# Please help me to derive that #((delS)/(delT))_P = C_P/T# and #((delS)/(delT))_V = C_V/T#?

##### 1 Answer

Okay, so I will assume you know or can figure out the Maxwell relations. Those are a common starting point. Furthermore, you have by definition that

Recall that

#dG = dH - d(TS)#

#= dH - SdT - TdS#

(use product rule)

Now, for the first derivation, you can divide through by

#((delG)/(delT))_P = ((delH)/(delT))_P - cancel(S((delT)/(delT))_P)^(1) - T((delS)/(delT))_P#

where the third term goes to

#1# because#T# does not change with respect to#T# ; it is one-to-one with itself.

By definition,

Next, recall that for the **natural variables**

#dG = -SdT + VdP# #" "" "" "bb((3))#

We do not yet know what

#((delG)/(delT))_P = -S# #" "" "" "bb((4))#

Therefore. plugging

#cancel(-S) = C_P - cancel(S) - T((delS)/(delT))_P#

It follows that if

A similar process follows for deriving

#A = U - TS# ,

take the derivative to get:

#dA = dU - SdT - TdS#

Almost like before, divide by

#((delA)/(delT))_V = ((delU)/(delT))_V - Scancel(((delT)/(delT))_V)^(1) - T((delS)/(delT))_V#

We know by definition that **natural variables**

So from the Maxwell relation:

#dA = -SdT - PdV#

we have that:

#((delA)/(delT))_V = -S# ,

so plugging back into the main equation, we get:

#cancel(-S) = C_V - cancel(S) - T((delS)/(delT))_V#

Since