Exercises — Kinetic theory — pressure derivation, temperature as mean KE

WHAT: direct use of $P=\frac{1}{3}Nm\overline{v^2}/V$. WHY: we are handed $\overline{v^2}$ directly, so no square-rooting is needed. $P=\frac{1}{3}\cdot \frac{(3.0\times 10^{23})(4.7\times 10^{-26})(2.4\times 10^5)}{2.0\times 10^{-3}}$ Numerator inside: $(3.0\times 10^{23})(4.7\times 10^{-26})=1.41\times 10^{-2}$. Times $2.4\times 10^5$ gives $3.384\times 10^3$. $P=\frac{1}{3}\cdot \frac{3.384\times 10^3}{2.0\times 10^{-3}}=\frac{1}{3}(1.692\times 10^6)\approx 5.64\times 10^5\text{Pa}$

WHAT: use $\overline{E_k}=\frac{3}{2}{k}_{B}T$. WHY: mean KE depends only on temperature, not on what gas it is. $\overline{E_k}=\frac{3}{2}(1.38\times 10^{-23})(350)\approx 7.25\times 10^{-21}\text{J}$

WHAT: molar form $v_{rms}=\sqrt{3RT/M}$. WHY: we know molar mass $M$, not the single-molecule mass $m$, so use the form with $R$ and $M$. $v_{rms}=\sqrt{\frac{3(8.314)(400)}{0.032}}=\sqrt{\frac{9976.8}{0.032}}=\sqrt{3.118\times 10^5}\approx 558\text{m/s}$

WHAT: invert $v_{rms}=\sqrt{3RT/M}$ to get $T$. WHY: square both sides to remove the root, then isolate $T$. $v_{rms}^2=\frac{3RT}{M}\Rightarrow T=\frac{M{v}_{rms}^2}{3R}$ $T=\frac{(0.004)(1360{)}^2}{3(8.314)}=\frac{(0.004)(1.8496\times 10^6)}{24.942}=\frac{7398.4}{24.942}\approx 297\text{K}$

WHAT: use proportionalities, not full numbers. WHY: ratios cancel all the constants, so we only track how each quantity scales with $T$. (a) $v_{rms}\propto \sqrt{T}$. Going $300\to 600$ multiplies $T$ by $2$, so $v_{rms}$ scales by $\sqrt{2}\approx 1.414$. (b) At constant $V$ and $N$, $PV=N{k}_{B}T\Rightarrow P\propto T$. So pressure doubles (factor $2$). (c) $\overline{E_k}=\frac{3}{2}{k}_{B}T\propto T$, so it also doubles (factor $2$). Look at Maxwell-Boltzmann Speed Distribution: the whole speed curve stretches to the right by $\sqrt{2}$ — see the figure below.

WHAT: ratio of two rms speeds at equal $T$. WHY: at equal $T$ the $3RT$ factor is identical, so only the masses differ. $\frac{v_{rms}(H_2)}{v_{rms}(O_2)}=\sqrt{\frac{3RT/M_{H_2}}{3RT/M_{O_2}}}=\sqrt{\frac{M_{O_2}}{M_{H_2}}}=\sqrt{\frac{0.032}{0.002}}=\sqrt{16}=4$ Hydrogen molecules move exactly 4× faster — yet both gases carry identical mean KE. See the figure.

(a) Macroscopic route. $P=\frac{nRT}{V}=\frac{(0.5)(8.314)(290)}{0.01}=\frac{1205.5}{0.01}=1.2055\times 10^5\text{Pa}$ (b) Microscopic route. First $N=n{N}_A=(0.5)(6.02\times 10^{23})=3.01\times 10^{23}$. The molar mass $M=m{N}_A=(6.6\times 10^{-26})(6.02\times 10^{23})=0.03973\text{kg/mol}$. $v_{rms}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3(8.314)(290)}{0.03973}}=\sqrt{\frac{7233.2}{0.03973}}=\sqrt{1.821\times 10^5}\approx 426.7\text{m/s}$ Now the microscopic pressure: $P=\frac{1}{3}\frac{Nm{v}_{rms}^2}{V}=\frac{1}{3}\frac{(3.01\times 10^{23})(6.6\times 10^{-26})(426.7^2)}{0.01}$ $(3.01\times 10^{23})(6.6\times 10^{-26})=0.019866$. Times $v_{rms}^2=1.821\times 10^5$: $=3617.6$. $P=\frac{1}{3}\cdot \frac{3617.6}{0.01}=\frac{1}{3}(3.6176\times 10^5)\approx 1.206\times 10^5\text{Pa}✓$ The two routes agree — that is the whole point of kinetic theory: microscopic motion reproduces the experimental law.

WHAT: total KE $=N\cdot \overline{E_k}=\frac{3}{2}N{k}_{B}T$. Equivalently $\frac{3}{2}nRT$. WHY: the mole form is cleaner because $N{k}_B=nR$. $E_{tot}=\frac{3}{2}nRT=\frac{3}{2}(0.5)(8.314)(290)=\frac{3}{2}(1205.5)\approx 1808\text{J}$

WHAT: two different averages of the same numbers. WHY: pressure and energy need $\overline{v^2}$ (then its root), not the plain mean — squaring weights the fast molecules more heavily. Mean speed: $\bar{v}=\frac{300+400+500+600}{4}=\frac{1800}{4}=450\text{m/s}$ Mean of squares: $\overline{v^2}=\frac{300^2+400^2+500^2+600^2}{4}=\frac{90000+160000+250000+360000}{4}=\frac{860000}{4}=215000$ $v_{rms}=\sqrt{215000}\approx 463.7\text{m/s}$ Indeed $463.7>450$. The gap exists because squaring inflates large speeds before averaging — a real, general inequality, not a coincidence.

WHAT: use the single-molecule result $F_1=m{v}_x^2/L$ from the parent derivation Step 3. WHY: with just one molecule there is no crowd to average over, but the time-average of its repeated kicks is still well defined. $F_1=\frac{m{v}_x^2}{L}=\frac{(5\times 10^{-26})(500{)}^2}{0.1}=\frac{(5\times 10^{-26})(2.5\times 10^5)}{0.1}=\frac{1.25\times 10^{-20}}{0.1}=1.25\times 10^{-19}\text{N}$ Assigned pressure over the wall area $L^2=0.01{\text{m}}^2$: $P=\frac{F_1}{L^2}=\frac{1.25\times 10^{-19}}{0.01}=1.25\times 10^{-17}\text{Pa}$ The physics point: this "pressure" is a time average of isolated whacks separated by $\Delta t=2L/v_x=4\times 10^{-4}$ s. A real gauge would read essentially zero except for individual pings. The smooth, steady pressure of the parent note only emerges when $N$ is astronomically large so kicks overlap into a continuous push. One molecule shows the limit where the macroscopic concept dissolves.

Which speed goes into the pressure formula directly? ::: $\overline{v^2}$ (mean of squares), not $v_{rms}$ and not $\bar{v}$. If $T$ doubles at constant $V$, what happens to $P$ and to $v_{rms}$? ::: $P$ doubles; $v_{rms}$ grows by $\sqrt{2}$. Ratio of rms speeds of two gases at equal $T$? ::: $\sqrt{M_2/M_1}$ (heavier is slower). Which constant pairs with $N$, and which with $n$? ::: $k_B$ with $N$; $R$ with $n$; and $R=N_A{k}_B$. Is $v_{rms}$ ever equal to $\bar{v}$? ::: Only if all molecules share one identical speed.