# Chemical Potential

1. Oct 20, 2015

### Crush1986

1. The problem statement, all variables and given/known data
I just have a question about chemical potential for ideal monatomic gas. I see that by definition $$\mu = \frac{\partial U}{\partial N}$$

2. Relevant equations
$$\mu = \frac{\partial U}{\partial N}$$

3. 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
$$\mu = -kT \ln({\frac{V}{N}(\frac{4 \pi mU}{3h^2}})^\frac{3}{2})$$

Last edited: Oct 20, 2015
2. Oct 20, 2015

### 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?

3. Oct 21, 2015

### Staff: Mentor

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

4. Oct 21, 2015

### Crush1986

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?

5. Oct 21, 2015

### Staff: Mentor

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

6. Oct 21, 2015

### Crush1986

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.

7. Oct 21, 2015

### Staff: Mentor

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.