What is wrong with this reasoning on chemical potential?

Question

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 mole 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 dU/dn 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?

Answer 1

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.