Question bank — Kinetic theory — pressure derivation, temperature as mean KE

True. $\overline{E_k}=\frac{3}{2}{k}_{B}T$ depends only on $T$, not on mass; same $T$ forces the same average translational KE for every ideal gas.

False. Equal KE with unequal mass means $v_{rms}=\sqrt{3k_{B}T/m}$ differs — the lighter hydrogen is much faster, since only the product $\frac{1}{2}m\overline{v^2}$ is fixed.

False. $v_{rms}\propto \sqrt{T}$, so doubling $T$ multiplies rms speed by $\sqrt{2}\approx 1.41$; you must quadruple $T$ to double the speed.

False. $v_{rms}=\sqrt{\overline{v^2}}$ always exceeds $\bar{v}$ because squaring gives extra weight to the fast molecules; they are two genuinely different averages (see Maxwell-Boltzmann Speed Distribution).

True. From $PV=N{k}_{B}T$ with $N,V$ fixed, $P\propto T$. Note $T$ must be in kelvin — doubling °C does not double $P$.

False. Pressure is $P=N{k}_{B}T/V$ — no mass appears. Heavier molecules hit harder but move proportionally slower, and the two effects cancel exactly.

False. It is the number of spatial dimensions: isotropy gives $\overline{v_x^2}=\frac{1}{3}\overline{v^2}$. In a 2D gas the factor would be $\frac{1}{2}$; it is pure geometry.

False. The wall is rigid and stationary and the collision is elastic, so only the direction of $v_x$ flips — the speed, and thus the KE, is unchanged (see Elastic Collisions and Momentum).

Partly true and harmless. Intermolecular collisions merely reshuffle velocities among molecules; because they conserve total momentum and energy, the population averages $\overline{v^2}$ and $\overline{v_x^2}$ are untouched, so the derived pressure is the same.

True within the model. $\overline{E_k}=\frac{3}{2}{k}_{B}T\to 0$ as $T\to 0$, so classically all motion stops — real gases liquefy or show quantum effects long before, but the ideal law predicts this limit.

False. Temperature is a statistical property of the whole population's average KE. A single molecule has a speed and a KE, but "temperature" only makes sense for the ensemble.

Error: it uses $\bar{v}$ instead of $\overline{v^2}$. Momentum transfer per unit time scales with $v_x^2$, so pressure depends on the mean of the squares, giving $v_{rms}$, not $\bar{v}$.

Error: it forgets the rebound. After an elastic bounce the momentum is $-m{v}_x$, so the change is $m{v}_x-(-m{v}_x)=2m{v}_x$ — the factor of 2 is essential.

Error: it uses a one-way trip. The molecule must travel to the opposite wall and back before hitting the same wall again, so $\Delta t=2L/v_x$.

Error: it drops the sum. Since $v^2=v_x^2+v_y^2+v_z^2$, the three equal parts add: $\overline{v^2}=3\overline{v_x^2}$, so $\overline{v_x^2}=\frac{1}{3}\overline{v^2}$.

Error: each translational degree of freedom carries $\frac{1}{2}{k}_{B}T$, not $k_{B}T$. Three of them give $\frac{3}{2}{k}_{B}T$ (see Boltzmann Constant and Equipartition).

Error: mixing per-molecule and per-mole quantities. With $k_B$ you must use molecular mass $m$; with molar mass $M$ you must use $R$: $v_{rms}=\sqrt{3RT/M}$.

Both come from the three spatial dimensions. $\overline{v_x^2}=\frac{1}{3}\overline{v^2}$ splits the speed among $x,y,z$; equipartition then hands $\frac{1}{2}{k}_{B}T$ to each of those three directions, giving $\frac{3}{2}{k}_{B}T$.

Because a sum of squared components equals $N$ times their average by definition of the mean: $\sum v_{xi}^2=N\overline{v_x^2}$. This is just repackaging, no approximation.

With $\sim 10^{23}$ molecules hitting billions of times per second, the discrete kicks blur into a constant average force — the fluctuations are utterly negligible at everyday scales.

Because $PV=\frac{2}{3}N\overline{E_k}$ and $\overline{E_k}=\frac{3}{2}{k}_{B}T$ depends only on temperature. The mass cancels since heavier molecules simply move slower to keep the same KE.

Because the physics — momentum flux to the wall — is built from $v_x^2$, so the average that emerges is $\overline{v^2}$, whose square root is $v_{rms}$ by construction. The other averages answer different questions (see Maxwell-Boltzmann Speed Distribution).

The pressure on the $x$-wall involves $\overline{v_x^2}$, but energy involves the full $\overline{v^2}$. Isotropy lets us swap one for the other via $\overline{v_x^2}=\frac{1}{3}\overline{v^2}$, connecting the wall force to molecular energy.

The pressure derivation only tracks centre-of-mass motion ($v_x,v_y,v_z$). Rotational or vibrational energy exists but pushes on no wall, so it never enters $P$ — it does affect heat capacity though (see Degrees of Freedom and Molar Heat Capacity).

$P$ doubles, since $P=N{k}_{B}T/V$ and $P\propto 1/V$. Physically, molecules cross the box in half the time, doubling their hit frequency.

Nothing — $\overline{E_k}=\frac{3}{2}{k}_{B}T$ depends only on $T$. The molecules hit the walls more often (higher $P$) but each still carries the same average KE.

With only two directions, $\overline{v_x^2}=\frac{1}{2}\overline{v^2}$, so pressure carries a $\frac{1}{2}$, and equipartition gives $\overline{E_k}=\frac{2}{2}{k}_{B}T=k_{B}T$ (one $\frac{1}{2}{k}_{B}T$ per dimension).

Not directly — with $v_x=0$ it never reaches the $x$-wall. But this is one molecule out of $10^{23}$; the population average $\overline{v_x^2}$ is what sets the pressure, and it stays $\frac{1}{3}\overline{v^2}$.

$v_{rms}\to \infty$, so vanishingly light molecules would move arbitrarily fast at fixed $T$ — this is why the lightest gases (hydrogen, helium) are the fastest and escape planetary atmospheres most easily.

No. The relation demands absolute temperature; at $0{}^{\circ }\text{C}$ the KE is not zero (that's $273\text{K}$ of motion). Only Kelvin makes $\overline{E_k}$ vanish at $T=0$.