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

##### 1 Answer

Okay, considering that

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

Now, if we divide by

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

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

The bottom of the derivative shows **natural variables** of the Gibbs' free energy, so from the Maxwell relation:

#dG = -SdT + VdP# ,

we can either recall the **cyclic relation**

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

showing that

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

meaning that

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

Therefore, from

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

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

Finally, recall that

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