Kinetic Theory of Gases

Helium and xenon are held at the same temperature. Which has the greater average kinetic energy per molecule, and which moves faster?

Same kinetic energy. Xenon moves far slower.

$\frac{1}{2}m{v}_{rms}^2=\frac{3}{2}{k}_{B}T$

Temperature fixes the energy, not the speed. Xenon is ~33 times heavier than helium, so its molecules move $\sqrt{33}\approx 5.7$ times slower to carry the identical energy.

That equation is the whole chapter. Everything before it builds it; everything after uses it.

1. The ideal gas

$PV=nRT=N{k}_{B}T$

$R=8.314$ J mol⁻¹K⁻¹ — the same for every gas, which is itself a strong hint that gases share one mechanism. $k_B=R/N_A=1.38\times 10^{-23}$ J K⁻¹ is just $R$ per molecule.

Every empirical gas law is a special case: Boyle ($T$, $n$ fixed), Charles ($P$ fixed), Gay-Lussac ($V$ fixed), Avogadro (equal volumes hold equal numbers).

Temperature must always be absolute. Using Celsius is the commonest arithmetic error here, and it never produces an obviously silly answer — which is what makes it dangerous.

Avogadro's number is the bridge between the two halves of that equation. $N_A=6.022\times 10^{23}$ per mole is the count that converts a laboratory quantity you can weigh into a molecular quantity you can only reason about, and it is what turns $nR$ into $N{k}_B$.

Since 2019 it is a defined constant, fixed at exactly $6.02214076\times 10^{23}$, so the mole is now specified by counting rather than by a reference kilogram of carbon.

Illustration 1

A sealed rigid vessel of gas is at $27°\text{C}$ and $2$ atm. It is heated to $127°\text{C}$. Find the new pressure. If instead the gas were allowed to expand at constant pressure, by what factor would the volume grow?

Absolute temperatures first — this is where the marks are lost:

$T_1=300\text{K},T_2=400\text{K}$

Rigid vessel means $V$ fixed, so $\frac{P}{T}$ is constant:

$P_2=2\left(\frac{400}{300}\right)=2.67\text{atm}$

At constant pressure instead, $\frac{V}{T}$ is constant, so the volume grows by $\frac{400}{300}=1.33$ times.

The trap: using $27$ and $127$ directly gives a factor of $4.7$ rather than $1.33$ — a wrong answer that still looks perfectly plausible, which is exactly why it is set.

2. Work done on a gas

The gas laws describe a gas; work is how it exchanges energy with the world. When a gas expands against a piston it pushes through a distance, so it does work:

$W={\int }_{V_1}^{V_2}PdV$

Read that as an area under the $P$–$V$ curve, exactly as $\int Fdx$ was an area under $F$–$x$.

Sign convention. $W$ positive means the gas expanded and did work on its surroundings. Compressing a gas means work is done on it, so $W$ is negative. Half the errors in this topic are sign errors, not integration errors.

For an isochoric process the piston does not move, so no work is done however much the pressure changes. That is why a rigid vessel is the simplest case in the table.

The energy bookkeeping that connects this work to heat and internal energy is the first law, treated in the Thermodynamics chapter. Here we need only the mechanical half.

Illustration 2

One mole of an ideal gas at $300\text{K}$ expands isothermally from $10\text{L}$ to $20\text{L}$. Find the work done by the gas. Compare it with the work if the same expansion happened at constant pressure equal to the initial pressure.

Isothermal:

$W=nRT\ln \frac{V_2}{V_1}=1(8.314)(300)\ln 2=2494(0.693)=1729\text{J}$

Isobaric at the initial pressure $P_1=\frac{nRT}{V_1}=\frac{2494}{0.01}=2.494\times 10^5\text{Pa}$:

$W=P_1\Delta V=2.494\times 10^5(0.01)=2494\text{J}$

Why the isobaric route does more work: holding the pressure up as the gas expands means pushing harder over the whole stroke, whereas in the isothermal case the pressure falls as the volume grows. The ratio is exactly $1/\ln 2=1.44$.

3. Assumptions

1. Very many identical molecules in constant random motion, obeying Newton's laws.
2. Point masses — their own volume is negligible against the container.
3. No intermolecular forces except during collisions; straight-line flight between.
4. Collisions perfectly elastic and of negligible duration.
5. $N$ large enough that statistical averages mean something.

Note which two fail first. Assumption 2 fails at high pressure; assumption 3 fails at low temperature. Section 10 returns to exactly this.

4. Deriving the pressure

One molecule of mass $m$ bounces elastically off a wall $\perp$ to $x$:

• Momentum change: $+m{v}_x\to -m{v}_x$, so the wall receives $2m{v}_x$ per hit.
• In a cube of side $L$, it returns after travelling $2L$, so time between hits $=2L/v_x$.
• Average force from this molecule $=\frac{2m{v}_x}{2L/v_x}=\frac{m{v}_x^2}{L}$

Sum over $N$ molecules, divide by area $L^2$: $P=\frac{Nm⟨v_x^2⟩}{L^3}$

Randomness means no direction is special, so $⟨v_x^2⟩=\frac{1}{3}⟨v^2⟩$:

$P=\frac{1}{3}\frac{Nm}{V}⟨v^2⟩=\frac{1}{3}\rho v_{rms}^2$

The density form is the useful one — it gives molecular speeds from bulk measurements alone, with no need for molecular mass.

Rewritten with total translational KE: $PV=\frac{2}{3}E$. That is the cleanest statement of the derivation.

Illustration 3

A cubical box of side $0.1\text{m}$ holds $3\times 10^{22}$ nitrogen molecules with an rms speed of $500\text{m/s}$. Find the pressure. ($M=0.028$ kg/mol.)

Mass of one molecule: $m=\frac{0.028}{6.022\times 10^{23}}=4.65\times 10^{-26}\text{kg}$

Density: $\rho =\frac{Nm}{V}=\frac{3\times 10^{22}(4.65\times 10^{-26})}{10^{-3}}=1.40{\text{kg/m}}^3$

$P=\frac{1}{3}\rho v_{rms}^2=\frac{1}{3}(1.40)(2.5\times 10^5)=1.16\times 10^5\text{Pa}$

Check by a completely different route. The same rms speed fixes the temperature:

$T=\frac{m{v}_{rms}^2}{3k_B}=\frac{4.65\times 10^{-26}(2.5\times 10^5)}{3(1.38\times 10^{-23})}=281\text{K}$

with $n=\frac{3\times 10^{22}}{6.022\times 10^{23}}=0.0498$ mol, so $P=\frac{nRT}{V}=\frac{0.0498(8.314)(281)}{10^{-3}}=1.16\times 10^5$ Pa.

The mechanical route and the equation of state agree — which is the whole point of the derivation.

Dalton's law falls out free. Nothing above required the molecules to be identical — each species bounces independently, so each contributes its own share:

$P_{total}=P_1+P_2+\dots$

Kinetic theory makes this obvious rather than empirical: with no interaction between collisions, one species has no mechanism to alter another's contribution. In practice, partial pressure = mole fraction × total pressure.

5. What temperature actually is

Compare the two expressions for $PV$: kinetic theory gives $\frac{2}{3}E$, the equation of state gives $N{k}_{B}T$. Setting them equal and dividing by $N$:

$\frac{1}{2}m{v}_{rms}^2=\frac{3}{2}{k}_{B}T$

Temperature is not correlated with molecular motion. It is the average translational kinetic energy per molecule, up to the constant $\frac{3}{2}{k}_B$.

• $\frac{3}{2}{k}_{B}T$ per molecule regardless of the gas. Helium and xenon match exactly.
• Absolute zero is where translational motion would cease — which is why Kelvin has a genuine zero and Celsius does not.
• For a monatomic ideal gas, $U=\frac{3}{2}nRT$: a function of temperature alone, exactly as thermodynamics assumed.

Illustration 4

Find the average translational kinetic energy of one molecule and of one mole of any ideal gas at $300\text{K}$. To what temperature must the gas be raised to double the rms speed?

Per molecule: $\frac{3}{2}{k}_{B}T=1.5(1.38\times 10^{-23})(300)=6.21\times 10^{-21}\text{J}$

Per mole: $\frac{3}{2}RT=1.5(8.314)(300)=3741\text{J}$

Neither answer mentions which gas it is, because neither depends on it.

For the speed: $v_{rms}\propto \sqrt{T}$, so doubling the speed needs four times the absolute temperature:

$T=4(300)=1200\text{K}$, i.e. $927°\text{C}$

The trap: "double the speed" tempts $600$ K. The square root is the entire content of the question.

6. Molecular speeds

Three averages, and questions routinely test whether you can tell them apart:

$v_{mp}<v_{avg}<v_{rms}\text{in the ratio }\sqrt{2}:\sqrt{8/\pi }:\sqrt{3}$

They differ because the distribution is asymmetric — a symmetric one would put all three at the same place.

The commonest numerical slip. $M$ must be in kg/mol. Oxygen is 0.032, not 32. Getting this wrong is a factor of $\sqrt{1000}\approx 31.6$ out.

All three go as $\sqrt{T}$ and as $1/\sqrt{M}$. Quadrupling the temperature only doubles the speed.

Graham's law follows immediately: rate of diffusion $\propto 1/\sqrt{M}$.

Illustration 5

Find the rms speed of nitrogen at $300\text{K}$. Take $M=28\text{g/mol}$, $R=8.314$.

The whole question is the units. $M$ must be in kg/mol:

$v_{rms}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3(8.314)(300)}{0.028}}=\sqrt{2.67\times 10^5}\approx 517\text{m/s}$

Using $M=28$ instead would give $5.17\text{m/s}$ — slower than a jogger, and out by exactly $\sqrt{1000}\approx 31.6$.

Sanity check it against something you know: this is comfortably above the speed of sound in air (~340 m/s), which it must be, since sound propagates by molecular collisions and cannot outrun the molecules carrying it.

Illustration 6

Equal volumes of two gases effuse through the same pinhole. Gas A takes 30 s, gas B takes 45 s. If $M_A=32$, find $M_B$.

Graham's law gives the rate as $\propto 1/\sqrt{M}$, so the time for a fixed volume is $\propto \sqrt{M}$:

$\frac{t_B}{t_A}=\sqrt{\frac{M_B}{M_A}}\Rightarrow \frac{45}{30}=1.5=\sqrt{\frac{M_B}{32}}$

$M_B=32(1.5{)}^2=72\text{g/mol}$

The trap is the inversion. Writing $t\propto 1/\sqrt{M}$ gives $M_B=32/2.25=14.2$, and the heavier gas would come out lighter. Slower effusion means a heavier molecule, so always sanity-check the direction before trusting the algebra.

7. Equipartition

A degree of freedom is an independent way a molecule can store energy.

A monatomic atom has no rotational energy — the mass sits in an essentially point-like nucleus. A diatomic gets only two rotational modes, because spinning about the bond axis itself contributes nothing for the same reason.

Vibration adds two, not one, because a vibrating bond stores energy as both $\frac{1}{2}m{v}^2$ and $\frac{1}{2}k{x}^2$.

Equipartition: each quadratic degree of freedom carries $\frac{1}{2}{k}_{B}T$ per molecule, or $\frac{1}{2}RT$ per mole. The word quadratic is doing real work — the energy must go as the square of a coordinate or a momentum, which is exactly why vibration counts twice.

Illustration 7

Find the internal energy of 2 mol of oxygen at $300\text{K}$, treating it as a rigid diatomic. How much of it is rotational?

$f=5$, so

$U=\frac{f}{2}nRT=\frac{5}{2}(2)(8.314)(300)=12,471\text{J}$

Each degree of freedom carries an equal share, so the split is simply $3:2$:

Cross-check: the translational part must be $\frac{3}{2}nRT=1.5(2)(8.314)(300)=7483$ J, whatever the molecule is. It matches, because translation always carries exactly three degrees of freedom.

8. Specific heats — and where the theory breaks

$C_V=\frac{f}{2}R,C_P=\left(\frac{f}{2}+1\right)R,\gamma =1+\frac{2}{f}$

$\gamma$ falls as $f$ rises — extra modes soak up heat without contributing to the pressure that does the expansion work.

Now the honest part. Measured $C_V$ for nitrogen at room temperature is close to $2.5R$. The vibrational mode contributes essentially nothing — even though the molecule certainly can vibrate.

Classical equipartition cannot explain that. It predicts every available mode is active at every temperature, so nitrogen should show $3.5R$ always.

The resolution is quantum. Vibrational energy levels are widely spaced, so at 300 K almost no molecule can reach the first excited vibrational state. The mode is frozen out.

Hydrogen's measured $C_V$ therefore climbs in steps as it is heated — about $1.5R$ at very low temperature (translation only), $2.5R$ across a wide middle range (rotation switches on), approaching $3.5R$ only at high temperature. That staircase was one of the earliest clear signs that classical physics was incomplete.

Illustration 8

A mixture holds 2 mol of helium and 3 mol of oxygen (rigid diatomic). Find $C_V$ and $\gamma$ for the mixture.

Specific heats add by moles — $\gamma$ does not.

$C_V=\frac{2(1.5R)+3(2.5R)}{5}=\frac{3R+7.5R}{5}=2.1R$

$C_P=C_V+R=3.1R\Rightarrow \gamma =\frac{3.1}{2.1}\approx 1.48$

Averaging the gammas gives $\frac{2(1.67)+3(1.40)}{5}=1.51$ — wrong, and close enough to sit in the options as a distractor.

9. Mean free path

$\lambda =\frac{1}{\sqrt{2}\pi d^{2}n}=\frac{k_{B}T}{\sqrt{2}\pi d^{2}P}$

$d$ is the molecular diameter, $n$ the number density. The $\sqrt{2}$ accounts for the other molecules moving too, rather than sitting still.

$\lambda \propto T$ at fixed pressure, and $\lambda \propto 1/P$ at fixed temperature. Pumping to high vacuum lengthens it enormously — which is what makes electron beams and thin-film deposition possible.

For air at ordinary conditions: $\lambda \approx 70$ nm, roughly 200 molecular diameters, with several billion collisions per second.

That collision rate resolves an old puzzle. Molecules travel at ~500 m/s, so a smell should cross a room instantly — yet diffusion takes minutes. The path is a random walk of billions of tiny steps, not a straight line.

Illustration 9

Estimate the mean free path of nitrogen at $300\text{K}$ and $1$ atm, taking the molecular diameter as $3.7\times 10^{-10}\text{m}$. Then find the collision frequency, given $v_{rms}=517\text{m/s}$.

Number density first, from the equation of state in the form $P=n{k}_{B}T$:

$n=\frac{P}{k_{B}T}=\frac{1.013\times 10^5}{1.38\times 10^{-23}(300)}=2.45\times 10^{25}{\text{m}}^{-3}$

$\lambda =\frac{1}{\sqrt{2}\pi d^{2}n}=\frac{1}{1.414(3.1416)(1.37\times 10^{-19})(2.45\times 10^{25})}=6.7\times 10^{-8}\text{m}$

That is 67 nm, about 180 molecular diameters — so a molecule really does fly a long way, in its own terms, between collisions.

Collision frequency:

$f=\frac{v_{rms}}{\lambda }=\frac{517}{6.7\times 10^{-8}}=7.7\times 10^9{\text{s}}^{-1}$

Read it back: nearly eight billion collisions every second is what turns a 500 m/s molecule into a smell that takes minutes to cross a room.

10. Real gases

Real gases follow $PV=nRT$ closely at low pressure and high temperature, and deviate at high pressure and low temperature. The two failures map exactly onto the two assumptions:

Those are precisely the two corrections in the van der Waals equation:

$\left(P+\frac{a{n}^2}{V^2}\right)(V-nb)=nRT$

The $b$ term subtracts the volume the molecules themselves occupy, so the space actually available is smaller than $V$. The $a$ term adds back the pressure lost because molecules approaching the wall are pulled inward by their neighbours. Setting $a=b=0$ returns the ideal law exactly.

A gas behaves most ideally when it is far from liquefying, which is why helium and hydrogen are closest to ideal at ordinary conditions and water vapour is among the furthest.

Illustration 10

One mole of carbon dioxide occupies $0.5\text{L}$ at $300\text{K}$. Compare the ideal-gas pressure with the van der Waals pressure. Take $a=3.59$ atm L² mol⁻², $b=0.0427$ L/mol, $R=0.0821$ L atm mol⁻¹K⁻¹.

Ideal: $P=\frac{RT}{V}=\frac{0.0821(300)}{0.5}=49.3\text{atm}$

Van der Waals, rearranged for $P$:

$P=\frac{RT}{V-b}-\frac{a}{V^2}=\frac{24.63}{0.4573}-\frac{3.59}{0.25}=53.9-14.4=39.5\text{atm}$

The two corrections pull in opposite directions, and here the attraction wins:

The real pressure is 20% below ideal. At $300\text{K}$ carbon dioxide is not far above its critical temperature of $304\text{K}$, so attraction dominates — exactly the "low temperature" row of the table. Heat the same sample hard enough and the $b$ term takes over instead, and the gas becomes harder to compress than ideal.

Summary

• $PV=nRT=N{k}_{B}T$. Temperature must be absolute, always.
• $N_A=6.022\times 10^{23}$ is the bridge between the mole you can weigh and the molecule you cannot; it is what turns $nR$ into $N{k}_B$.
• $W=\int PdV$, the area under the $P$–$V$ curve. Isobaric $P\Delta V$, isothermal $nRT\ln (V_2/V_1)$, isochoric zero.
• $P=\frac{1}{3}\rho v_{rms}^2$, derived from momentum transfer at a wall. Equivalently $PV=\frac{2}{3}E$.
• $\frac{1}{2}m{v}_{rms}^2=\frac{3}{2}{k}_{B}T$ — temperature is average translational kinetic energy, the same for every gas.
• $v_{mp}<v_{avg}<v_{rms}$ as $\sqrt{2}:\sqrt{8/\pi }:\sqrt{3}$. All go as $\sqrt{T/M}$. Use $M$ in kg/mol.
• The three differ only because the Maxwell distribution is asymmetric.
• $f$: monatomic 3, rigid diatomic 5, rigid polyatomic 6. Vibration adds two.
• $C_V=\frac{f}{2}R$, $\gamma =1+\frac{2}{f}$. For mixtures combine $C_V$, never $\gamma$.
• Equipartition fails because modes freeze out — a quantum effect, and hydrogen's stepped $C_V$ is the evidence.
• $\lambda =k_{B}T/\sqrt{2}\pi d^{2}P$. Diffusion is slow because the path is a random walk.
• Ideal behaviour breaks at high $P$ (molecular volume) and low $T$ (attraction) — the two van der Waals corrections.
• Van der Waals: $\left(P+\frac{a{n}^2}{V^2}\right)(V-nb)=nRT$. The two corrections push $P$ in opposite directions, and which wins depends on the conditions.