Kinetic Theory of Gases

Beginner

A balloon looks still, yet the air inside it is a swarm of molecules racing about.

Step 1: what a gas is. In a gas they are far apart, with mostly empty space between them, and nothing holds them together. Each one flies in a straight line until it hits another molecule or a wall, bounces, and flies on. The empty space is why you can squeeze air in a bicycle pump but hardly squeeze water at all, and why the smell of toast reaches every room of a house within minutes.

Step 2: where pressure comes from. Every molecule that hits a wall gives it a tiny shove. Think of rain on a tin roof: a few drops make separate taps, but a downpour blends into one steady roar. Molecules are so numerous that their hits blend into a steady push on every bit of wall: the gas pressure.

Step 3: more gas, more pressure. Pump more air into a tyre and more molecules strike each patch of wall every second, so the pressure rises.

Step 4: temperature is motion. The temperature of a gas measures how fast its molecules move on average. Heat it and they speed up, so they hit the walls harder and more often: two reasons for the pressure to rise. At absolute zero, 0 K or −273 °C, the motion would be the smallest possible.

Check yourself: a sealed can of air warms in the sun. Does the number of molecules inside change? Does the pressure? Answer: the number does not change; the pressure rises, because the same molecules move faster.

In the animation, yellow flashes mark hits on the walls. When the flame comes on, molecules that touch the hot floor turn orange and speed up, and the pressure needle climbs.

Intermediate

The ideal gas and what we measure

A gas presses steadily on its walls, yet its molecules touch them only in separate hits. Kinetic theory pictures a gas as an enormous number $N$ of molecules, tiny compared with the gaps between them, flying straight in random directions. They bounce elastically off the walls and off each other and exert no forces between collisions. This model is the ideal gas; air at room conditions follows it to within about 0.1 %.

Four quantities describe the gas as a whole. The pressure $p$ is the force per unit area on the walls, in pascals ($1\mathrm{Pa}=1N/{\mathrm{m}}^2$); the air around you is at about 101 kPa. The volume $V$ is in cubic metres (1 L = 0.001 m³). The temperature $T$ must be in kelvins: add 273.15 to the Celsius reading, so 27 °C is 300 K. The amount is the number of molecules $N$ or of moles $n=N/N_A$, with the Avogadro constant $N_A=6.022\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$.

The ideal gas law

Experiments on many gases and the molecular model lead to the same equation of state, together with the link between its two constants:

• $pV=nRT=NkT$
• $R=N_{A}k$

Here $R=8.314J/(\mathrm{mol}\cdot \mathrm{K})$ is the gas constant for a mole and $k=1.38\times 10^{-23}J/K$ is the Boltzmann constant, the same constant for a single molecule. The law does not care which gas it is: at 300 K and 100 kPa a mole of helium or of nitrogen fills $V=RT/p=24.9\mathrm{L}$.

Hold two variables fixed and the older gas laws appear: at fixed $V$ and $n$, $p/T$ is constant; at fixed $T$, $pV$ is constant (Boyle's law); at fixed $p$, $V/T$ is constant (Charles's law), so a balloon taken out into the frost shrinks. A shortcut: with $V$ in litres, $nRT/V$ comes out in J/L, which is kPa; two moles at 300 K in a rigid 10 L tank are at $2\cdot 8.314\cdot 300/10\approx 499\mathrm{kPa}$.

Pressure and temperature, molecule by molecule

Pressure is the combined effect of impacts. A molecule of mass $m$ that hits a wall with velocity component $v_x$ comes back with $-v_x$, handing the wall a momentum $2m{v}_x$. A faster molecule pushes harder at each hit and returns sooner, so its share grows as $v_x^2$. Adding up all the molecules, with random motion sharing $v^2$ equally among three directions, gives
$p=\frac{1}{3}\frac{N}{V}m⟨v^2⟩$
where $⟨v^2⟩$ is the mean of the squared speeds.

Now compare this with $pV=NkT$. Since $\frac{1}{3}m⟨v^2⟩=\frac{2}{3}⟨\frac{1}{2}m{v}^2⟩$, the two agree only if
$⟨\frac{1}{2}m{v}^2⟩=\frac{3}{2}kT$
Temperature measures the mean kinetic energy of the molecules' motion from place to place, and nothing else. At 300 K the molecules of every gas carry on average $\frac{3}{2}\cdot 1.38\times 10^{-23}\cdot 300=6.2\times 10^{-21}\mathrm{J}$.

Solving for the speed gives the root-mean-square speed, the square root of $⟨v^2⟩$:
$v_{\mathrm{rms}}=\sqrt{\frac{3kT}{m}}=\sqrt{\frac{3RT}{M}}$
with $M$ the molar mass in kg/mol. For nitrogen, $M=0.028kg/mol$ and $v_{\mathrm{rms}}=517m/s$ at 300 K, faster than sound; helium, with $M=0.004kg/mol$, reaches 1370 m/s at the same temperature. Equal energies mean that lighter molecules move faster, as $1/\sqrt{m}$, and heating raises every speed as $\sqrt{T}$.

Worked example: heating the gas in the animation

The box holds nitrogen at 27 °C ($T_1=300\mathrm{K}$) and $p_1=100\mathrm{kPa}$, and the clamped piston keeps $V$ and $n$ fixed. Once the flame is on, a molecule bouncing off the hot floor for the first time leaves with its kinetic energy doubled and its speed multiplied by $\sqrt{2}$. When all have done so, the mean kinetic energy, and with it the kelvin temperature, has doubled:
$T_2=2T_1=600\mathrm{K},p_2=p_1\frac{T_2}{T_1}=100\cdot \frac{600}{300}=200\mathrm{kPa}$
$v_{\mathrm{rms}}=517\cdot \sqrt{2}\approx 731m/s$
The mean kinetic energy per molecule rises from $6.2\times 10^{-21}$ to $1.24\times 10^{-20}\mathrm{J}$.

Follow the readouts: the orange $T$ and the blue $p$ climb together, their ratio fixed at $\frac{1}{3}kPa/K$. Each molecule turns orange when it first touches the floor, and the purple arrow $\vec{v}$ of a heated molecule grows by about 41 %. The run ends before every molecule has reached the floor: two of the sixteen are still blue, so the readouts stop at about 586 K and 195 kPa, with the gauge needle just below its 200 kPa mark.

Worked example: a tyre warming up

On a cold morning at 7 °C a tyre gauge reads 230 kPa; after an hour's driving the air inside is at 42 °C. If the volume hardly changes, what does the gauge read now, and how much air does the 25 L tyre hold?

A tyre gauge shows the excess over the atmosphere, so convert to absolute pressure and kelvins: $p_1=230+101=331\mathrm{kPa}$, $T_1=280\mathrm{K}$, $T_2=315\mathrm{K}$. At fixed volume
$p_2=p_1\frac{T_2}{T_1}=331\cdot \frac{315}{280}\approx 372\mathrm{kPa}$
so the gauge reads $372-101=271\mathrm{kPa}$. The temperature rose by 12.5 % but the gauge reading by 18 %, because the atmosphere's share stays the same; that is why tyre pressures are specified cold.

For the amount, kilopascals times litres give joules: $n=pV/(RT)=331\cdot 25/(8.314\cdot 280)\approx 3.55\mathrm{mol}$, about $2.1\times 10^{24}$ molecules. With a molar mass of 0.029 kg/mol for air, that is roughly 100 g of air carrying a quarter of the car's weight.

Reading the pressure–temperature line

Plot the pressure in a sealed, rigid container against temperature and you get a straight line. Against kelvins it passes through the origin with slope $nR/V$: for the animation's gas, from 100 kPa at 300 K to 200 kPa at 600 K, a slope of $\frac{1}{3}kPa/K$. Against degrees Celsius it reaches zero pressure at −273 °C, and the lines for different amounts of gas, with different slopes, all point to that one temperature. Such extrapolations gave the first estimates of absolute zero, though real gases liquefy before reaching it.

At fixed temperature the pressure–volume graph is a hyperbola instead, $p=nRT/V$, lying higher for a hotter gas: halve the volume and the pressure doubles.

Common mistakes

• Using degrees Celsius. Heating from 27 °C to 327 °C at fixed volume doubles the pressure to 200 kPa, because 300 K becomes 600 K; the ratio 327/27 would give an absurd 1211 kPa.
• Using the gauge reading as the pressure. The tyre showing 230 kPa is really at 331 kPa; with the gauge value the amount of air comes out about 30 % too small.
• Putting the molar mass in grams per mole. With $M=28$ instead of 0.028, $\sqrt{3RT/M}$ gives 16 m/s for nitrogen instead of 517 m/s.
• Thinking that equal temperatures mean equal speeds. Helium and nitrogen at 300 K share the same mean kinetic energy, but their rms speeds are about 1370 m/s and 517 m/s.
• Picturing pressure as molecules at rest pushing apart. It comes from impacts, and an elastic bounce still gives the wall $2m{v}_x$: elastic means that no kinetic energy is lost, not that nothing is exchanged.

Where you meet it

Tyres, bicycle pumps and weather balloons swelling as they rise into thinner air all follow $pV=nRT$. A hot-air balloon floats because air at 100 °C holds about 20 % fewer molecules per cubic metre than air at 27 °C at the same pressure, since $N/V=p/(kT)$. First Law of Thermodynamics and Heat Engines builds on this molecular energy, and The Second Law of Thermodynamics and Entropy explains why heat flows from hot to cold.

Advanced

Pressure as a flow of momentum

A gas pushes on its container because its molecules keep striking the walls. Each strike hands the wall a little momentum, and pressure is nothing more than the rate at which momentum arrives per unit area. Following that flow step by step links a quantity we can measure, the pressure $p$, to one we cannot see, the speeds of the molecules.

Legend

$A$	Area of the patch of wall, drawn in perspective.	${\mathrm{m}}^2$
$v_x$	Velocity component of a molecule perpendicular to the wall, towards it.	$m/s$
$\Delta t$	A short time interval.	$\mathrm{s}$
$v_x\Delta t$	Distance such a molecule covers in $\Delta t$: the length of the slab, whose volume is $A{v}_x\Delta t$.	$\mathrm{m}$
$m$	Mass of one molecule.	$\mathrm{kg}$
$m{v}_x$	Momentum of the molecule before the hit; afterwards it is $-m{v}_x$, so the wall receives $2m{v}_x$.	$\mathrm{kg}\cdot m/s$
$F$	Force of all the hits on the patch: the momentum handed over per unit time.	$\mathrm{N}$

Figure 1 looks at a patch of wall of area $A$, perpendicular to the $x$ axis. For the push only the velocity component $v_x$ matters, the part of a molecule's velocity aimed straight at the wall; motion along the wall hands it no momentum. So sort the molecules into groups by their $v_x$, and follow one group with a given $v_x>0$, heading for the wall. Say it has $\Delta N$ of the $N$ molecules in the volume $V$, so $\Delta N/V$ of them in each unit of volume.

Which molecules of this group reach the patch during a short time $\Delta t$? Each covers $v_x\Delta t$ towards the wall in that time, so it arrives only if it starts within that distance of the patch. These are the molecules inside the shaded slab of Figure 1, of length $v_x\Delta t$ and volume $A{v}_x\Delta t$. Molecules farther back arrive later, and those moving away never arrive. The slab therefore holds $(\Delta N/V)A{v}_x\Delta t$ molecules of the group, and all of them hit the patch.

The enlarged hit below the slab shows what each arrival does. The bounce is elastic, so the velocity component turns from $v_x$ into $-v_x$ and the momentum from $m{v}_x$ into $-m{v}_x$, where $m$ is the mass of one molecule. Momentum is conserved, so the wall receives the difference, $2m{v}_x$.

Now multiply the number of hits by the momentum per hit: in $\Delta t$ the group delivers $(\Delta N/V)A{v}_x\Delta t\cdot 2m{v}_x$. Notice that $v_x$ appears twice. A faster molecule arrives more often and also strikes harder, so the push grows as $v_x^2$. The force $F$ is momentum per unit time, so divide by $\Delta t$; the pressure is force per unit area, $p=F/A$, so divide by $A$. The group's share of the pressure is $2m(\Delta N/V)v_x^2$.

Adding the shares of all the groups heading for the wall gives
$p=\frac{2m}{V}{\sum }_{v_x>0}\Delta N{v}_x^2$
The motion is random, so for every group with $+v_x$ there is an equally large group with $-v_x$. The sum over the half moving towards the wall is therefore half the sum over all $N$ molecules, that is $\frac{1}{2}N⟨v_x^2⟩$, where angle brackets mean the average over all the molecules. This half cancels the factor 2 of the bounce:
$p=\frac{N}{V}m⟨v_x^2⟩$
Last, the squared speed is $v^2=v_x^2+v_y^2+v_z^2$, and random directions favour no axis, so each component carries a third of it: $⟨v_x^2⟩=⟨v_y^2⟩=⟨v_z^2⟩=⟨v^2⟩/3$. Hence
$p=\frac{1}{3}\frac{N}{V}m⟨v^2⟩$
which is the formula the expert layer of the animation writes.

Three features of this result are easy to overlook. First, it needs only the mean of $v^2$, not any particular distribution of speeds. Second, collisions between molecules do not change it: a collision only passes momentum from one molecule to another, and the momentum still reaches the wall in the end. Third, in a mixture each kind of molecule adds its own term, so the partial pressures of the components add up to the total (Dalton's law).

To put in numbers, use the measured gas law $pV=NkT$, where $k=1.38\times 10^{-23}J/K$ is the Boltzmann constant. Nitrogen at 300 K and 100 kPa holds $N/V=p/(kT)=2.4\times 10^{25}$ molecules per cubic metre. Each square centimetre of wall takes about $3\times 10^{23}$ hits per second (the rate is $\frac{1}{4}(N/V)⟨v⟩$, with the mean speed $⟨v⟩$ found below). With so many separate impacts, the push feels perfectly steady.

Temperature and equipartition

The pressure formula becomes far more than a formula once it meets experiment, because it tells us what temperature is. Multiply it by $V$ and write $m⟨v^2⟩$ as twice the mean kinetic energy:
$pV=\frac{2}{3}N⟨\frac{1}{2}m{v}^2⟩$
Experiment says $pV=NkT$. Comparing the two,
$⟨\frac{1}{2}m{v}^2⟩=\frac{3}{2}kT$
Temperature measures the mean kinetic energy of the molecules' motion from place to place: at 300 K it is $6.2\times 10^{-21}\mathrm{J}$, or 0.039 eV, per molecule. Since the three directions share $⟨v^2⟩$ equally, each direction holds on average $\frac{1}{2}kT$.

This is the simplest case of equipartition: in classical equilibrium every term of a molecule's energy that is quadratic in a velocity or a coordinate, such as $\frac{1}{2}m{v}_x^2$, averages $\frac{1}{2}kT$. A diatomic molecule such as nitrogen can also rotate about the two axes perpendicular to its bond, which adds two more quadratic terms, so its mean energy is $\frac{5}{2}kT$. (Spinning about the bond itself does not count: nearly all the mass lies on that axis, so its moment of inertia is tiny and, like the vibration discussed under the limits below, this motion is frozen out.)

Per mole, multiply by the Avogadro number $N_A$; with the gas constant $R=N_{A}k=8.314J/(\mathrm{mol}\cdot \mathrm{K})$ the internal energy of nitrogen is $\frac{5}{2}RT$. Raising the temperature by 1 K at constant volume therefore takes the molar heat capacity $C_V=\frac{5}{2}R=20.8J/(\mathrm{mol}\cdot \mathrm{K})$, against $\frac{3}{2}R=12.5J/(\mathrm{mol}\cdot \mathrm{K})$ for helium or argon, whose atoms only move from place to place.

Why heavy and light molecules share the energy

Equipartition makes a bold claim: in a mixture, a heavy molecule and a light one carry the same mean kinetic energy, so the heavy one must move more slowly. The mechanism is the collisions between them, and one collision drawn the right way shows how it works.

Legend

$O$	Origin of velocity space: every velocity arrow starts here.
$m_1$	Mass of the heavy molecule, drawn in gold.	$\mathrm{kg}$
$m_2$	Mass of the light molecule, drawn in blue; here $m_1=2m_2$.	$\mathrm{kg}$
${\vec{v}}_1$	Velocity of the heavy molecule before the collision.	$m/s$
${\vec{v}}_2$	Velocity of the light molecule before the collision.	$m/s$
$\vec{w}$	Relative velocity ${\vec{v}}_1-{\vec{v}}_2$: the arrow from the tip of ${\vec{v}}_2$ to the tip of ${\vec{v}}_1$.	$m/s$
$\vec{V}$	Centre-of-mass velocity $(m_1{\vec{v}}_1+m_2{\vec{v}}_2)/(m_1+m_2)$: it ends on $\vec{w}$, nearer the heavy tip, at distances from the tips in the inverse ratio of the masses.	$m/s$
${\vec{w}}^{\prime }$	Relative velocity after the collision: the same length as $\vec{w}$, in a new direction.	$m/s$

Figure 2 is drawn in velocity space: every velocity is an arrow from one common origin $O$, so the motion of a molecule is fully given by where its arrow ends. The heavy molecule, of mass $m_1=2m_2$, has the gold velocity ${\vec{v}}_1$; the light one, of mass $m_2$, has the blue ${\vec{v}}_2$. The green arrow from the tip of ${\vec{v}}_2$ to the tip of ${\vec{v}}_1$ is their relative velocity, $\vec{w}={\vec{v}}_1-{\vec{v}}_2$.

The orange arrow is the centre-of-mass velocity $\vec{V}=(m_1{\vec{v}}_1+m_2{\vec{v}}_2)/M$, with $M=m_1+m_2$. It is an average of ${\vec{v}}_1$ and ${\vec{v}}_2$ weighted by mass, so its tip lies on $\vec{w}$, nearer the heavier tip: the distances to the two tips are in the inverse ratio of the masses, $(m_2/M)|\vec{w}|$ to the heavy tip and $(m_1/M)|\vec{w}|$ to the light one. With $m_1=2m_2$ it ends one third of the way along $\vec{w}$ from the heavy tip. Reading the figure from the end of $\vec{V}$,
${\vec{v}}_1=\vec{V}+\frac{m_2}{M}\vec{w},{\vec{v}}_2=\vec{V}-\frac{m_1}{M}\vec{w}$

What can a collision change? The total momentum $M\vec{V}$ is conserved, so $\vec{V}$ stays put. Substituting the two velocities above, the kinetic energy splits into $\frac{1}{2}M{V}^2+\frac{1}{2}(m_1{m}_2/M)w^2$, so an elastic collision, which conserves it, also keeps the length of $\vec{w}$. Only the direction of $\vec{w}$ can change: it turns into ${\vec{w}}^{\prime }$. In Figure 2 the two tips therefore swing about the end of $\vec{V}$ along the dashed arcs, the heavy tip on the small one, of radius $(m_2/M)|\vec{w}|$, and the light tip on the large one, of radius $(m_1/M)|\vec{w}|$.

Now the step to equilibrium. Each molecule collides billions of times a second with partners arriving from all sides, so after a few collisions the direction of $\vec{w}$ has forgotten where it started and bears no relation to $\vec{V}$. The two are uncorrelated, and the average of their dot product vanishes: $⟨\vec{V}\cdot \vec{w}⟩=0$. Multiplying out the product gives
$M\vec{V}\cdot \vec{w}=m_1{v}_1^2-m_2{v}_2^2+(m_2-m_1){\vec{v}}_1\cdot {\vec{v}}_2$
The velocities of two molecules about to meet are independent and point in random directions, so $⟨{\vec{v}}_1\cdot {\vec{v}}_2⟩=0$ as well. Averaging leaves $m_1⟨v_1^2⟩=m_2⟨v_2^2⟩$: the same mean kinetic energy. The heavier molecule is slower by exactly the factor that keeps its energy equal; with $m_1=2m_2$ its root-mean-square speed is lower by $\sqrt{2}$.

Squeezing a gas: the moving piston

Push a piston into a gas without letting heat flow and the gas gets hotter. Kinetic theory shows exactly where the piston's work goes: into faster molecules, one bounce at a time.

Legend

$u$	Speed at which the piston moves into the gas; each hit adds $2u$ to the molecule's speed.	$m/s$
$v_x$	Speed of the molecule towards the piston before the hit, measured in the box.	$m/s$
$v_x+2u$	Speed with which the molecule leaves the piston, measured in the box.	$m/s$

In Figure 3 the piston moves inwards at a speed $u$ much smaller than the molecular speeds, and one molecule approaches it with the speed $v_x$ (upper arrow). The bounce is easiest to see from the piston's own frame, where the wall is at rest. There the molecule approaches at $v_x+u$, because the piston is coming towards it, and a bounce off a resting wall reverses the velocity without changing its size, so it leaves at $v_x+u$. Back in the box's frame we add the piston's own speed $u$ in the direction of the rebound, so the molecule leaves with $v_x+2u$ instead of $v_x$. In the lower arrow of Figure 3 the blue part is the old $v_x$ and the orange part is the extra $2u$ that the moving wall adds.

Each hit therefore raises the molecule's energy by
$\frac{1}{2}m(v_x+2u{)}^2-\frac{1}{2}m{v}_x^2=2m{v}_{x}u+2m{u}^2\approx 2m{v}_{x}u$
where the last term is dropped because $u\ll v_x$.

Count the hits as in Figure 1: a group strikes the piston face, of area $A$, $(\Delta N/V)A{v}_x$ times per second, and each strike adds $2m{v}_{x}u$. Summed over the groups this is $uA$ times the very sum that gave the pressure, so the gas gains energy at the rate $pAu$. Meanwhile the piston sweeps out the volume $Au$ each second, so the volume changes at the rate $-Au$. In other words $dU=-pdV$: the piston's work appears, in full, as faster molecules.

With no heat flowing in, and $U=\frac{3}{2}NkT$ for point molecules that only move from place to place, put $p=NkT/V$ into $dU=-pdV$:
$\frac{3}{2}NkdT=-\frac{NkT}{V}dV$
Dividing by $NkT$ gives $\frac{3}{2}dT/T=-dV/V$, and integrating gives $\frac{3}{2}\ln T+\ln V=\mathrm{const}$. Since $T$ is proportional to $pV$, this is
$T{V}^{2/3}=\mathrm{const},p{V}^{5/3}=\mathrm{const}$

The same steps work for any molecule. With $f$ quadratic energy terms, $U=\frac{f}{2}NkT$ and $p{V}^{\gamma }=\mathrm{const}$ with $\gamma =1+2/f$: 5/3 for atoms ($f=3$) and 7/5 for nitrogen and oxygen ($f=5$). A diesel engine squeezes air about eighteen-fold; starting from 300 K this heats it to $300\cdot 18^{0.4}$, some 950 K, hot enough to ignite the fuel without a spark.

Light in a box with mirrored walls obeys the same counting of hits and momentum. A photon carries momentum $E/c$ instead of $mv$ and moves at $c$, so where the derivation multiplied momentum by speed, $m{v}^2$ becomes the photon energy $E$. The pressure is then a third of the energy per unit volume $\epsilon$, $p=\epsilon /3$. With $U=\epsilon V=3pV$, the rule $dU=-pdV$ gives $p{V}^{4/3}=\mathrm{const}$: an adiabatic exponent of 4/3.

The Boltzmann factor

So far only averages have appeared. To find how the speeds themselves are spread, we need a broader rule of statistical mechanics: in equilibrium at temperature $T$, a state of energy $E$ has a probability proportional to the Boltzmann factor $e^{-E/kT}$. States with energy much larger than $kT$ are rare.

Applied to potential energy, a molecule at height $h$ has the extra energy $mgh$, so the number of molecules per unit volume falls with height as $e^{-mgh/kT}$. It drops by a factor $e$ every $kT/(mg)$ of height, which is 9.1 km for nitrogen at 300 K. The real atmosphere is colder higher up, which makes the fall steeper; on top of Mount Everest the pressure is about a third of its sea-level value.

Applied to kinetic energy, $\frac{1}{2}m(v_x^2+v_y^2+v_z^2)$, the exponential of a sum is a product of exponentials. Each velocity component is therefore independent of the others and follows its own Gaussian weight, $e^{-m{v}_x^2/(2kT)}$, centred on zero: for any single component, values near zero are the most likely.

The speed distribution

Each velocity component is most likely near zero, and yet hardly any molecule is nearly at rest. Turning the weight of each velocity into a distribution of speeds resolves this, and gives the three characteristic speeds of a gas.

Legend

$v$	Speed of a molecule.	$m/s$
$f(v)$	Distribution of speeds: the fraction of molecules per unit speed, so a strip of width $dv$ holds $f(v)dv$.	$s/m$
$dv$	Width of the strip: a small range of speeds.	$m/s$
$v_p$	Most probable speed, at the peak of $f(v)$.	$m/s$
$⟨v⟩$	Mean speed: the average of $v$ over all the molecules.	$m/s$
$v_{rms}$	Root-mean-square speed, $\sqrt{⟨v^2⟩}$: the speed of a molecule with the mean kinetic energy.	$m/s$
$T_1$	Temperature of the blue curve.	$\mathrm{K}$
$T_2$	Higher temperature of the orange curve, here $2T_1$: the curve is lower and wider but encloses the same area.	$\mathrm{K}$

Figure 4 plots the speed distribution $f(v)$. Its meaning is set by the thin strip: $f(v)dv$ is the fraction of molecules with speeds between $v$ and $v+dv$, which is the area of the strip of width $dv$ under the curve. Every molecule has some speed, so the whole area under each curve is 1.

To build $f(v)$, collect every velocity whose length lies between $v$ and $v+dv$. In velocity space these arrows end in a thin spherical shell of radius $v$ and thickness $dv$, whose volume is $4\pi v^{2}dv$. Each of them has the Gaussian weight $e^{-m{v}^2/(2kT)}$. Multiplying the two and fixing the constant so that the area is 1 gives the Maxwell–Boltzmann distribution
$f(v)=4\pi {\left(\frac{m}{2\pi kT}\right)}^{3/2}{v}^2{e}^{-m{v}^2/(2kT)}$
The two factors pull in opposite directions. The factor $v^2$ empties the region near zero speed, because there are few velocities of small length, even though each component on its own is most likely near zero. The exponential cuts off the fast tail. The peak lies between the two.

Three speeds summarise the curve. The most probable speed $v_p$ is at the peak, where the slope of $v^2{e}^{-m{v}^2/(2kT)}$ is zero: $2v-(m/kT)v^3=0$, so $v_p^2=2kT/m$. The root-mean-square speed $v_{\mathrm{rms}}=\sqrt{⟨v^2⟩}$ follows at once from $⟨\frac{1}{2}m{v}^2⟩=\frac{3}{2}kT$. The mean speed $⟨v⟩$ is the average of $v$ weighted by $f(v)$, an integral that gives
$v_p=\sqrt{\frac{2kT}{m}}<⟨v⟩=\sqrt{\frac{8kT}{\pi m}}<v_{\mathrm{rms}}=\sqrt{\frac{3kT}{m}}$
All three are proportional to $\sqrt{T/m}$, in the fixed ratio 1 : 1.128 : 1.225, and the dashed lines of Figure 4 mark them in this true proportion. For nitrogen at 300 K they are 422, 476 and 517 m/s.

The orange curve is the same gas at the higher temperature $T_2=2T_1$. All three speeds lie farther right by the factor $\sqrt{2}$, so the curve is stretched sideways by $\sqrt{2}$. Its area must stay equal to 1, so its peak is lower by the same factor.

Reading the histogram

The expert layer of the animation draws $f(v)$ as eight bars, each 200 m/s wide, from 0 to 1600 m/s. Their heights come from the formula at the temperature on the readout, not from counting the sixteen molecules on the screen, which would be far too few. At 300 K the tallest bar, from 400 to 600 m/s, holds 36 % of the molecules, and the dashed $v_{\mathrm{rms}}$ marker stands at 517 m/s. At 586 K the peak has moved to $v_p=590m/s$, the 400–600 and 600–800 m/s bars are almost equal at 26 % each, and the marker and the $v_{\mathrm{rms}}$ readout show 723 m/s, larger by $\sqrt{586/300}=1.40$.

Because the area under $f(v)$ stays equal to 1, a widening curve must drop, as the orange curve of Figure 4 does: the height of the peak falls as $1/\sqrt{T}$. The fast tail changes most. The fraction of molecules faster than 1000 m/s grows from 1.1 % at 300 K to 12.5 % at 586 K, more than tenfold for less than a doubling of $T$. Evaporation and chemical reactions need a minimum energy, which only the molecules in this tail have, so their rates also climb steeply with temperature.

Worked problem: heating the gas two ways

Take one mole of the animation's nitrogen at 300 K and 100 kPa, so its volume is $V_1=R{T}_1/p_1=24.9\mathrm{L}$. (a) How much heat must the flame supply to bring it to 600 K with the piston clamped? (b) What compression, with no heat exchanged, reaches the same 600 K, and at what final pressure and cost in work?

(a) With the piston clamped the volume does not change, so the gas does no work and all the heat goes into internal energy. Real nitrogen stores $\frac{5}{2}kT$ per molecule, or $\frac{5}{2}RT$ per mole, so
$Q=\frac{5}{2}R\Delta T=2.5\cdot 8.314\cdot 300\approx 6.24\mathrm{kJ}$
Only 3.74 kJ of it, the $\frac{3}{2}R\Delta T$ of motion from place to place, shows up in the speeds and in the pressure, which doubles to 200 kPa because the temperature doubles at fixed volume. The other 2.49 kJ sets the molecules spinning, a motion the animation's point molecules cannot show.

(b) With no heat exchanged, $T{V}^{\gamma -1}$ stays constant, with $\gamma =7/5$ for nitrogen. Solving for the volume ratio,
$\frac{V_2}{V_1}={\left(\frac{T_1}{T_2}\right)}^{1/(\gamma -1)}=2^{-5/2}\approx 0.177$
and the gas law $pV=nRT$ then gives the final pressure:
$p_2=p_1\frac{T_2}{T_1}\frac{V_1}{V_2}=100\cdot 2^{7/2}\approx 1131\mathrm{kPa}$
The gas must be squeezed from 24.9 L to 4.4 L, and it ends at more than five times the pressure reached in (a). No heat flows, so the piston's work equals the rise in internal energy, again 6.24 kJ, because the final temperature is the same. A monatomic gas, with $\gamma =5/3$, would need only $V_2/V_1=2^{-3/2}\approx 0.354$, ending at 566 kPa after 3.74 kJ of work.

Limits of validity

The ideal gas assumes molecules that are small and far apart. That holds well for nitrogen at 300 K and 100 kPa: the mean spacing $(V/N{)}^{1/3}$ is 3.5 nm, about nine molecular diameters, and the molecules themselves fill only 0.06 % of the volume.

The van der Waals equation $(p+a{n}^2/V^2)(V-nb)=nRT$, for $n$ moles, corrects for both effects: the constant $a$ for the attraction between molecules, which lowers the pressure, and $b$ for their own volume, which raises it. Under these conditions each correction is about 0.2 % and the two nearly cancel. At high density or near the boiling point, 77 K, they grow and no longer cancel; below the critical temperature the equation even describes condensation into a liquid.

Equipartition is a classical result, and its failures were early signs of quantum mechanics. A vibrating bond would add another $kT$ per molecule (one quadratic term for its motion, one for its stretch) and raise $C_V$ of nitrogen to $\frac{7}{2}R$, yet at room temperature it is $\frac{5}{2}R$. The reason: vibrational energy comes in quanta of 0.29 eV, eleven times $kT=0.026\mathrm{eV}$, so the Boltzmann factor leaves hardly any molecule excited and the vibration is frozen. Hydrogen below a few tens of kelvins loses even its rotations.

The classical picture fails completely once the thermal de Broglie wavelength $h/\sqrt{2\pi mkT}$, the quantum size of a molecule at temperature $T$, approaches the spacing between molecules. For nitrogen at 300 K it is 0.019 nm, far below 3.5 nm, but not so in liquid helium near 2 K or for the electrons in a metal.

Collisions between molecules, left out of the animation, do not alter an ideal gas's pressure but set its other properties. With a molecular diameter $d\approx 0.37\mathrm{nm}$, a nitrogen molecule at room conditions travels on average a mean free path $\lambda =kT/(\sqrt{2}\pi d^{2}p)\approx 68\mathrm{nm}$ between collisions; at its mean speed of 476 m/s that is about $7\times 10^9$ collisions a second. These collisions bring any starting distribution to the Maxwell–Boltzmann form within nanoseconds, and they govern diffusion, viscosity and heat conduction; the animation's bars show the equilibrium they would establish. Relativity plays no part: $v_{\mathrm{rms}}/c\approx 2\times 10^{-6}$.

Where this leads

The energy $U=\frac{f}{2}nRT$, the work $-pdV$ and the adiabat are the working parts of First Law of Thermodynamics and Heat Engines. In The Second Law of Thermodynamics and Entropy, counting the ways of sharing energy among molecules, the idea behind the Boltzmann factor, becomes entropy. Blackbody Radiation and the Quantum shows equipartition failing again, for the modes of light in a cavity, and Semiconductors, Diodes and LEDs uses $e^{-E/kT}$ to count conducting electrons. The speed of sound, $\sqrt{\gamma RT/M}=353m/s$ for nitrogen at 300 K, with $M$ here the molar mass, links back to Sound and the Doppler Effect, and the mean free path explains why the viscosity of a gas, behind the drag in Drag and Terminal Velocity, hardly depends on its pressure.