Pressure of an ideal gas

Open a bicycle pump valve for a fraction of a second and you feel a hiss of air pushing out — a tangible force from something you cannot see. A gas has no fixed shape, yet it pushes outward on every wall of its container with a steady, measurable pressure. Where does that push come from, if there is nothing solid inside the container at all, only a swarm of tiny, fast, invisible molecules flying around in empty space? The kinetic theory of gases answers this with a beautifully simple picture. Air in a sealed tyre is not a continuous fluid pressing outward — it is billions upon billions of nitrogen and oxygen molecules, each moving in a straight line until it collides with a wall or another molecule, then bouncing off like a tiny elastic ball. Every single collision with the wall delivers a microscopic, momentary kick. No one collision is felt. But add up the kicks from an astronomical number of molecules, every second, over every square centimetre of wall, and the result is the steady force per unit area we call pressure. This article turns that qualitative story into a precise formula, $P = \tfrac13\dfrac{Nm\overline{v^2}}{V} = \tfrac13\rho\overline{v^2}$ , connecting a macroscopic quantity you can read off a gauge to a microscopic average of molecular speeds. We derive it from momentum transfer in a single wall collision, work through two numerical problems (one board-style, one competitive-style), and clear up the single most common mix-up students make between mean square speed and the square of the mean speed.

Key takeaways

Gas pressure arises from molecules colliding elastically with the container walls and transferring momentum on every impact.
The central result is $P = \tfrac13\dfrac{Nm\overline{v^2}}{V} = \tfrac13\rho\,\overline{v^2}$ , linking macroscopic pressure to microscopic mean square speed.
The factor of $\tfrac13$ comes from splitting molecular motion equally among the three spatial directions: $\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2} = \tfrac13\overline{v^2}$ .
$\overline{v^2}$ (mean square speed, average first squared then, since squares are never negative, always at least as large as $(\bar v)^2$ ) is not the same as $(\bar v)^2$ (square of the mean speed) — mixing these up is the number-one exam error.
$\rho = Nm/V$ is mass density; it is not the same as number density $n = N/V$ , which differs by a factor of the molecular mass $m$ .
The formula is a statistical result about a huge population of molecules ( $N$ of order $10^{23}$ , set by Avogadro's number $N_A = 6.022\times10^{23}$ ); it has no meaning applied to a single molecule.
Key idea: gas pressure is nothing but the accumulated rate of momentum transfer from countless elastic molecular collisions with the walls, condensed into the single formula P = (1/3)ρ v-squared-bar, where the one-third comes from sharing motion equally among three perpendicular directions.

The core insight

Pressure is the macroscopic footprint of microscopic momentum transfer: summing the momentum each molecule delivers per collision over the huge number of collisions per second on a wall gives $P = \tfrac13\dfrac{Nm\overline{v^2}}{V} = \tfrac13\rho\overline{v^2}$ , with the factor $\tfrac13$ arising because only one of the three perpendicular velocity components drives momentum transfer to any one wall.

Where it fits in the syllabus

This article covers the pressure-of-an-ideal-gas result inside the CBSE Class 11 Kinetic Theory chapter, immediately after the kinetic theory assumptions (elastic collisions, negligible molecular size, no intermolecular forces except during collision) and immediately before the kinetic interpretation of temperature. The formula $P = \tfrac13\frac{Nm\overline{v^2}}{V} = \tfrac13\rho\overline{v^2}$ , together with its derivation from momentum transfer during wall collisions, is squarely in the prescribed 2025-26 CBSE scope and is a favourite for both short-answer derivations and numericals. The Maxwell–Boltzmann speed distribution is explicitly out of scope at this stage — this article only ever needs the single averaged quantity $\overline{v^2}$ , never the full spread of molecular speeds.

The physical picture

Picture $N$ identical gas molecules, each of mass $m$ , sealed inside a cubical container of volume $V$ and side length $l$ (so $V = l^3$ ). Kinetic theory treats each molecule as a tiny hard sphere that travels in a straight line at constant velocity until it strikes a wall or another molecule, and every collision is assumed perfectly elastic, so no kinetic energy is lost overall. Because the molecules are in random, ceaseless motion, at any instant equal numbers move, on average, in every direction — there is no preferred direction, and no wall is struck more often than any other in the long run.

P = \frac{1}{3}\frac{Nm\overline{v^2}}{V}

where: $P$ is the pressure exerted on any wall (pascal, Pa), $N$ is the total number of molecules in the container, $m$ is the mass of one molecule (kg), $V$ is the container volume (m $^3$ ), and $\overline{v^2}$ is the mean square speed — the average of $v^2$ taken over all $N$ molecules, not the square of their average speed. The physical reading is direct: pressure grows if you pack in more molecules ( $N$ larger), if each molecule is heavier ( $m$ larger), if the molecules move faster ( $\overline{v^2}$ larger), or if you shrink the container ( $V$ smaller) — all of which match everyday intuition about pumping up a tyre or heating a sealed flask. Since $Nm$ is the total mass of gas in the container, $Nm/V$ is simply the ordinary mass density $\rho$ (kilograms of gas per cubic metre), so the same result is often written more compactly as

P = \frac{1}{3}\rho\,\overline{v^2}

Both forms say the same thing. The first is useful when you know the molecule count and mass individually (as in many textbook numericals); the second is useful when you are handed the bulk density of the gas directly, as measured in a lab.

Figure

A cubical box of gas with molecules shown as dots moving in random directions and one molecule bouncing off the right wall, its velocity arrow shown before and after the bounce with the perpendicular component reversed and the parallel component unchanged

Combining $P = \tfrac13\rho\overline{v^2}$ with the equation of state $PV = nRT$ (studied in the earlier part of this chapter) lets us relate the mean square speed of molecules directly to the absolute temperature of the gas — a connection developed fully in the next article on the kinetic interpretation of temperature.

Derivation ADeriving the pressure formula from a single wall collision

The goal of this derivation is to build the pressure formula $P = \tfrac13\dfrac{Nm\overline{v^2}}{V}$ from nothing more than Newton's laws applied to one elastic collision, then scale that single event up to the trillions of collisions happening every second across a whole wall of the container. The strategy has three parts: find the momentum transferred in one collision, count how many such collisions happen per second on one wall, and then average honestly over all $N$ molecules moving in every possible direction, since molecules do not all travel perpendicular to the wall.

Step 1. Consider one molecule of mass $m$ moving with velocity component $v_x$ perpendicular to a wall of area $A$ (taken as the face $x=l$ of a cubical box of side $l$ , so $V=l^3$ ). Before it strikes the wall its momentum along $x$ is $mv_x$ ; because the collision is perfectly elastic and the wall is rigid, the molecule rebounds with its $x$ -component exactly reversed to $-mv_x$ , while its $y$ and $z$ components are unchanged (a wall exerts no force parallel to itself). The change in the molecule's momentum is therefore $-mv_x - mv_x = -2mv_x$ , so by Newton's third law the momentum delivered to the wall in this one collision is $2mv_x$ .

Step 2. Find how often this molecule hits that particular wall. After rebounding, the molecule travels across the box, bounces off the opposite wall, and returns — a round trip of distance $2l$ at speed $v_x$ , taking time $2l/v_x$ . So this single molecule strikes the wall once every $2l/v_x$ seconds, delivering momentum $2mv_x$ each time. The average force this one molecule exerts on the wall is therefore momentum delivered divided by time between hits: $f = \dfrac{2mv_x}{2l/v_x} = \dfrac{mv_x^2}{l}$ .

Step 3. Sum over all $N$ molecules. Every molecule in the box contributes a force of this form, but with its own $v_x$ value, so the total force on the wall is $F = \dfrac{m}{l}\sum v_x^2 = \dfrac{mN\overline{v_x^2}}{l}$ , where $\overline{v_x^2}$ is the mean square of the $x$ -velocity component averaged over all $N$ molecules. Pressure is force per unit area, and the wall's area is $A = l^2$ , so $P = F/A = \dfrac{mN\overline{v_x^2}}{l\cdot l^2} = \dfrac{mN\overline{v_x^2}}{l^3} = \dfrac{mN\overline{v_x^2}}{V}$ .

Step 4. Remove the artificial restriction to one direction. Molecular motion is random and has no preferred direction, so by symmetry the mean square speed splits equally among the three perpendicular axes: $\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2}$ . Since $\overline{v^2} = \overline{v_x^2} + \overline{v_y^2} + \overline{v_z^2}$ by Pythagoras applied to each molecule's velocity and then averaged, this gives $\overline{v_x^2} = \tfrac13\overline{v^2}$ . Substituting into Step 3, $P = \dfrac{Nm}{V}\cdot\dfrac{1}{3}\overline{v^2} = \dfrac{1}{3}\dfrac{Nm\overline{v^2}}{V}$ , which is exactly the pressure formula, and writing $\rho = Nm/V$ gives the equivalent density form $P = \tfrac13\rho\overline{v^2}$ .

Step 5. Sanity-check the pieces. Every symbol traces to something physical: $2mv_x$ is a momentum kick (kg·m/s), dividing by a time gives a force (newton), dividing by an area gives a pressure (pascal) — the algebra is dimensionally consistent at every stage. The derivation used only Newton's laws for one elastic collision plus a symmetry argument for splitting directions; it never assumed anything about the distribution of speeds among molecules, which is why the result is exact even though real molecular speeds vary enormously from one molecule to the next.

Worked examples

Board level - rms speed from density and pressure

A gas has a density of $1.2\ \text{kg/m}^3$ and exerts a pressure of $1.0\times10^5\ \text{Pa}$ on its container. Find the root-mean-square speed of the gas molecules, $v_{rms} = \sqrt{\overline{v^2}}$ .

Step 1. Write the pressure formula in density form, since density is what is given: $P = \tfrac13\rho\overline{v^2}$ .

Step 2. Rearrange to isolate the mean square speed: $\overline{v^2} = \dfrac{3P}{\rho} = \dfrac{3\times1.0\times10^5}{1.2} = 2.5\times10^5\ \text{m}^2/\text{s}^2$ .

Step 3. Take the square root to get the rms speed: $v_{rms} = \sqrt{\overline{v^2}} = \sqrt{2.5\times10^5} \approx 500\ \text{m/s}$ . So the rms speed of the gas molecules is about 500 m/s, a typical order of magnitude for air-like gases near ordinary atmospheric conditions — comparable to the speed of sound in air, which is no coincidence, since sound propagates through exactly these molecular collisions.

Competitive level - pressure from molecule count, mass and mean square speed

A closed vessel of volume $2.0\ \text{m}^3$ contains $5.0\times10^{25}$ molecules of a gas, each of mass $5.0\times10^{-26}\ \text{kg}$ , with mean square speed $\overline{v^2} = 2.7\times10^5\ \text{m}^2/\text{s}^2$ . Find (a) the pressure exerted by the gas and (b) the corresponding rms speed, and check the density is physically reasonable.

Step 1. Use the molecular form of the pressure formula, since $N$ , $m$ and $\overline{v^2}$ are all given directly: $P = \dfrac{1}{3}\dfrac{Nm\overline{v^2}}{V}$ .

Step 2. Compute the total mass of gas first, to keep the arithmetic manageable: $Nm = 5.0\times10^{25}\times5.0\times10^{-26} = 2.5\ \text{kg}$ . This is a genuine physical quantity — the total mass of gas inside the $2.0\ \text{m}^3$ vessel — and it is worth pausing to sanity-check it: a density of $2.5/2.0 = 1.25\ \text{kg/m}^3$ is indeed close to the density of air at room conditions, so the numbers are physically sensible before proceeding.

Step 3. Multiply by the mean square speed: $Nm\overline{v^2} = 2.5\times2.7\times10^5 = 6.75\times10^5$ .

Step 4. Divide by volume and by 3: $P = \dfrac{1}{3}\times\dfrac{6.75\times10^5}{2.0} = \dfrac{1}{3}\times3.375\times10^5 = 1.125\times10^5\ \text{Pa}$ . So the pressure exerted is about $1.125\times10^5\ \text{Pa}$ , close to one standard atmosphere.

Step 5. For part (b), the rms speed follows directly: $v_{rms} = \sqrt{\overline{v^2}} = \sqrt{2.7\times10^5} \approx 520\ \text{m/s}$ , again a plausible molecular speed. Both the pressure and the speed independently land in physically reasonable ranges, which is a strong consistency check that no arithmetic slip occurred. Timing note: in a competitive exam, do not multiply $Nm\overline{v^2}$ out fully in one shot — compute $Nm$ first (Step 2) as a clean intermediate number, since it is often reusable (density, moles, or a second sub-question), and only then multiply by $\overline{v^2}$ ; this staged approach both reduces arithmetic slips under time pressure and gives you a free sanity-check (the density) partway through, at essentially zero extra time cost.

Common traps

Confusing mean square speed $\overline{v^2}$ (average the squares) with the square of the mean speed $(\bar v)^2$ (average first, then square) — these are generally not equal, and in fact $\overline{v^2} \ge (\bar v)^2$ always; the pressure formula strictly requires $\overline{v^2}$ .
Forgetting the factor of $\tfrac13$ altogether, or misremembering it as $\tfrac12$ — the $\tfrac12$ belongs to kinetic energy formulas ( $\tfrac12 mv^2$ ), not to this pressure formula.
Mixing up mass density $\rho = Nm/V$ (kilograms per cubic metre) with number density $n = N/V$ (molecules per cubic metre) — the two differ by a factor of the molecular mass $m$ , and using the wrong one silently changes the units of the answer.
Applying the wall-collision argument to a single molecule and expecting it to "have" a pressure — pressure is a statistical, population-averaged quantity that only makes sense for the astronomically large $N$ (of order Avogadro's number) that real gas samples contain; a lone molecule bouncing in a box exerts a fluctuating, not steady, force.
Forgetting to convert to SI units before substituting — density in $\text{g/cm}^3$ or pressure in atmospheres plugged directly into the formula gives an answer wrong by orders of magnitude; always convert to kg/m $^3$ and pascals first.

You should be able to…

State the pressure formula $P = \tfrac13\frac{Nm\overline{v^2}}{V} = \tfrac13\rho\overline{v^2}$ from memory and correctly identify every symbol including its SI unit.
Reproduce the wall-collision derivation in outline: momentum transfer per hit, hits per second, summing over molecules, then splitting $\overline{v^2}$ equally among three directions.
Explain in one or two sentences why the factor $\tfrac13$ appears, without hand-waving.
Clearly distinguish mean square speed $\overline{v^2}$ from the square of mean speed $(\bar v)^2$ , and state which one belongs in the pressure formula.
Solve numericals for $P$ , $\rho$ , $N$ , $m$ , or $\overline{v^2}$ given the other quantities, converting units to SI first.
Combine the pressure formula with $PV = nRT$ to connect a microscopic average to a macroscopic, measurable temperature.

Ask the tutor: Want a slower, line-by-line walkthrough of the momentum-transfer derivation with a diagram, or a worked example that starts from moles and molar mass instead of molecule count directly? Ask the AI tutor for a guided walkthrough.

Why does only the perpendicular velocity component matter for pressure on a wall?

A wall can only push or be pushed along the direction perpendicular to its own surface — it cannot exert or feel a force sideways along its own face, because a rigid, frictionless wall has nothing to grip a passing molecule with in that direction. When a molecule bounces off elastically, its perpendicular velocity component reverses (delivering a momentum kick to the wall) while its parallel component sails through completely unchanged, contributing nothing to the push. That is exactly why the derivation tracks $v_x$ alone for the wall at $x=l$ , and why the final formula ends up needing only one-third of the total mean square speed rather than all of it.

Does this formula still work if the container is not a cube?

Yes. The cube is chosen purely to make the geometry of "time between successive hits on the same wall" easy to compute (a clean round trip of distance $2l$ ). For any container shape, the same momentum-transfer reasoning and the same symmetry argument (random motion splits equally among three perpendicular directions) still apply, and a more careful treatment using calculus over an arbitrary volume element gives the identical result $P = \tfrac13\rho\overline{v^2}$ . CBSE only expects the cubical-box version of the argument, but it is worth knowing the result is not an artifact of that particular shape.

Why is pressure the same on every wall of the container, not just the one we analysed?

The derivation singled out one wall (perpendicular to the x-axis) only to make the bookkeeping concrete, but nothing in the physics favours that wall over any other. Molecular motion is assumed completely random, with no preferred direction, so the same argument applied to a wall perpendicular to y or to z gives exactly the same formula with exactly the same value of $\overline{v^2}$ (since $\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2}$ by that same symmetry). This is also why a gas pushes equally hard in every direction inside a sealed container at rest — pressure has no direction of its own, unlike a force.

Is this the same pressure as in P V equals n R T, and can I use both formulas together?

Yes, it is exactly the same pressure — the kinetic theory formula $P = \tfrac13\rho\overline{v^2}$ derives pressure from molecular motion, while the ideal gas equation $PV = nRT$ describes how pressure, volume and temperature relate macroscopically; both describe the same physical quantity $P$ for the same gas sample, so you can set them equal to link microscopic motion to temperature. Doing exactly that is the subject of the next article, on the kinetic interpretation of temperature, where combining the two equations gives $\tfrac12 m\overline{v^2} = \tfrac32 k_BT$ .