# What is wrong with this reasoning on chemical potential?

One definition of the chemical potential is the change of internal energy of the system with respect to a particle of the added substance with the system entropy, molar volume, and all other species particle numbers remaining constant.

A thought experiment: Let's suppose I have a completely rigid container with $1\,\rm mol$ of argon in it with an infinite amount of insulation surrounding the container. Therefore, I have zero heat transfer and the molar volume is constant. I admit one atom of argon into the container in a reversible manner (same $P$ and $T$). The pressure of the system must increase differentially according to the ideal gas law.

Hence the entropy will decrease differentially except that it must stay constant and so temperature will have to increase for this process to be isentropic. But for an ideal gas, internal energy depends only on temperature. So, $\frac{\mathrm{d}U}{\mathrm{d}n}$ is positive. Yet, when I look up values for argon as a gas at STP, its value is $0$ (like all the elements in their standard states).

Where is my reasoning wrong?

Your argument purports to show the chemical potential is positive. You look it up and find that it's zero. In fact, the chemical potential for an ideal gas is negative!

First, I'm not following the step where you conclude that the entropy must go down because the pressure goes up. The entropy of an ideal gas is given by the Sackur-Tetrode equation: $$S=Nk\left[\ln\left({V\over N}\left(4\pi m U\over 3Nh^2\right)^{3/2}\right)+{5\over 2}\right]$$ Incidentally, I got this particular form of the equation from the textbook by Schroeder, which is the one I like best for this sort of thing.

Under the circumstances you describe (increasing $N$ by 1 while holding $U/N$ constant), this reduces to $$S=Nk\left[\ln(V/Nv_Q)+ {5\over 2}\right],$$ where $v_Q$, the quantum volume per particle, is constant and $Nv_Q\ll V$. (The latter is the condition for the gas to be nondegenerate.) The derivative of this with respect to $N$ is positive: upon adding a particle at constant $T,V$, the entropy goes up by $$\Delta S=k\left[\ln(V/nv_Q)+{3/2}\right]>0.$$

Second, where are you looking up the value for $\mu$ for argon at STP? Is it possible that the table you're looking in lists values relative to STP? It's certainly not true that the chemical potential of an ideal gas at STP is zero.

One way to get the chemical potential of an ideal gas is by imagining adding the new particle in such a way that the total energy (not the temperature) remains fixed. The relevant identity then is $$\mu=-T\left(\partial S\over\partial N\right)_{U,V},$$ which comes from $dU=T\,dS-P\,dV+\mu\,dN$. Taking the derivative of the above expression for $S$, we get $$\mu=-kT\ln\left[{V\over N}\left(2\pi mkT\over h^2\right)^{3/2}\right].$$ Again, for gases that are far from degenerate, the quantity inside the logarithm is large, so the chemical potential is negative.

• My source for values of chemical potential is: job-stiftung.de/index.php?id=54,0,0,1,0,0 My background is as a chemical engineer, not a physicist so I will have to study your answer for a bit. Thanks for the quick help! – Jason Waldrop Jul 6 '11 at 14:53
• That page does say "$\mu = 0$ for $e_{\rm gas}$ under standard conditions." I'm not sure exactly what that means, but it sounds like it might mean that the chemical potentials are measured relative to the values at STP. The table also gives temperature coefficients, which are presumably $d\mu/dT$. We can check that for consistency with the formula I gave. I differentiated $\mu$ with respect to $T$, plugged in the appropriate numbers for the various constants, and got $d\mu/dT=-154$ J/mol, which agrees quite well with the $-154.73$ J/mol given in that table. So I think everything's consistent. – Ted Bunn Jul 6 '11 at 17:28
• Units on those last two numbers should be J/(mol K), not J/mol. I'd edit the comment, but I was already right up against the character limit! – Ted Bunn Jul 6 '11 at 17:30