Question bank — Maxwell-Boltzmann distribution — full derivation

True. $g(v_x)\propto e^{-m{v}_x^2/2k_{B}T}$ has no $v_x^2$-prefactor, so it is a pure Gaussian whose maximum sits at $v_x=0$ — the most likely value of any one component is zero motion along that axis.

False. $f(v)$ carries the extra factor $v^2$, which vanishes at $v=0$; the peak is pushed out to the most probable speed $v_p=\sqrt{2k_{B}T/m}$. Zero speed means a shell of zero radius, which holds no velocity vectors.

False. $v_p=\sqrt{2k_{B}T/m}\propto \sqrt{T}$, so doubling $T$ multiplies $v_p$ by $\sqrt{2}\approx 1.41$, not by $2$. Speed scales with the square root of temperature.

False. Heavier means smaller $v_p\propto 1/\sqrt{m}$, so the curve shifts left and becomes narrower and taller (it is squeezed toward low speeds, not spread out). Lighter gases are the broad, fast ones.

True. Their ratio $\sqrt{2}:\sqrt{8/\pi }:\sqrt{3}$ is fixed and independent of $T$ and $m$; the long high-speed tail always drags the mean speed and rms speed to the right of the peak.

False. The area is always ${\int }_0^{\infty }f(v)dv=1$ (it is a probability). Heating moves and reshapes the curve but never changes the total area — a taller peak forces a narrower one and vice versa.

False. $⟨v_x⟩=0$ only says leftward and rightward motion cancel by symmetry. The speed is never zero on average: the mean speed $⟨v⟩=\sqrt{8k_{B}T/\pi m}>0$ because speed cannot be negative.

False. $g(v_x)$ is symmetric about $0$, but $f(v)$ lives only on $v\ge 0$ and has a long right tail — it is skewed, which is exactly why the three characteristic speeds differ.

Missing the $v^2$ geometric factor. The Boltzmann factor gives the weight of one velocity state; speed lumps together the $4\pi v^2$ states of that magnitude, so $f(v)\propto v^2{e}^{-m{v}^2/2k_{B}T}$.

The rms speed is $\sqrt{⟨v^2⟩}$, the square root of the mean of the squares, not the mean itself. Since squaring emphasises large values, $\sqrt{⟨v^2⟩}>⟨v⟩$ for any spread-out distribution.

Units are mixed. Use per-molecule mass $m$ (in kg) with $k_B$, or molar mass $M$ (in kg/mol) with the gas constant $R=N_A{k}_B$: $v_p=\sqrt{2RT/M}$. Never pair $k_B$ (a per-molecule constant) with a per-mole mass.

We do not integrate them away; we multiply the three component densities and then re-express $d{v}_{x}d{v}_{y}d{v}_z$ as the shell volume $4\pi v^{2}dv$. Integration comes only later, to find averages.

$v_p$ is the mode (peak location), not the median. Because of the right tail, the median lies above $v_p$, so more than half of the molecules move faster than $v_p$ — the question's premise is wrong. (The mean speed $⟨v⟩$ sits even further right, above the median.)

A negative $b$ gives $e^{+b{v}_x^2}$, which grows without bound and cannot be normalized (its integral diverges). $b>0$ is forced by the requirement $\int gd{v}_x=1$.

Independence forces $g(v_x)g(v_y)g(v_z)$ to depend only on the sum $v_x^2+v_y^2+v_z^2$; taking logs, $\ln g(v_x)$ must be linear in $v_x^2$. A $v_x^4$ term would not add up into a function of that sum. See Gaussian integrals.

Components are statistically independent and isotropic, which lets the joint distribution factorize — a huge simplification. Speed mixes all three, so there is no factorization to exploit at the start.

The functional form leaves one unknown, $b$. Equipartition pins each translational degree of freedom to $\frac{1}{2}{k}_{B}T$, giving $⟨v_x^2⟩=k_{B}T/m$ and hence $b=m/2k_{B}T$ — it is what injects the temperature.

The $v^2$ shell-volume factor is the killer: at $v=0$ the sphere of radius $v$ has no surface, so there is no "room" for velocity vectors, regardless of the Boltzmann weight. See Kinetic theory of gases.

For the same $T$, smaller $m$ gives larger mean speed $⟨v⟩\propto 1/\sqrt{m}$, and effusion rate scales with mean speed — this is the root of Graham's law.

It assumes each velocity state carries the Boltzmann weight $e^{-E/k_{B}T}$ with no occupation limit. When particles crowd into states, quantum statistics take over — see Maxwell-Boltzmann vs Fermi-Dirac vs Bose-Einstein and Boltzmann factor and partition function.

Both speeds carry the same $\sqrt{k_{B}T/m}$ factor, which cancels in the ratio, leaving the pure number $\sqrt{3/2}$. The shape of MB is universal; only its horizontal scale depends on $T$ and $m$.

The curve collapses toward $v=0$: $v_p\propto \sqrt{T}\to 0$ and the distribution becomes an infinitely narrow spike at zero speed — all molecules frozen, consistent with vanishing thermal energy.

The peak $v_p\propto \sqrt{T}$ runs off to arbitrarily high speed and the curve flattens and broadens indefinitely; every finite speed becomes vanishingly probable as probability spreads to infinity.

No. For every $v>0$ both $v^2>0$ and $e^{-m{v}^2/2k_{B}T}>0$, so $f(v)>0$ everywhere on $(0,\infty )$ — there is always a nonzero chance of any positive speed, however large.

Exactly zero, like any single point of a continuous distribution. Only ranges have nonzero probability: $f(v)dv$ is a fraction over an interval, not at a point.

The shell becomes a ring of circumference $2\pi v$, so the geometric factor is $v^1$ not $v^2$, giving $f(v)\propto v{e}^{-m{v}^2/2k_{B}T}$. The peak and characteristic speeds shift because the dimension of the "room" factor changed.

No — the factorization into three identical Gaussians relied on isotropy. With a preferred axis, $g(v_z)$ would differ from $g(v_x)$ and the clean $4\pi v^2$ shell (which assumed direction-independence) would no longer apply.