Gibbs Free Energy -- Connection between V, P, N & T

In summary, the given expression for Gibbs free energy is used to obtain expressions for entropy (S), the equation of state (V = kTN/P), and internal energy (U). The equation of state is derived from the relationship dG = -SdT + VdP.
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cake-jake1
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Homework Statement


For a Particular system the following expression for Gibbs free energy is known:

G = -kTN ln (a T^(5/2) / P)

where a is a constant (whose dimensions make the argument of the logarithm dimensionless). Obtain expressions for

a) The entropy, S
b) The connection between V, P, N and T (Called the equation of state)
c) The Internal energy U

Homework Equations


a) For entropy it is known that S = - (dG/dT)p where p denotes constant pressure, I understand that this gives a derivative and I believe I have this section correct.

b) Main confusion comes from where to begin but would assume it is is PV=NKT equation.

c) calculating internal energy. It is known that G = U -TS + PV

The Attempt at a Solution


Nothing besides the problem statement was provided in the question and my main issue is with calculating part b.
As far as I am aware there are many equations of state, but have read here that V= (dG/dP) due to the relationship dG = -SdT + VdP.
Does this differential for V supply me with the correct answer?

I realize it seems that I may have answered my own question, however I just wish to be certain as its a university exam question with little guidance in our supplied texts.
 
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  • #2
Well, what is the value of [itex]V[/itex] you get from [itex]V = \frac{dG}{dP}[/itex]?
 
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  • #3
I ended up getting a Value of V = kTN/P
 
  • #4
cake-jake1 said:
I ended up getting a Value of V = kTN/P

And that's exactly right.
 

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