Worked examples — Kinetic theory — pressure derivation, temperature as mean KE

Everything here rests on three boxed facts from the parent, restated in plain words so nothing is assumed:

See the Ideal Gas Law PV=nRT for the macroscopic law we keep comparing against, and the Maxwell-Boltzmann Speed Distribution for where $v_{rms}$, $\bar{v}$ and the most-probable speed part ways.

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The scenario matrix

Each cell below is a class of question. The "Example" column tells you which worked example proves you can handle it.

Cell	What's special about it	Example
A. T → speed	Given temperature, find $v_{rms}$	Ex 1
B. speed → T	Given a speed, find temperature	Ex 2
C. micro → P	Given $N,m,V,v$, find pressure	Ex 3
D. Two gases, same T	Compare speeds/KE across masses	Ex 4
E. Ratio / scaling	$T$ or $m$ changes by a factor — no numbers plugged	Ex 5
F. Degenerate $T\to 0$	Limiting/zero input, what breaks	Ex 6
G. Real-world word problem	Tyre / balloon style, translate then solve	Ex 7
H. Exam twist	$v_{rms}$ vs $\bar{v}$ trap, or heavier≠higher-P	Ex 8

Figure of the whole landscape first, so you can see how the cells relate:

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Example 1 — Cell A: temperature to speed

Step 1. Pick the molar form $v_{rms}=\sqrt{3RT/M}$. Why this step? We were handed $M$ (per mole), not $m$ (per molecule). Using the form that already contains $M$ saves a division by $N_A$.

Step 2. Plug in: $v_{rms}=\sqrt{\frac{3(8.314)(300)}{0.032}}=\sqrt{\frac{7482.6}{0.032}}=\sqrt{233831}$

Step 3. Take the root: $v_{rms}\approx 483.6\text{m/s}$ Why this step? The square root undoes the "mean-square"; it returns us to something with units of speed.

Verify: Slightly below nitrogen's $517\text{m/s}$ because oxygen is heavier ($32$ vs $28$), and $v\propto 1/\sqrt{M}$ so the ratio should be $\sqrt{28/32}=0.935$, giving $517\times 0.935\approx 483$. ✓ Units: $\sqrt{\text{J/mol}÷\text{kg/mol}}=\sqrt{\text{J/kg}}=\sqrt{{\text{m}}^2/{\text{s}}^2}=\text{m/s}$. ✓

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Example 2 — Cell B: speed back to temperature

Step 1. Start from $\frac{1}{2}m\overline{v^2}=\frac{3}{2}{k}_{B}T$ and solve for $T$: $T=\frac{m\overline{v^2}}{3k_B}=\frac{m{v}_{rms}^2}{3k_B}$ Why this step? We invert the mean-KE relation because we know the speed and want temperature — the exact reverse of Example 1.

Step 2. Plug numbers ($v_{rms}^2=1370^2=1.8769\times 10^6$): $T=\frac{(6.64\times 10^{-27})(1.8769\times 10^6)}{3(1.38\times 10^{-23})}$

Step 3. Compute top and bottom: $T=\frac{1.2463\times 10^{-20}}{4.14\times 10^{-23}}\approx 301\text{K}$

Verify: About room temperature — consistent with helium at $300\text{K}$. Reverse-check: at $301\text{K}$, $v_{rms}=\sqrt{3(1.38\times 10^{-23})(301)/6.64\times 10^{-27}}\approx 1370\text{m/s}$. ✓

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Example 3 — Cell C: microscopic numbers to pressure

Step 1. Use $P=\frac{1}{3}\frac{Nm{v}_{rms}^2}{V}$ (since $\overline{v^2}=v_{rms}^2$). Why this step? This is the direct microscopic pressure formula — every quantity in it is given.

Step 2. Numerator pieces: $Nm=(3.0\times 10^{24})(5.3\times 10^{-26})=0.159\text{kg}$, and $v_{rms}^2=480^2=230400{\text{m}}^2/{\text{s}}^2$.

Step 3. Combine: $P=\frac{1}{3}\cdot \frac{(0.159)(230400)}{0.010}=\frac{1}{3}\cdot \frac{36633.6}{0.010}=\frac{1}{3}(3.66336\times 10^6)$ $P\approx 1.22\times 10^6\text{Pa}\approx 12\text{atm}$

Verify: Units: $\frac{\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}{{\text{m}}^3}=\frac{\text{N}\cdot \text{m}}{{\text{m}}^3}=\frac{\text{N}}{{\text{m}}^2}=\text{Pa}$. ✓ ($1\text{atm}=1.013\times 10^5\text{Pa}$, so $\approx 12\text{atm}$ — reasonable for a dense box.)

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Example 4 — Cell D: two gases at one temperature

Step 1 (a). $\overline{E_k}=\frac{3}{2}{k}_{B}T$ depends only on $T$, not on mass: $\frac{\overline{E_{k,H_2}}}{\overline{E_{k,O_2}}}=\frac{\frac{3}{2}{k}_{B}T}{\frac{3}{2}{k}_{B}T}=1$ Why this step? This is the deep content of "temperature = mean KE" — at a shared $T$, every species carries the identical average translational energy (see Boltzmann Constant and Equipartition).

Step 2 (b). Equal KE but different mass forces different speed. Since $\frac{1}{2}m{v}_{rms}^2$ is equal: $\frac{v_{rms,H_2}}{v_{rms,O_2}}=\sqrt{\frac{M_2}{M_1}}=\sqrt{\frac{0.032}{0.002}}=\sqrt{16}=4$ Why this step? Fixing energy and letting mass vary makes speed scale as $1/\sqrt{M}$; the lighter gas is faster.

Verify: Hydrogen is $16\times$ lighter → $\sqrt{16}=4\times$ faster, energies equal. Cross-check energy: $\frac{1}{2}{M}_1{v}_1^2$ vs $\frac{1}{2}{M}_2{v}_2^2$: $(0.002)(4^2)=0.032=(0.032)(1^2)$. ✓

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Example 5 — Cell E: scaling with no numbers plugged

Step 1 (a) pressure. At fixed $N,V$: $PV=N{k}_{B}T\Rightarrow P\propto T$. So $P\to 4P$ (factor $4$). Why this step? Rigid box = constant $V$; sealed = constant $N$. Only $T$ moves, and $P$ tracks it linearly.

Step 2 (b) rms speed. $v_{rms}=\sqrt{3k_{B}T/m}\propto \sqrt{T}$. So $v_{rms}\to \sqrt{4}{v}_{rms}=2v_{rms}$ (factor $2$). Why this step? Speed lives under a square root of $T$, so a $\times 4$ in $T$ is only a $\times 2$ in speed.

Step 3 (c) mean KE. $\overline{E_k}=\frac{3}{2}{k}_{B}T\propto T$. So $\overline{E_k}\to 4\overline{E_k}$ (factor $4$).

Verify: Consistency: KE $\propto v_{rms}^2$, and indeed $(2{)}^2=4$. ✓ Pressure and energy scale like $T$ (factor 4); speed like $\sqrt{T}$ (factor 2). No number ever entered — pure scaling.

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Example 6 — Cell F: the degenerate limit $T\to 0$

Step 1. As $T\to 0$: $v_{rms}=\sqrt{3k_{B}T/m}\to \sqrt{0}=0$, and $\overline{E_k}=\frac{3}{2}{k}_{B}T\to 0$, and $P=\frac{1}{3}\rho \overline{v^2}\to 0$. Why this step? Every quantity carries a factor of $T$ (directly, or under a root), so all vanish together at absolute zero.

Step 2. Sanity of the limit: zero speed means molecules frozen still, zero pressure means no wall hits — internally consistent (see Internal Energy of Ideal Gas).

Step 3. Where it breaks: the ideal model assumes point particles with only translational motion and no quantum effects. Real gases liquefy/solidify before $0\text{K}$, and quantum "zero-point" motion forbids a molecule from being truly at rest. Why this step? A good scientist states the domain of validity: this is a degenerate input where the formula still returns a clean $0$, but the physics has left ideal-gas territory.

Verify: At exactly $T=0$, plugging in gives literally $v_{rms}=0$, $\overline{E_k}=0$, $P=0$ — a self-consistent (if unphysical) prediction. ✓

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Example 7 — Cell G: a real-world word problem

Step 1. Translate the words. "Rigid, no leaks" $\Rightarrow V$ and $N$ constant. So $PV=N{k}_{B}T$ collapses to $P\propto T$, i.e. $\frac{P_1}{T_1}=\frac{P_2}{T_2}$. Why this step? Identifying what stays fixed turns a paragraph into a proportionality — the microscopic reason: hotter molecules hit the wall harder and more often, and $P/T$ is constant.

Step 2. Solve for $P_2$: $P_2=P_1\cdot \frac{T_2}{T_1}=2.20\times 10^5\cdot \frac{330}{290}$

Step 3. Compute: $P_2=2.20\times 10^5\times 1.1379\approx 2.50\times 10^5\text{Pa}$

Verify: Rise of $\frac{330-290}{290}=13.8\%$; pressure rose $\frac{2.50-2.20}{2.20}=13.6\%$ (rounding). ✓ This is exactly why tyre pressure should be checked cold.

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Example 8 — Cell H: the exam trap ($v_{rms}$ vs $\bar{v}$, and heavier ≠ more pressure)

Step 1 (a). Ordinary average — add and divide: $\bar{v}=\frac{200+400+400+600+800}{5}=\frac{2400}{5}=480\text{m/s}$ Why this step? $\bar{v}$ is the plain arithmetic mean — useful for things like average flight distance, not pressure.

Step 2 (b). rms — square, average, root: $\overline{v^2}=\frac{200^2+400^2+400^2+600^2+800^2}{5}=\frac{40000+160000+160000+360000+640000}{5}$ $=\frac{1360000}{5}=272000\Rightarrow v_{rms}=\sqrt{272000}\approx 521.5\text{m/s}$ Why this step? Pressure came from momentum-transfer $\propto v_x^2$, so the mean of the squares is the physically relevant average.

Step 3 (c). Use $v_{rms}$ in $P=\frac{1}{3}\rho \overline{v^2}$. And $v_{rms}\ge \bar{v}$ always, because squaring gives extra weight to the fast molecules; equality only if every molecule has identical speed. Why this step? This is the classic trap the parent note warned about — the two "averages" are different tools for different jobs. Bonus trap: at equal $T$ a heavier gas is slower by exactly enough that pressure is unchanged, since $\frac{1}{2}m\overline{v^2}=\frac{3}{2}{k}_{B}T$ has no mass in the final $P=N{k}_{B}T/V$.

Verify: $v_{rms}=521.5>\bar{v}=480$, as required. ✓ Gap arises because the spread of speeds is nonzero.

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Active recall

Recall Cover the answers and self-test

Which speed goes into the pressure formula, $\bar{v}$ or $v_{rms}$? ::: $v_{rms}$, because pressure comes from $\overline{v^2}$ (mean of squares), not the mean of speeds. At the same temperature, which is bigger for two gases — the mean KE or is it equal? ::: Equal; $\overline{E_k}=\frac{3}{2}{k}_{B}T$ depends only on $T$. If $T$ quadruples at fixed $V,N$, what happens to $P$ and to $v_{rms}$? ::: $P\to 4P$ (linear in $T$); $v_{rms}\to 2v_{rms}$ (square root of $T$). At constant $V,N$, what is the constant ratio? ::: $P/T$ is constant. Why does $v_{rms}=0$ at $T=0$ signal a model limit, not a fact? ::: Real gases condense and quantum zero-point motion forbids true rest; the ideal formula still returns $0$ but the physics no longer applies.

Return to the parent topic · related: Degrees of Freedom and Molar Heat Capacity, Pressure as Force per Area, Elastic Collisions and Momentum.