Question

Please help show that `((delT)/(delP))_S = (ValphaT)/C_P`?

Answer 1

Okay, considering that `((delT)/(delP))_S` contains `S`, `T`, and `P`, I would start from `S = S(T,P)` and take the total derivative:

`dS = ((delS)/(delT))_PdT + ((delS)/(delP))_TdP` `" "" "bb((1))`

Now, if we divide by `delP` at a constant `S`, we'd get:

`cancel(((delS)/(delP))_S)^(0) = ((delS)/(delT))_P((delT)/(delP))_S + ((delS)/(delP))_Tcancel(((delP)/(delP))_S)^(1)`

`" "" "bb((2))`

since a constant `S` forces a derivative of `S` with respect to any variable to go to `0`, and `(dP)/(dP) = 1` no matter what else is held constant.

From this question, we can refer back and recall that

`((delS)/(delT))_P = 1/T ((delH)/(delP))_T = C_P/T`.

`bb((3))`

Now we need to figure out what `((delS)/(delP))_T` is.

The bottom of the derivative shows `T` and `P`, which are the natural variables of the Gibbs' free energy, so from the Maxwell relation:

`dG = -SdT + VdP`,

we can either recall the cyclic relation `((delS)/(delP))_T = -((delV)/(delT))_P`, or rederive it. Again, from the total derivative of `G(T,P)`:

`dG = ((delG)/(delT))_PdT + ((delG)/(delP))_TdP`,

showing that `-S = ((delG)/(delT))_P` and `V = ((delG)/(delP))_T`. Since the order of second partial differentiation doesn't matter,

`((del^2G)/(delPdelT))_(P,T) = ((del^2G)/(delTdelP))_(T,P)`,

meaning that

`-((delS)/(delP))_T = ((delV)/(delT))_P`. `" "" "bb((4))`

Therefore, from `bb((2))`, subtract over `((delS)/(delP))_T`, plug in `bb((3))` and `bb((4))`, and we have:

`((delV)/(delT))_P = C_P/T((delT)/(delP))_S`

`=> ((delT)/(delP))_S = T/C_P ((delV)/(delT))_P`

Finally, recall that `alpha = 1/V((delV)/(delT))_P`, so that we have:

`=> color(blue)(((delT)/(delP))_S = (ValphaT)/C_P)`