- #1
knut-o
- 17
- 0
Homework Statement
I know that Gibbs free energy: G=H-TS, and therefor [tex]\frac{dG}{dT}=-S[/tex] , and that [tex]dS=C_P\frac{dT}{T}[/tex]
Now, I am to show that more generally, [tex]dS=C_P\frac{dT}{T}-\frac{dV}{dT}dP(1)[/tex]
(assuming that the difference between delta and d is mostly the same (symbolwise). The hint I have is using a maxwell relation.
And at last, show that [tex]S(T,P)=S(T_i,P)+C_Pln(\frac{T}{T_i})[/tex];
[tex]G(T,P)=G(T_i,P)-S(T_i,P)(T-T_i)+C_P(T-T_i)-C_PTln(\frac{T}{T_i}[/tex]
Where subindex i represent initial value > 0.
Homework Equations
Seeing the solution, I assume the maxwell relation I need is [tex]-\frac{dS}{dP}=\frac{dV}{dT}(2)[/tex]
The Attempt at a Solution
The problem is how do isolate dS alone, generally combining (1) and (2), unless I can simply just call them dS' and add it to both sides and say that dS=2dS' ? Otherwise I am only able eliminate dS from the equation, but that doesn't really help me :3
And for the integrationpart, the [tex]C_P\frac{dT}{T}[/tex] from [tex]T_i[/tex] to T is simple enough, but the other link, integrate from [tex]P_i[/tex] to P and just call it [tex]S(T_i,P)[/tex]? Doesn't sound intuitive to me.
And then show the G(T,P)-part, woa, I can accept the 2nd part (ST basically), and the last one. And to some degree the first part, but to argue for the first part I am unsure except that it must be there.. And the third part, makes no sense what it does there :( .
Halp please :3