Kinetic Theory of Gases

A perfume bottle is opened at one end of a still room. Its molecules leave at roughly $400$ m s${}^{-1}$ — faster than a passenger aircraft. How long before the scent reaches someone $5$ m away?

By pure diffusion, about ten days.

The molecule is fast but it does not travel. At atmospheric pressure it covers only about $60$ nm before colliding, and each collision sends it off in a fresh random direction. It is performing a random walk, and in a random walk the net displacement after $N$ steps is not $N\lambda$ but

$d=\lambda \sqrt{N}$

To cover $5$ m in steps of $6\times 10^{-8}$ m needs $N={\left(5/6\times 10^{-8}\right)}^2\approx 7\times 10^{15}$ collisions. At roughly $8\times 10^9$ collisions per second, that is $9\times 10^5$ s.

Real rooms are far quicker than this, because convection carries whole parcels of air rather than individual molecules. But the calculation makes the point that drives this chapter: at Advanced level the interesting quantities are the distribution and the collisions, not the average. Main asks for $v_{rms}$. Advanced asks how many molecules are near it, how far one gets between collisions, and what happens when the classical picture stops working.

1. Mean free path and collision frequency

If molecules of diameter $d$ move with number density $n$, one molecule sweeps a cylinder of cross-section $\pi d^2$ and collides with anything whose centre falls inside. Treating the others as stationary gives $\lambda =1/n\pi d^2$ — but they are not stationary. Averaging over the relative speed of two molecules introduces a factor $\sqrt{2}$:

$\lambda =\frac{1}{\sqrt{2}n\pi d^2}=\frac{kT}{\sqrt{2}\pi d^{2}P}$

$z=\frac{\bar{v}}{\lambda }=\sqrt{2}\pi d^2\bar{v}n$

Read the second form of $\lambda$ carefully. At constant temperature $\lambda \propto 1/P$. At constant volume, $n$ is fixed, so heating a sealed vessel does not change $\lambda$ at all — though it does raise the collision frequency, because the molecules cover the same distance faster.

Illustration 1

Find the mean free path and collision frequency for nitrogen at $300$ K and $1$ atm. Take $d=3.7\times 10^{-10}$ m.

$n=\frac{P}{kT}=\frac{1.01\times 10^5}{1.38\times 10^{-23}\times 300}=2.44\times 10^{25}$ m${}^{-3}$

$\lambda =\frac{1}{\sqrt{2}\left(2.44\times 10^{25}\right)\pi {\left(3.7\times 10^{-10}\right)}^2}=6.7\times 10^{-8}$ m

With $\bar{v}=476$ m s${}^{-1}$, $z=\frac{476}{6.7\times 10^{-8}}=7.1\times 10^9$ s${}^{-1}$.

Compare $\lambda$ with $d$: the free path is about $180$ molecular diameters, which is why a gas at ordinary pressure is still describable as mostly empty space with occasional collisions.

Illustration 2

A sealed rigid vessel of gas is heated from $300$ K to $1200$ K. By what factors do $\lambda$, $\bar{v}$ and $z$ change?

$n$ is fixed, so $\lambda$ is unchanged.

$\bar{v}\propto \sqrt{T}$, so it doubles.

$z=\bar{v}/\lambda$ therefore doubles as well.

The contrast with an isothermal compression is worth holding on to. There, $\bar{v}$ is fixed and $\lambda$ falls, so $z$ rises for the opposite reason. Naming which of $n$ and $T$ is held fixed settles every question of this type.

2. The Maxwell speed distribution

The number of molecules with speeds between $v$ and $v+dv$ is $Nf(v)dv$, where

$f(v)=4\pi {\left(\frac{m}{2\pi kT}\right)}^{3/2}{v}^2{e}^{-m{v}^2/2kT}$

The product of a rising $v^2$ and a falling exponential makes a curve that starts at zero, peaks, and has a long tail. Three speeds are extracted from it, and they are always in the same order:

$v_p=\sqrt{\frac{2RT}{M}},\bar{v}=\sqrt{\frac{8RT}{\pi M}},v_{rms}=\sqrt{\frac{3RT}{M}}$

$v_p:\bar{v}:v_{rms}=1:1.128:1.225$

Two features carry most of the marks. Raising $T$ or lowering $M$ shifts the peak right and flattens the curve, because the total area is fixed at the number of molecules. And because the tail is exponential, the fraction of molecules above a high threshold is extraordinarily sensitive to temperature — which is the kinetic-theory reason why reaction rates and evaporation depend so sharply on it.

Illustration 3

Find $v_p$, $\bar{v}$ and $v_{rms}$ for oxygen at $300$ K, with $M=32$ g mol${}^{-1}$.

$v_p=\sqrt{\frac{2\left(8.314\right)\left(300\right)}{0.032}}=\sqrt{1.559\times 10^5}=395$ m s${}^{-1}$

$\bar{v}=1.128\times 395=446$ m s${}^{-1}$

$v_{rms}=1.225\times 395=484$ m s${}^{-1}$

Only compute one of them. The ratio $1:1.128:1.225$ is fixed for every gas at every temperature, so the other two follow by multiplication.

Illustration 4

At what temperature does hydrogen have the same $v_{rms}$ as oxygen at $300$ K?

$\frac{3R{T}_H}{2}=\frac{3R(300)}{32}\Rightarrow T_H=\frac{2}{32}\times 300$

$T_H=18.75$ K

This is why hydrogen and helium have escaped the Earth's atmosphere while nitrogen has not. At any shared temperature the lightest molecules are the fastest, and enough of them exceed escape velocity to be lost over geological time.

Illustration 5

Compare the number of molecules near $2v_p$ with the number near $v_p$, at the same temperature.

$\frac{f(2v_p)}{f(v_p)}=\frac{{\left(2v_p\right)}^2{e}^{-4}}{v_p^2{e}^{-1}}=4e^{-3}$

$=4\times 0.0498=0.199$

Only a fifth as many, at twice the most probable speed. The exponential beats the $v^2$ handily once past the peak, and this is what makes the high-speed tail so thin.

3. Momentum flux at the wall

Pressure is momentum delivered per second per unit area. A molecule with velocity component $v_x$ rebounds elastically carrying $2m{v}_x$, and the number reaching area $A$ in time $dt$ is $nA{v}_{x}dt$ for that component. Multiplying and averaging,

$P=nm⟨v_x^2⟩=\frac{1}{3}nm⟨v^2⟩$

The one third is an isotropy factor, nothing more. A directed beam of the same density and speed, all molecules moving squarely at the wall, would deliver $2nm{v}^2$ — six times as much, because none of the molecules is travelling the wrong way and each rebounds through the full $2mv$.

Averaging the same flux over all directions in a hemisphere gives the rate at which molecules arrive at unit area:

$\Phi =\frac{1}{4}n\bar{v}$

Two consequences follow at once. Different species contribute independent momentum fluxes, so their partial pressures simply add — that is Dalton's law, derived rather than asserted. And if the wall is retreating at speed $u$, each molecule returns slower by $2u$, which is the microscopic picture of a gas cooling as it expands adiabatically.

Illustration 6

How many nitrogen molecules strike each square centimetre of a wall per second at $300$ K and $1$ atm?

$\Phi A=\frac{1}{4}n\bar{v}A=\frac{1}{4}\left(2.44\times 10^{25}\right)\left(476\right)\left(10^{-4}\right)$

$=2.9\times 10^{23}$ per second

That is roughly half a mole every second on a fingernail-sized patch. The steadiness of pressure is a statistical illusion produced by the sheer number of impacts, and the fluctuations go as $1/\sqrt{N}$.

4. Effusion and Graham's law

Molecules strike unit area of a wall at the rate $\frac{1}{4}n\bar{v}$. Punch a hole small compared with $\lambda$ and that same expression becomes the effusion rate, since molecules leave one at a time without disturbing the distribution:

$\text{rate}=\frac{1}{4}n\bar{v}A\propto \frac{1}{\sqrt{M}}$

which is Graham's law. The escaping gas is slightly enriched in the lighter component, and repeating the step many times is how uranium isotopes were separated.

Illustration 7

Natural uranium hexafluoride contains ${}^{235}$UF${}_6$ ($M=349$) and ${}^{238}$UF${}_6$ ($M=352$). Find the enrichment factor per effusion stage.

$\frac{r_{235}}{r_{238}}=\sqrt{\frac{352}{349}}=\sqrt{1.0086}=1.0043$

A gain of $0.43\%$ per stage. Thousands of stages in cascade are needed, which is precisely why isotope separation is an industrial undertaking rather than a laboratory one.

Illustration 8

Two gases effuse through identical pinholes under identical conditions, and gas A takes $20$ s to deliver the same number of moles that gas B delivers in $30$ s. If $M_B=64$, find $M_A$.

Rate is inversely proportional to time, so $\frac{r_A}{r_B}=\frac{30}{20}=1.5=\sqrt{\frac{64}{M_A}}$

$M_A=\frac{64}{2.25}=28.4$ g mol${}^{-1}$

The square root halves every discrepancy. A gas must be more than four times lighter to effuse twice as fast, which is why Graham's law separates so slowly.

5. Degrees of freedom, and where equipartition fails

Equipartition assigns $\frac{1}{2}kT$ to every quadratic term in the energy. A vibrational mode contributes two such terms, kinetic and potential, so it carries $kT$ rather than $\frac{1}{2}kT$. With $f$ effective degrees of freedom,

$C_V=\frac{f}{2}R,\gamma =1+\frac{2}{f}$

But $f$ is not a fixed property of a molecule. Rotational and vibrational modes are quantised, and a mode contributes nothing until $kT$ is comparable to its energy spacing. Hydrogen shows the whole staircase.

Below about $100$ K only translation is active and $C_V=\frac{3}{2}R$; at room temperature rotation has joined and $C_V=\frac{5}{2}R$; above a few thousand kelvin vibration contributes and $C_V\to \frac{7}{2}R$. Classical equipartition has no mechanism for this — it was one of the first clear failures of classical physics, and the resolution is quantum.

Illustration 9

Find $\gamma$ for a diatomic gas at a temperature high enough that vibration is fully active.

$f=3\text{translational}+2\text{rotational}+2\text{vibrational}=7$

$C_V=\frac{7}{2}R,C_P=\frac{9}{2}R,\gamma =\frac{9}{7}=1.286$

Note the direction of the change. Every mode that switches on raises $C_V$ and pushes $\gamma$ towards 1, which is why hot gases are always closer to isothermal in their behaviour than cold ones.

Illustration 10

A vessel contains a gas whose measured $\gamma$ is $1.4$ at $300$ K but $1.33$ at $2000$ K. Interpret the change.

At $300$ K, $\gamma =1+2/f$ gives $f=5$: three translational and two rotational modes, a rigid diatomic.

At $2000$ K, $f=2/0.33=6$: one extra degree of freedom.

A vibrational mode is partially excited. Because vibration contributes two quadratic terms, $f$ would reach $7$ if it were fully active; the intermediate value $6$ means the mode is switching on but not yet saturated.

6. The Boltzmann factor and the atmosphere

Equipartition tells you the energy per mode; the Boltzmann factor tells you how many molecules have a given energy at all. For any energy $E$,

$n\propto e^{-E/kT}$

Applied to gravitational potential energy $mgh$, this gives the isothermal atmosphere:

$n(h)=n_0{e}^{-mgh/kT},H=\frac{kT}{mg}=\frac{RT}{Mg}$

The scale height $H$ is the altitude over which the density falls by a factor $e$. For air at $273$ K it is about $8$ km, which is why aircraft need pressurisation and why the highest mountains sit near the limit of unaided breathing.

Illustration 11

Find the ratio of atmospheric density at $16$ km to that at sea level, assuming a uniform temperature of $273$ K and $M=29$ g mol${}^{-1}$.

$H=\frac{RT}{Mg}=\frac{8.314\times 273}{0.029\times 9.8}=7990$ m

$\frac{n}{n_0}=e^{-16000/7990}=e^{-2.0}=0.135$

Just $13.5\%$ of sea-level density. The real atmosphere falls off faster still, because temperature drops with altitude and the isothermal assumption then overestimates $H$.

The same factor drives a gas centrifuge

Replace gravity by the centrifugal potential $-\frac{1}{2}m{\omega }^2{r}^2$ and the Boltzmann factor gives

$n(r)=n(0)e^{M{\omega }^2{r}^2/2RT}$

so heavier molecules crowd towards the rim. The separation factor between two isotopes of molar mass difference $\Delta M$ is $\exp \left(\Delta M{\omega }^2{r}^2/2RT\right)$, and because $\omega r$ can reach several hundred metres per second this beats effusion decisively.

Illustration 12

A gas centrifuge spins UF${}_6$ with a rim speed of $500$ m s${}^{-1}$ at $300$ K. Find the separation factor per stage and compare it with effusion.

$\frac{\Delta M{\left(\omega r\right)}^2}{2RT}=\frac{3\times 10^{-3}\times 2.5\times 10^5}{2\times 8.314\times 300}=\frac{750}{4988}=0.150$

Separation factor $=e^{0.150}=1.162$, a gain of $16\%$ per stage.

Against $0.43\%$ for effusion, that is nearly forty times better, which is why every modern enrichment plant is a centrifuge cascade and the enormous diffusion plants of the 1940s were abandoned.

7. Real gases: van der Waals and the critical point

Two corrections turn the ideal gas law into the van der Waals equation. Molecules occupy volume, reducing the space available by $nb$; and they attract one another, reducing the pressure on the wall by $a{n}^2/V^2$:

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

At the critical point the isotherm has an inflection with a horizontal tangent, so $\frac{\partial P}{\partial V}=\frac{{\partial }^{2}P}{\partial V^2}=0$. Solving both conditions together gives

$V_c=3b,T_c=\frac{8a}{27Rb},P_c=\frac{a}{27b^2}$

Two further quantities are standard. The compressibility factor $Z=PV/nRT$ equals $1$ for an ideal gas; at the critical point every van der Waals gas gives $Z_c=3/8$. And the Boyle temperature $T_B=a/Rb$ is where the two corrections cancel, so the gas behaves ideally over a wide pressure range.

Illustration 13

For carbon dioxide, $a=3.6$ atm L${}^2$ mol${}^{-2}$ and $b=4.3\times 10^{-2}$ L mol${}^{-1}$. Find $T_c$ and $P_c$.

$T_c=\frac{8a}{27Rb}=\frac{8\times 3.6}{27\times 0.0821\times 0.043}=\frac{28.8}{0.0953}=302$ K

$P_c=\frac{a}{27b^2}=\frac{3.6}{27\times 1.85\times 10^{-3}}=72$ atm

Against measured values of $304$ K and $73$ atm, which is close enough to confirm that two crude corrections capture most of what non-ideality does.

Illustration 14

A gas has $Z=0.4$ at some pressure and temperature, and $Z=1.6$ at another. What dominates in each case?

$Z<1$ means the gas is more compressible than ideal, so the attractive term $a/V^2$ dominates — typical at moderate pressures and lower temperatures.

$Z>1$ means it resists compression, so the excluded-volume term $b$ dominates — typical at high pressures where molecules are crowded.

At the Boyle temperature the two effects cancel, giving $Z\approx 1$ over a usefully wide range, which is why that temperature is quoted for gases used as pressure standards.

Summary

• Molecules are fast but travel slowly: a random walk gives net displacement $\lambda \sqrt{N}$, not $N\lambda$.
• $\lambda =\frac{1}{\sqrt{2}n\pi d^2}=\frac{kT}{\sqrt{2}\pi d^{2}P}$; the $\sqrt{2}$ comes from averaging over relative speeds.
• At constant $T$, $\lambda \propto 1/P$. At constant $V$, $\lambda$ is unchanged by heating, though $z=\bar{v}/\lambda$ rises.
• Maxwell distribution $f(v)\propto v^2{e}^{-m{v}^2/2kT}$: a rising $v^2$ against a falling exponential.
• $v_p:\bar{v}:v_{rms}=1:1.128:1.225$ always. Compute one and scale.
• Raising $T$ or lowering $M$ shifts the peak right and flattens the curve, since the area is fixed.
• $P=\frac{1}{3}nm⟨v^2⟩$; the one third is isotropy. A directed beam of the same density delivers six times as much.
• Arrival rate at a wall is $\frac{1}{4}n\bar{v}$, and independent momentum fluxes give Dalton's law.
• Effusion rate $=\frac{1}{4}n\bar{v}A\propto 1/\sqrt{M}$ — Graham's law, and the basis of isotope separation.
• A vibrational mode carries $kT$, not $\frac{1}{2}kT$, because it has both kinetic and potential quadratic terms.
• $C_V=\frac{f}{2}R$ and $\gamma =1+\frac{2}{f}$, but $f$ grows with temperature as quantised modes switch on: hydrogen goes $\frac{3}{2}R\to \frac{5}{2}R\to \frac{7}{2}R$.
• Every mode that activates pushes $\gamma$ towards $1$.
• Boltzmann factor $n\propto e^{-E/kT}$; the isothermal atmosphere has scale height $H=RT/Mg\approx 8$ km.
• The same factor with a centrifugal potential gives $n\propto e^{M{\omega }^2{r}^2/2RT}$ — a centrifuge separates isotopes about forty times better than effusion.
• Van der Waals: $\left(P+a{n}^2/V^2\right)\left(V-nb\right)=nRT$, with $V_c=3b$, $T_c=8a/27Rb$, $P_c=a/27b^2$ and $Z_c=3/8$.
• $Z<1$ means attraction dominates; $Z>1$ means excluded volume dominates; at $T_B=a/Rb$ they cancel.