5.2 Why Is the Chemical Potential Negative?
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It is worth noting that the volume dependence of both the internal energy and the
Helmholtz energy resides entirely with their respective translational contributions.
5.2 Why Is the Chemical Potential Negative?
To understand why the chemical potential μ for an ideal classical gas is negative,
we shall consider explicitly what happens for a monatomic gas, where we can
make use of the Sackur–Tetrode equation for the entropy. We shall also utilize the
thermodynamic relation
μ =
∂U
∂N
S,V
(5.2.1)
obtained from the extension of the combined first and second laws (of thermodynamics) expression d U = T d S −P d V for closed systems to open systems namely,
dU = T d S − P d V + μd N .
For convenience, we shall also write the Sackur–Tetrode equation for the entropy in
the (equivalent) form [2]
S ≡ S(T , V ; N) = Nk B
ln
V
N
+
3
2
ln
4πmU
3Nh 2
+
5
2
,
(5.2.2)
with U =
3
2 Nk B T . The key to understanding why the chemical potential for an
ideal classical gas is inherently negative lies in asking how the internal energy U
changes with particle number while holding the volume and entropy of the system
constant.
Let us add a particle to the system, so that N becomes N + 1, and let the change
in the internal energy associated with the addition of this particle be μ, so that the
internal energy U becomes U + μ. To see what value μ must take, let us calculate
the entropy S(N + 1) corresponding to N + 1 particles by employing the Sackur–
Tetrode equation (5.1.7): we obtain
S(N + 1) = (N + 1)k B
ln
V
N + 1
+
3
2
ln
4πm(U + μ)
3(N + 1)h 2
+
5
2
.
However, our thermodynamic definition of μ requires that S(N + 1) has the same
(i.e., unchanged) value of S(N): this requirement constrains μ. To see how the
constraint works, let us write