Derivation of Chemical Potential

Mbaboy
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Homework Statement


Derive the following:
<br /> \mu_i=T\left( \frac{\partial S}{\partial n_i}\right)_{U,V,n_j\not=i}<br />

\mu_i is the chemical potential of the ith component
G is the Gibbs free energy

Homework Equations


dU = TdS - PdV + \sum_i \mu_i dn_i
\mu_i = \left(\frac{\partial G}{\partial n_i}\right)_{p,T,n_j\not=i}

The Attempt at a Solution


I've tried to work it out but really haven't gotten anywhere. This is an engineering course, and up until now we haven't done anything like this or even used calculus. This is only a very small part of the question, but I figure if I can get this I'll be on the right track. I have little experience with multivariable calculus, so I guess that is my main problem. I'm sure there are other equations that could be used, but this was all that was given.

Thanks!
 
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We're dealing with partial derivatives, which we write as

\left(\frac{\partial S}{\partial n_i}\right)_{x,y,\dots}

where x, y, etc. are the variables we're holding constant; the partial derivative doesn't mean anything unless we state these constraints.

In your equation for dU, you could differentiate with respect to n_i, but you need to specify what variables are being held constant (e.g., volume).

I don't see any way of getting to \mu_i=T(\partial S/\partial n_i)_{x,y,\dots}, though. Are you sure there's not a minus sign missing somewhere?
 
Ok, so I updated the original post and defined the equations a little better.

Eventually in the problem we are suppose to prove

<br /> \mu_i=T\left( \frac{\partial S}{\partial n_i}\right)_{U,V,n_j\not=i}<br /> =\left( \frac{\partial U}{\partial n_i}\right)_{S,V,n_j\not=i}<br /> =\left( \frac{\partial A}{\partial n_i}\right)_{V,T,n_j\not=i}<br /> =\left( \frac{\partial H}{\partial n_i}\right)_{p,S,n_j\not=i}<br />
 
OK, this really just requires application of the definition of a partial derivative with certain variables held constant. Again though, check for a minus sign on that T(\partial S/\partial n_i)_{U,V,n_j\neq i} term. It's not correct as written.
 
Ok, I think I get it. Thanks a lot. And I agree there should be a negative in there.
 
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