Exercises — Kinetic molecular theory — derivation of P = (1 - 3)ρv²_rms

Use $P=\frac{1}{3}\rho v_{rms}^2$, rearranged to $v_{rms}=\sqrt{\frac{3P}{\rho }}$. Why: this form links only bulk, directly-measurable quantities ($P$ and $\rho$). The form $\sqrt{3RT/M}$ needs the temperature $T$ and the molar mass $M$ — neither is given here, so it is unusable. The two forms are the same physics: substituting $\rho =Nm/V$ and $PV=N{k}_{B}T$ turns one into the other.

The molecule reverses: its $x$-momentum goes from $+m{v}_x$ to $-m{v}_x$, a change of $-2m{v}_x$. By Newton's third law the wall receives the opposite: $+2m{v}_x$. The magnitude is $\boxed{2m{v}_x}$. The factor of 2 is the whole point — the molecule does not stop, it bounces.

Pressure comes from momentum delivery, and force per molecule scaled as $v_x^2$ (fast molecules hit harder and more often). Averaging brings in $\overline{v^2}$, whose square root is by definition ==$v_{rms}=\sqrt{\overline{v^2}}$==. Because $\sqrt{\overline{v^2}}\ne \overline{v}$ (squaring weights the fast ones more), only $v_{rms}$ is correct here.

Convert molar mass to SI: $M=0.028\text{kg/mol}$. $v_{rms}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3(8.314)(300)}{0.028}}=\sqrt{2.673\times 10^5}\approx 517\text{m/s}$ Why kg/mol: $R$ is in joules $={\text{kg m}}^2{\text{s}}^{-2}$; using grams would make the answer $\sqrt{1000}\approx 31.6\times$ too small.

$v_{rms}=\sqrt{\frac{3P}{\rho }}=\sqrt{\frac{3(2.0\times 10^5)}{1.60}}=\sqrt{3.75\times 10^5}\approx 612\text{m/s}$ Why: no $T$ or $M$ needed — the bulk form connects them for us.

At equal $T$, $v_{rms}\propto 1/\sqrt{M}$: $\frac{v_{rms}(\text{He})}{v_{rms}(\text{Ar})}=\sqrt{\frac{M_{\text{Ar}}}{M_{\text{He}}}}=\sqrt{\frac{40}{4}}=\sqrt{10}\approx 3.16$ This is exactly the reasoning behind Graham's Law of Diffusion: lighter gas moves (and effuses) faster.

Since $v_{rms}\propto \sqrt{T}$ (same gas), $\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}$. Doubling speed needs ${\left(\frac{v_2}{v_1}\right)}^2=4$, so $T_2=4T_1=4(300)=\boxed{1200\text{K}}$. Why square: speed grows as the root of $T$, so to double it you multiply $T$ by $2^2=4$.

From the parent result $K{E}_{avg}=\frac{3}{2}{k}_{B}T$ (see Average Kinetic Energy and Temperature): $K{E}_{avg}=\frac{3}{2}(1.381\times 10^{-23})(300)=6.21\times 10^{-21}\text{J}$ Why it's independent of the gas: the $\frac{3}{2}{k}_{B}T$ depends only on $T$, not on $m$ — heavy molecules simply move slower to carry the same energy.

Use $PV=\frac{1}{3}Nm\overline{v^2}$, so $P=\frac{Nm\overline{v^2}}{3V}$: $P=\frac{(2.0\times 10^{25})(5.3\times 10^{-26})(2.4\times 10^5)}{3(0.020)}=\frac{2.544\times 10^5}{0.06}\approx 4.24\times 10^6\text{Pa}$ Why $\frac{1}{3}$: we sum all $N$ pushes but only one third of the motion drives any single wall (3 directions share the speed equally).

Rearrange $v_{rms}=\sqrt{3RT/M}$ to $M=\frac{3RT}{v_{rms}^2}$: $M=\frac{3(8.314)(300)}{(1360{)}^2}=\frac{7482.6}{1.8496\times 10^6}=4.05\times 10^{-3}\text{kg/mol}\approx 4.0\text{g/mol}$ That is helium ($He$). Why square $v_{rms}$: the speed enters the formula squared, so we must square it before dividing.

Set the two right-hand sides equal (same $PV$): $\frac{1}{3}Nm\overline{v^2}=N{k}_{B}T\Rightarrow \overline{v^2}=\frac{3k_{B}T}{m}\Rightarrow v_{rms}=\sqrt{\frac{3k_{B}T}{m}}$ This is the same as $\sqrt{3RT/M}$ because $k_B=R/N_A$ and $M=m{N}_A$. Numerically: $v_{rms}=\sqrt{\frac{3(1.381\times 10^{-23})(300)}{5.31\times 10^{-26}}}=\sqrt{2.341\times 10^5}\approx 484\text{m/s}$ Matches Example 1 of the parent note — consistency check passed. See Ideal Gas Equation PV = nRT.

Effusion rate $\propto v_{rms}\propto 1/\sqrt{M}$, and time $\propto 1/\text{rate}\propto \sqrt{M}$: $\frac{t_{H_2}}{t_{C{O}_2}}=\sqrt{\frac{M_{H_2}}{M_{C{O}_2}}}=\sqrt{\frac{2}{44}}=\sqrt{0.04545}=0.2132$ $t_{H_2}=0.2132\times 88\approx 18.8\text{s}$ Why: lighter $H_2$ is faster, so it escapes sooner — shorter time. Foundation of Graham's Law of Diffusion.

Take the average of $v^2=v_x^2+v_y^2+v_z^2$: $\overline{v^2}=\overline{v_x^2}+\overline{v_y^2}+\overline{v_z^2}=3\overline{v_x^2}\Rightarrow \overline{v_x^2}=\frac{1}{3}\overline{v^2}$

Set $v_{rms}=\sqrt{3RT/M}=v_{esc}$, solve for $T=\frac{M{v}_{esc}^2}{3R}$: $T=\frac{(0.002)(1.12\times 10^4{)}^2}{3(8.314)}=\frac{(0.002)(1.2544\times 10^8)}{24.942}=\frac{2.5088\times 10^5}{24.942}\approx 1.006\times 10^4\text{K}$ So $T\approx 10,060\text{K}$. Interpretation: the average $H_2$ molecule needs ~10,000 K, far above real temperatures — yet the fast tail of the Maxwell-Boltzmann Speed Distribution always has a few molecules above escape speed, so hydrogen slowly leaks into space over geological time.

$\frac{r_{235}}{r_{238}}=\sqrt{\frac{M_{238}}{M_{235}}}=\sqrt{\frac{352}{349}}=\sqrt{1.008596}\approx 1.00429$ Interpretation: the lighter isotope effuses only ~0.43% faster per stage — so thousands of stages are cascaded to enrich uranium. A tiny speed difference, amplified enormously. This is Graham's Law of Diffusion taken to industrial extreme.

Magnitude of momentum given to wall per bounce ::: $2m{v}_x$ $v_{rms}$ in terms of $P,\rho$ ::: $\sqrt{3P/\rho }$ $v_{rms}$ in terms of $T,M$ ::: $\sqrt{3RT/M}$ To double $v_{rms}$ you multiply $T$ by ::: $4$ Effusion time scales as ::: $\sqrt{M}$ (heavier = slower = longer) Source of the $\frac{1}{3}$ ::: 3 dimensions share the speed equally, $\overline{v_x^2}=\frac{1}{3}\overline{v^2}$