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

Question

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?

Answer 1

For continuous systems, the definition slightly changes. I will quote the following definitions from some lecture notes on soft condensed matter (they are not online sadly):

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 $H(\Gamma)<E$. 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}

See also this question for the distinction between the different kinds of entropy.