Can a gas molecule theoretically have zero velocity? According to Maxwell's speed distribution law gas molecule can have speed which lies between zero to infinity. But in the graph of the distribution curve it seems to touch zero velocity. So can a gas particle practically have exactly zero velocity or is that just in theoretic sense?
 A: There are several way to look at why the answer is effectively "no". The Botlzman factor is the probability $P$ of a degree of freedom having energy $E$ at temperature $T$:
$$ P(E) \propto e^{-E/kT}$$
which, for an ideal gas atom with mass $m$ moving in the $x$-direction is:
$$  P(E_x) \propto e^{-mv_x^2/2kT} $$
which is finite as $v_x \rightarrow 0$, where $v_x$ is the velocity in the $x$-direction, with $-\infty<v_x<+\infty$.
The problem is that $v_x=0$ is not enough. For $$||\vec v||=0$$ we must also have $v_y=v_z=0$, and that is clearly much less probable.
As pointed out in the comments, one must consider an infinitesimal region $dv_x$ to get a non zero result, so the combination of $x$, $y$, and $z$ looks like:
$$P(\vec v)d^3v \propto e^{-mv_x^2/2kT}e^{-mv_y^2/2kT}e^{-mv_z^2/2kT}=e^{-mv^2/2kT}d^3v$$
where the Boltzman factor is again finite at $||v||=0$. The problem is $d^3v$, which when converted into a speed via spherical coordinates is:
$$d^3v \rightarrow v^2 dv$$
and is interpreted as the differential volume of a shell extending from $v$ to $v+dv$. This given the final form of the Maxwell-Boltzmann distribution:
$$P(v) \propto v^2 e^{-mv^2/2kT}$$
which is zero at $v=0$ because  there is no volume in phase space there.
