses the product rule to give:

`((del(PV))/(delP))_T = P((delV)/(delP))_T + V((delP)/(delP))_T`

`= P((delV)/(delP))_T + V`

Therefore, we get:

`((delH)/(delP))_T - cancel(P((delV)/(delP))_T) - V = [T((delP)/(delT))_V - cancel(P)] ((delV)/(delP))_T`

As a result, add the volume over to get:

`((delH)/(delP))_T = V + T((delP)/(delT))_V ((delV)/(delP))_T`

There is a cyclic relation that

`-1 = ((delx)/(dely))_z((dely)/(delz))_x((delz)/(delx))_y`

So if `x = V`, `y = T`, and `z = P`, then:

`-1 = ((delV)/(delT))_P ((delT)/(delP))_V ((delP)/(delV))_T`

That means:

`((delV)/(delT))_P = -1/[((delT)/(delP))_V ((delP)/(delV))_T]`

`= -((delP)/(delT))_V ((delV)/(delP))_T`

Therefore:

`color(blue)(((delH)/(delP))_T = - T((delV)/(delT))_P + V)`

I've also derived this in another way, so we know this is correct.