Relation between chemical potential and S,V,T,P

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

Chemical potential is defined as $\mu = Gibbs potential per particle$.

So, is the system consists of N particles, $\mu = \frac { G } {N }$.

Now, dG = VdP – SdT

$\frac {dG } { N } = \frac { VdP } { N } - \frac { SdT } { N}$

So, $d{\mu} = -sdT + v dP$.

Hence the correct option is (a).

Is this correct?

I didn't get the physical significane of this question.
How does the knowledge of above eqn help us in our study of thermodynamics?

 

[Useful nucleus]

Your answer is correct. This is the Gibbs-Duhem equation which tells us that for a simple system, the intensive parameters (T, P, μ) are not independent. Knowledge of 2 is enough to determine the 3rd up to an integration constant. Have a look at Wiki page too:
https://en.wikipedia.org/wiki/Gibbs–Duhem_equation

 

[Chestermiller]

I didn't get the physical significane of this question.
How does the knowledge of above eqn help us in our study of thermodynamics?

For the single component system you are considering, the chemical potential doesn't make much sense. But its value comes into play "big time" when you are considering multicomponent mixtures. It is used in describing multicomponent vapor-liquid equilibrium, multiphase equilibrium, and chemical reaction equilibrium. Be patient, and you will learn about this soon.

 

[Pushoam]

Thanks for the reply.
The following eqn uses extensive parameters.

dG = VdP – SdT

.
We have derived the following eqn from the above eqn. One of the significance of the following eqn is: it consists of only intensive parameters.

$d{\mu} = -sdT + v dP$