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

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)#