Chemical Potential Homework: Ideal Monatomic Gas

[Crush1986]

Homework Statement

I just have a question about chemical potential for ideal monatomic gas. I see that by definition \mu = \frac{\partial U}{\partial N}

Homework Equations

\mu = \frac{\partial U}{\partial N}

The Attempt at a Solution

I was wondering why it is wrong to use U=3/2NkT take the partial with respect to N and get \mu = 3/2kT.

I know this isn't right, but what is exactly wrong with it?

The correct equation for the chemical potential of a monatomic ideal gas by the way is
\mu = -kT \ln({\frac{V}{N}(\frac{4 \pi mU}{3h^2}})^\frac{3}{2})[/B]

 

[Crush1986]

I'm thinking there is something to do with the variables that need to be held fixed, they are entropy and volume. Does 3/2NkT somehow not keep these values fixed maybe?

 

[DrClaude]

I'm thinking there is something to do with the variables that need to be held fixed, they are entropy and volume. Does 3/2NkT somehow not keep these values fixed maybe?

Indeed. You need to start from the thermodynamic identity
$$
dU = T dS - PdV + \mu dN
$$
from which you will see that
$$
\mu = \left( \frac{\partial U}{\partial N} \right)_{S,V}
$$

 

[Crush1986]

Indeed. You need to start from the thermodynamic identity
$$
dU = T dS - PdV + \mu dN
$$
from which you will see that
$$
\mu = \left( \frac{\partial U}{\partial N} \right)_{S,V}
$$

Yes, I most definitely see that.

My biggest issue right now is why is it not correct to use U=\frac {f}{2}NkT take that partial with respect to N and obtain an answer? My thought is that this equation doesn't hold S and V constant inherently, but how would I see that? Just that fact that T is in the equation maybe?

 

[DrClaude]

Yes, you would have a term in $\partial T / \partial N$ in there.

 

[Crush1986]

Yes, you would have a term in $\partial T / \partial N$ in there.

Thank you very much. Little simple things like this keep holding me up a little. Hopefully my math skills tighten up in these last few years of my undergraduate studies.

 

[DrClaude]

You can take comfort in the fact that you had the right intuition as to why simply taking the derivative of U didn't work. The more you use math, the better your skills.