# Why is the number of microstates corresponding to a macrostate even finite for an ideal gas in a box? [duplicate]

I was looking at the Sackur-Tetrode equation which gives an exact formula for the entropy of an ideal gas. I tried to relate it to Boltzmann's famous $$S = k_B \ln(W)$$ where $$S$$ is the entropy and $$W$$ is the number of microstates which correspond to the macrostate.

The Sackur-Tetrode equation suggests that e.g. if I have 1 particle of an ideal gas in a $$1\text{m}^3$$ cubic box at $$300\text{K}$$ then this system has some finite entropy, say $$S^*$$. This then suggests that the number of microstates the gas particle inside the box can occupy is $$e^{S^*/k_B}$$, which is also finite.

However this goes against my intuition because I think I can place the atom at any physical position within the $$1\text{m}^3$$ box (of which there are an infinite number of choices of) and also I am free to choose the direction of its velocity (the magnitude of the velocity is fixed by the internal energy since we're at $$300\text{K}$$ but the direction is freely variable), which again gives me a choice of an infinite number of vectors for the direction.

Taken together to me this suggests there are an infinite number of microstates that the gas particle can be in that all correspond to the same macrostate, so therefore the entropy should be infinite (because $$\ln(x) \to \infty$$ as $$x \to \infty$$) and not a finite number like the equation says. Where is my intuition going wrong here?

• For the phase space, you should work with phase space volume..."number of microstates" is ill-defined, as you noted. This should be explained in any textbook. But this has been asked for sure on this site before. Commented Apr 22 at 9:13
• Right, but then that's not unit invariant (assuming you meant replacing $W$ with phase space volume). I can change my units of measure and after taking logs this will add some constant to the calculated entropy. Commented Apr 22 at 9:16
• It is defined up to an additive constant anyway. Commented Apr 22 at 9:16
• @naturallyInconsistent I disagree. Statistical mechanics/physics is, after all, a "toolbox", and does not care if the world is "classical" or "quantum". If you define all quantities properly, there is no contradiction or so at all. The term "number of microstates" is just not defined for the case OP is discussing. Commented Apr 22 at 10:26
• Quantum mechanics cannot "follow" from statistical mechanics (neither from these considerations nor from a common misunderstanding of Gibbs' paradox). I agree, though, that as far as we know, the world is quantum and not "classical". Commented Apr 22 at 11:05

Let $$\Gamma=(\mathbf r^N,\mathbf p^N)$$ denote a point in phase space, where each $$\mathbf r_i$$ is inside some volume $$V$$. The microcanonical probability distribution is given by
$$f(\Gamma)=\frac{\delta(E-H(\Gamma))}{\omega(E,V,N)}$$ where $$H(\Gamma)=\sum_{i=1}^{N}\frac {p_i^2}{2m}+U(\mathbf r^N)$$ and $$\omega(E,V,N)=\int\frac{\mathrm d\Gamma}{N!h^{3N}}\,\delta(E-H(\Gamma))$$ This last expression is an integral over the surface in phase space where $$H(\Gamma)=E$$ and $$h$$ is a constant with units of angular momentum that makes the expression unitless. Here $$N!$$ is required to have extensive dependence of $$N$$ in the thermodynamic limit, and can be interpreted as a correction-factor that corrects for overcounting identical particles. Now, define $$\Omega(E,V,N)=\int\frac{\mathrm d\Gamma}{N!h^{3N}}\,\Theta(E-H(\Gamma)).$$ This corresponds to the volume of all points in phase space such that $$H(\Gamma). We also have $$\omega=(\partial \Omega)/(\partial V)$$.
We can then define the 'volume entropy' as $$S_v(E,V,N)=k\log\Omega(E,V,N).$$ Or, we could define the more familiar Boltzmann entropy: \begin{align} S_B=k\log W=k\log\Delta E\,\omega. \end{align} Or the surface entropy: \begin{align} S_s=k\log\omega. \end{align}