Exercises — Ideal gas law PV = nRT — derivation from kinetic theory

WHAT: we want $P$, and we already know $n,T,V$ — every symbol in $PV=nRT$ is known except $P$. WHY this tool: the ideal gas law is the direct relation between exactly these four quantities; no kinetic detail is needed. $P=\frac{nRT}{V}=\frac{3\times 8.314\times 400}{0.02}=\frac{9976.8}{0.02}\approx 4.99\times 10^5\text{Pa}$ Units check: $\frac{\text{mol}\cdot ({\text{J mol}}^{-1}{\text{K}}^{-1})\cdot \text{K}}{{\text{m}}^3}=\frac{\text{J}}{{\text{m}}^3}=\text{Pa}$. ✓

• $P$ — pressure — pascal (Pa = N/m²)
• $V$ — volume — cubic metre (m³)
• $n$ — amount of substance — mole (mol)
• $R$ — universal gas constant — J mol⁻¹ K⁻¹
• $T$ — absolute temperature — kelvin (K)

The trap-free point: $T$ is Kelvin, never Celsius (see the L1 mistake below).

WHAT: solve $PV=nRT$ for $V$. WHY: three of four unknowns given, one target. $V=\frac{nRT}{P}=\frac{0.5\times 8.314\times 320}{1.5\times 10^5}=\frac{1330.24}{1.5\times 10^5}\approx 8.87\times 10^{-3}{\text{m}}^3$ That's about $8.9$ litres — a reasonable lab flask.

WHAT: use the per-mole speed formula. WHY this tool: rms speed follows from equating translational energy per mole $\frac{3}{2}RT$ to $\frac{1}{2}M⟨v^2⟩$ — that scaling is done in the parent's Example 1. $v_{rms}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3\times 8.314\times 300}{0.032}}=\sqrt{\frac{7482.6}{0.032}}=\sqrt{2.338\times 10^5}\approx 483.4\text{m/s}$ Heavier than the $N_2$ of the parent (517 m/s) because $M$ is larger → slower. ✓ See Maxwell-Boltzmann Speed Distribution for the full spread of speeds around this value.

WHAT: use $PV=N{k}_{B}T$ (molecule form). WHY this form: the question asks for $N$ directly, so use $k_B$ not $R$. $N=\frac{PV}{k_{B}T}=\frac{1.0\times 10^5\times 2.0\times 10^{-3}}{1.381\times 10^{-23}\times 273}=\frac{200}{3.770\times 10^{-21}}\approx 5.31\times 10^{22}$ Cross-check using $N_A$: first find the moles, $n=PV/(RT)=200/(8.314\times 273)\approx 0.0881$ mol. Then convert to molecules with Avogadro's number: $N=n{N}_A=0.0881\times 6.022\times 10^{23}\approx 5.31\times 10^{22}$. ✓ Both routes agree, which is exactly the statement $R=N_A{k}_B$.

WHAT: relate two states of the same gas. WHY a ratio: $n,V,R$ are unchanged, so $P/T=nR/V$ is constant. Dividing the two states cancels everything we don't know. $\frac{P_2}{P_1}=\frac{T_2}{T_1}\Rightarrow P_2=P_1\frac{T_2}{T_1}=2.0\times 10^5\times \frac{450}{300}=3.0\times 10^5\text{Pa}$ Microscopic reading: $⟨v^2⟩\propto T$, so at $1.5\times$ the temperature molecules smack the walls harder and more often → pressure rises by $1.5\times$.

WHAT: set the two rms speeds equal and solve for $T_{\text{He}}$. WHY squares matter: $v_{rms}^2=3RT/M$, so equal speeds means equal $T/M$ ratios. $\frac{3R{T}_{\text{He}}}{M_{\text{He}}}=\frac{3R{T}_{N_2}}{M_{N_2}}\Rightarrow T_{\text{He}}=T_{N_2}\frac{M_{\text{He}}}{M_{N_2}}=300\times \frac{0.004}{0.028}\approx 42.9\text{K}$ Helium is $7\times$ lighter, so it reaches the same speed at $\frac{1}{7}$ the temperature. Cold helium is as fast as warm nitrogen — surprising but correct.

WHAT: invert the density form of the kinetic pressure. WHY this form: here we're given $\rho =Nm/V$ (mass per volume), not $M$ or $N$ — the density form $P=\frac{1}{3}\rho ⟨v^2⟩$ is built exactly for that. $⟨v^2⟩=\frac{3P}{\rho }=\frac{3\times 10^5}{1.25}=2.4\times 10^5{\text{m}}^2/{\text{s}}^2$ $v_{rms}=\sqrt{2.4\times 10^5}\approx 489.9\text{m/s}$

Step A — volume. $V=L^3=(0.10{)}^3=1.0\times 10^{-3}{\text{m}}^3$. Step B — pressure (pure mechanics, no temperature yet): $P=\frac{1}{3}\frac{Nm⟨v^2⟩}{V}=\frac{1}{3}\cdot \frac{(1.0\times 10^{22})(4.65\times 10^{-26})(2.4\times 10^5)}{1.0\times 10^{-3}}$ Numerator inside: $1.0\times 10^{22}\times 4.65\times 10^{-26}=4.65\times 10^{-4}$; $\times 2.4\times 10^5=111.6$. Then $\frac{1}{3}\times 111.6/10^{-3}$: $P=\frac{37.2}{10^{-3}}=3.72\times 10^4\text{Pa}$ Step C — force on ONE wall (look at the red wall of the figure, area $L^2$): $F=P\cdot L^2=3.72\times 10^4\times (0.10{)}^2=3.72\times 10^4\times 0.01=372\text{N}$ Step D — temperature from the bridge: $T=\frac{PV}{N{k}_B}=\frac{3.72\times 10^4\times 10^{-3}}{1.0\times 10^{22}\times 1.381\times 10^{-23}}=\frac{37.2}{0.1381}\approx 269.4\text{K}$ So this gas is near freezing — consistent with a modest speed. Every arrow of the derivation used, end to end.

WHAT: the total number of moles is conserved; temperature is fixed. WHY count moles: $PV=nRT$ says $n=PV/(RT)$, and moles just add when the gases merge. Moles before: $n_A=\frac{P_A{V}_A}{RT}$, $n_B=\frac{P_B{V}_B}{RT}$. After opening, total volume $V_f=V_A+V_B$, total moles $n_A+n_B$, same $T$: $P_f(V_A+V_B)=(n_A+n_B)RT=P_A{V}_A+P_B{V}_B$ The $RT$ cancels — beautiful. Solving: $P_f=\frac{P_A{V}_A+P_B{V}_B}{V_A+V_B}=\frac{(3.0\times 10^5)(2.0)+(1.0\times 10^5)(3.0)}{2.0+3.0}$ Unit consistency note: although the litre factors cancel in this ratio, always carry a single consistent unit system through the whole calculation. If you convert to SI, use $V_A=2.0\times 10^{-3}{\text{m}}^3$ and $V_B=3.0\times 10^{-3}{\text{m}}^3$; the $10^{-3}$ then cancels top and bottom, giving the identical answer. Never mix litres in one term with m³ in another — in a problem where units don't cancel, that silent mismatch is a classic wrong answer. $P_f=\frac{6.0\times 10^5+3.0\times 10^5}{5.0}=\frac{9.0\times 10^5}{5.0}=1.8\times 10^5\text{Pa}$ It's just a volume-weighted average of the two pressures.

Step A — start from mechanics: $PV=\frac{1}{3}Nm⟨v^2⟩$. Step B — introduce energy by inserting a factor $\frac{1}{2}$: multiply and divide by 2, $PV=\frac{2}{3}N\left(\frac{1}{2}m⟨v^2⟩\right)=\frac{2}{3}N⟨E_k⟩.$ Step C — equate to the measured law $PV=N{k}_{B}T$: $\frac{2}{3}N⟨E_k⟩=N{k}_{B}T\Rightarrow \boxed{⟨E_k⟩=\frac{3}{2}{k}_{B}T}.$ WHY it depends only on $T$: mass $m$ and speed cancelled out; nothing about the type of molecule survives — only $T$. This is exactly the Equipartition Theorem result for 3 translational degrees of freedom. Numeric at 300 K: $⟨E_k⟩=\frac{3}{2}\times 1.381\times 10^{-23}\times 300=6.215\times 10^{-21}\text{J}.$ Consequence: helium and nitrogen at 300 K have identical $⟨E_k⟩=6.22\times 10^{-21}$ J. The light helium simply moves faster to carry the same energy — connecting back to Temperature and Internal Energy.

WHAT: compare ideal $P_{\text{ideal}}=nRT/V$ with the volume-corrected $P_{\text{corr}}=nRT/(V-nb)$. WHY (V − nb): point particles could use the whole box, but real molecules occupy space, so less room is free — see Real Gases and Van der Waals Equation. The ratio makes $nRT$ cancel: $\frac{P_{\text{corr}}}{P_{\text{ideal}}}=\frac{V}{V-nb}=\frac{1.0\times 10^{-3}}{1.0\times 10^{-3}-1\times 3.2\times 10^{-5}}=\frac{1.0\times 10^{-3}}{9.68\times 10^{-4}}\approx 1.03306$ So the corrected pressure is about $3.31\%$ higher: $\text{underestimate}=\left(\frac{V}{V-nb}-1\right)\times 100\%=\frac{nb}{V-nb}\times 100\%\approx 3.31\%.$ The ideal law under-predicts pressure here because it wrongly assumes molecules can access the full volume.

Symbols first. A molecule's velocity has three perpendicular components $(v_x,v_y,v_z)$, one along each axis of the box. $v_x$ is just how fast it moves in the $x$-direction; $⟨v_x^2⟩$ is the average of the square of that component over every molecule. The full speed obeys $v^2=v_x^2+v_y^2+v_z^2$. WHAT: relate the full-speed spread to the one-axis spread. WHY $\sqrt{3}$: isotropy gave $⟨v^2⟩=3⟨v_x^2⟩$ in Step 5 of the parent derivation, because no direction is special ($⟨v_x^2⟩=⟨v_y^2⟩=⟨v_z^2⟩$). Take square roots: $v_{rms}=\sqrt{⟨v^2⟩}=\sqrt{3⟨v_x^2⟩}=\sqrt{3}\sqrt{⟨v_x^2⟩}.$ For oxygen, $v_{rms}\approx 483.4$ m/s (Ex 2.2), so $\sqrt{⟨v_x^2⟩}=\frac{v_{rms}}{\sqrt{3}}=\frac{483.4}{1.732}\approx 279.1\text{m/s}.$ Each single axis carries $1/\sqrt{3}$ of the total speed spread — the geometric heart of the factor $\frac{1}{3}$.

Which form of the gas law uses molecule count $N$? ::: $PV=N{k}_{B}T$ (pairs with $k_B$). At fixed $V,n$, doubling $T$ does what to $P$? ::: doubles it ($P\propto T$). To double $v_{rms}$, temperature must rise by what factor? ::: four times ($v_{rms}\propto \sqrt{T}$). Average translational KE per molecule at temperature $T$? ::: $\frac{3}{2}{k}_{B}T$, independent of gas type. Merging two tanks at fixed $T$: what is conserved? ::: total moles, i.e. $PV$ adds.