Thermal physics (Gibbs free energy mostly)

[knut-o]

Homework Statement

I know that Gibbs free energy: G=H-TS, and therefor \frac{dG}{dT}=-S , and that dS=C_P\frac{dT}{T}

Now, I am to show that more generally, dS=C_P\frac{dT}{T}-\frac{dV}{dT}dP(1)
(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 S(T,P)=S(T_i,P)+C_Pln(\frac{T}{T_i});
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}
Where subindex i represent initial value > 0.

Homework Equations

Seeing the solution, I assume the maxwell relation I need is -\frac{dS}{dP}=\frac{dV}{dT}(2)

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 C_P\frac{dT}{T} from T_i to T is simple enough, but the other link, integrate from P_i to P and just call it S(T_i,P)? 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

 

[brollysan]

You must take the same class as me haha!

The first part I cannot help with as I barely know latex and I need partial derivatives, just know that if S= S(a,b) then dS = partial S with a at constant b + partial s with b at constant a

For the second part, since you are using Cp you are operating at constant pressure. So dP from (1) disappears. All you have to do now is integrate, with constant pressure. Also remember dN = 0 for calculating dG

 

[knut-o]

You must take the same class as me haha!

The first part I cannot help with as I barely know latex and I need partial derivatives, just know that if S= S(a,b) then dS = partial S with a at constant b + partial s with b at constant a

For the second part, since you are using Cp you are operating at constant pressure. So dP from (1) disappears. All you have to do now is integrate, with constant pressure. Also remember dN = 0 for calculating dG

Not impossible, :p . So confusing!

But I am supposed to show that (1) is true,
And if I put dP to 0, then the integration becomes the integral of ? Makes no sense :s . I assume then it becomes an integration constant that I somehow need to show is S(T_i,P)?

 

[brollysan]

For the second part, the integration becomes an integral of temperature. You will get dS = Cp 1/T dt. And yes it is an integration constant, we are operating at constant pressure but different temperatures, so it must have an entropy at its initial temperature Ti: S(Ti,P)

The final enthropy is therefore its initial entropy + any change dS integrated over the temperature range.

 

[knut-o]

So I end up with something like integrating dS from S_i to S(T,P) which gives me that S(T,P) - S_i=C_Pln(T/T_i)+C and just say that C=0 and move S_i over?

And how would it work to calculate \Delta G and \Delta S for difference between liquid water and water vapoir? I know I have C_P, T_i, but what else, what about T? and G(T_i,P)?
I am confusos.