Kinetic molecular theory — derivation of P = (1 - 3)ρv²_rms: Question bank

Definition Symbols used on this page

Every symbol below is built in the parent note; here is a quick roster so nothing is used undefined:

$v_{x}$ — the velocity component of a molecule along one axis; $\overline{v^{2}}$ — the mean of the squared speeds; $v_{r m s} = \overline{v^{2}}$ — the root-mean-square speed.
$ρ = N m / V$ — density ( $N$ molecules of mass $m$ in volume $V$ ); $\frac{1}{3}$ — the "three directions share the motion" factor.
$L$ — the side length of the cubic box (so $V = L^{3}$ ); a molecule crosses it and returns in a round trip of distance $2 L$ .
$k_{B}$ — the Boltzmann constant, the "per-molecule gas constant" ( $k_{B} = R / N_{A} \approx 1.38 \times 1 0^{- 23} J/K$ ), so $K E_{a v g} = \frac{3}{2} k_{B} T$ is the average kinetic energy of one molecule.
$\overline{v}$ — the ordinary average speed (mean of $v$ ); $v_{m p}$ — the most probable speed, the speed at the peak of the Maxwell–Boltzmann curve (the single most common value). These are compared in Maxwell-Boltzmann Speed Distribution.

True or false — justify

Only the

x

-component of velocity contributes to the force on a wall facing the

x

-direction.

True — the

y

and

z

motion slides along that wall and pushes nothing into it; only the component pointed at the wall carries momentum toward it.

If you double every molecule's speed, the pressure doubles.

False — pressure depends on

\overline{v^{2}}

, so doubling speed multiplies

v^{2}

by 4; pressure quadruples.

At the same temperature,

H_{2}

and

O_{2}

have the same average kinetic energy.

True —

K E_{a v g} = \frac{3}{2} k_{B} T

depends only on

T

, not on mass; lighter

H_{2}

just moves faster to carry that same energy.

At the same temperature,

H_{2}

and

O_{2}

have the same

v_{r m s}

False — equal KE with smaller mass means larger speed, so

v_{r m s} (H_{2}) = 32/2 = 4

times that of

O_{2}

The factor

\frac{1}{3}

comes from the collision being elastic.

False — it comes purely from 3-D isotropy (

\overline{v_{x}^{2}} = \frac{1}{3} \overline{v^{2}}

); elasticity is what gives the factor of 2 in

2 m v_{x}

, a separate thing.

Pressure would be the same on a wall of any orientation in the cube.

True — random motion has no preferred direction, so

\overline{v_{x}^{2}} = \overline{v_{y}^{2}} = \overline{v_{z}^{2}}

and every wall feels an equal one-third share.

In the derivation we assumed all molecules move at the same speed.

False — we summed

v_{x 1}^{2} + \dots + v_{x N}^{2}

over different speeds and only replaced the sum by

N \overline{v_{x}^{2}}

; the spread of speeds is fully allowed.

If intermolecular collisions were included, the final

P = \frac{1}{3} ρ v_{r m s}^{2}

would change.

False — collisions between molecules only reshuffle velocities among them; the total momentum delivered to the walls per second is unchanged, so the result stands.

v_{r m s}

equals the average speed

\overline{v}

False — squaring weights fast molecules more, so

v_{r m s} > \overline{v} > v_{m p}

; the ratio is

3 : 8/ π : 2 \approx 1.085 : 0.921 : 0.798

(in units of

R T / M

Spot the error

Δ p = m v_{x}

because the molecule delivers its momentum

m v_{x}

to the wall."

Wrong — the molecule reverses, going from

+ m v_{x}

- m v_{x}

, so

∣Δ p ∣ = 2 m v_{x}

; dropping the 2 halves the pressure.

"Time between hits is

L / v_{x}

because the wall is a distance

L

away."

Wrong — a hit on the same wall requires a round trip to the opposite wall and back, distance

2 L

(with

L

the box side), so

Δ t = 2 L / v_{x}

"Force per molecule is

2 m v_{x} / (2 L / v_{x})

... which simplifies to

m v_{x} / L

Arithmetic slip — the

v_{x}

in the numerator times

1/ (2 L / v_{x})

gives a second

v_{x}

, so

f_{x} = m v_{x}^{2} / L

(note

v_{x}

squared).

P V = N m \overline{v^{2}}

since each of

N

molecules contributes

m \overline{v^{2}} / L

of force."

Wrong — this forgot the isotropy step; you must use

\overline{v_{x}^{2}} = \frac{1}{3} \overline{v^{2}}

, giving

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

(three times smaller).

v_{r m s} = 3 R T / M

with

M = 32

for oxygen gives ~15 m/s."

Unit error —

M

must be in kg/mol (

0.032

), not g/mol; using grams makes the speed

1000 \approx 31.6 \times

too small.

"Since pressure comes from collisions, a gas with fewer molecules but the same speed has proportionally lower pressure."

Correct reasoning —

P \propto N / V

, so halving

N

at fixed

V

and speed halves

P

; this one is actually right, the trap is doubting a true statement.

"The wall gains kinetic energy from each elastic collision."

Wrong — in an elastic collision with a fixed (infinitely massive) wall, no kinetic energy is transferred; the molecule leaves with the same speed, only its direction flips.

Why questions

Why do we use

v_{r m s}

rather than the ordinary average speed

\overline{v}

Because the force per molecule scales as

v_{x}^{2}

, so pressure genuinely depends on the mean of squares, and

\overline{v^{2}} \neq = \overline{v}

— using

\overline{v}

would give the wrong physics.

Why does

v_{x}^{2}

(a square) appear in the single-molecule force, not just

v_{x}

Two separate factors of

v_{x}

enter: faster molecules deliver more momentum per hit (

2 m v_{x}

) and hit more often (

v_{x} /2 L

), and their product carries

v_{x}^{2}

Why can we ignore collisions between molecules in this derivation?

They conserve total momentum internally and only redistribute speeds; averaged over time they don't alter how much momentum reaches the walls each second.

Why must molar mass be in kg/mol inside

3 R T / M

R = 8.314

carries SI units (J = kg·m²/s²), so mass must be in kg for the units to cancel into m/s; mixing grams breaks the unit bookkeeping.

Why is the pressure steady rather than a series of jerks?

Individual collisions are discrete kicks, but billions occur per second over the whole wall, so their sum averages into one smooth, constant force.

Why does raising temperature raise pressure at fixed volume?

Higher

T

means higher

\overline{v^{2}}

(since

K E_{a v g} = \frac{3}{2} k_{B} T

with

k_{B}

the per-molecule gas constant), and

P = \frac{1}{3} ρ v_{r m s}^{2}

scales directly with that mean-square speed.

Why does the derivation need the molecules' volume to be negligible?

So each molecule travels the full

2 L

round trip (across the box side

L

) unobstructed and the "empty box" geometry (

Δ t = 2 L / v_{x}

) holds exactly.

Edge cases

What happens to the momentum delivered per hit if a molecule grazes the wall with

v_{x} \approx 0

The kick

2 m v_{x} \to 0

and the round-trip time

2 L / v_{x} \to \infty

, so that molecule contributes essentially no pressure — consistent, since motion parallel to a wall pushes nothing into it.

If a gas were confined to move only in one dimension (all motion along

x

), what replaces the

\frac{1}{3}

Then

\overline{v_{x}^{2}} = \overline{v^{2}}

with no direction-sharing, so the factor becomes

1

and

P V = N m \overline{v^{2}}

; the

\frac{1}{3}

is entirely a 3-D artefact.

For a 2-D gas confined to a plane, what factor appears?

Two equal directions share the motion, so

\overline{v_{x}^{2}} = \frac{1}{2} \overline{v^{2}}

and the prefactor is

\frac{1}{2}

, not

\frac{1}{3}

At absolute zero (

T = 0

), what does the model predict for pressure?

v_{r m s} = 3 R T / M \to 0

, so

P \to 0

— no motion means no collisions and no push, the ideal-gas limit of vanishing pressure.

Does

P = \frac{1}{3} ρ v_{r m s}^{2}

hold for a real gas at very high pressure?

Not exactly — molecular volume and intermolecular forces (ignored here) become significant, which is why we then move to the Real Gases and van der Waals Equation.

If two boxes have the same

ρ

and same

v_{r m s}

but different molar masses, do they have the same pressure?

Yes —

P = \frac{1}{3} ρ v_{r m s}^{2}

depends only on density and mean-square speed, so identical

ρ

and

v_{r m s}

force identical

P

regardless of what the gas is.

Connections

Average Kinetic Energy and Temperature — the $K E_{a v g} = \frac{3}{2} k_{B} T$ link behind the "same $T$ , same KE" traps.
Maxwell-Boltzmann Speed Distribution — why $v_{r m s} > \overline{v} > v_{m p}$ and where $v_{m p}$ , $\overline{v}$ come from.
Graham's Law of Diffusion — the light-vs-heavy speed ratio in action.
Newton's Laws — Impulse and Momentum — the $2 m v_{x}$ reversal is impulse.
Ideal Gas Equation PV = nRT — combine to pin down temperature.