Ideal gas law PV = nRT — derivation from kinetic theory: Question bank

True or false — justify

The gas law is false without the collision picture; each item forces you to say why.

Doubling the number of molecules

N

at fixed

V

and

T

doubles the pressure.

True —

P = N k_{B} T / V

is linear in

N

; twice as many molecules means twice as many wall-kicks per second, so twice the push. This is $P=Nk_BT/V$ read directly.

At the same temperature, heavy molecules and light molecules exert the same pressure (same

N

V

True — pressure depends only on

N

V

T

, not on mass. The heavier molecules move slower (

v_{r m s} \propto 1/ m

) but hit harder per collision; the two effects cancel exactly because

\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T

is the same for both.

⟨ v_{x} ⟩ = 0

for a gas in equilibrium in a still container.

True — equal numbers of molecules move in

+ x

and

- x

, so the average velocity component cancels. If it were non-zero the gas would be drifting as a wind, not sitting still.

Because

⟨ v_{x} ⟩ = 0

, the molecules exert zero average force on the wall.

False — force scales with

⟨ v_{x}^{2} ⟩

, not

⟨ v_{x} ⟩

. Squares are always positive, so they don't cancel; that is exactly why pressure survives.

If the collisions were slightly inelastic, pressure would still be constant forever.

False — inelastic collisions would bleed kinetic energy into the wall, cooling the gas and dropping the pressure over time. Elasticity is what keeps

⟨ v^{2} ⟩

(and thus

T

) steady.

The factor

\frac{1}{3}

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

is a fudge factor from fitting experiments.

False — it comes from pure geometry:

⟨ v^{2} ⟩ = ⟨ v_{x}^{2} ⟩ + ⟨ v_{y}^{2} ⟩ + ⟨ v_{z}^{2} ⟩

and isotropy makes the three equal, so each is

\frac{1}{3} ⟨ v^{2} ⟩

. It is exact, not empirical.

Raising the temperature increases pressure only because molecules move faster.

False — heating raises pressure two ways at once: each molecule hits harder (

2 m v_{x}

bigger) and hits more often (rate

v_{x} /2 L

bigger). Both factors carry a

v_{x}

, which is why the result depends on

v_{x}^{2} \propto T

The derivation assumes molecules constantly collide with each other.

False — the ideal-gas derivation assumes molecules only collide with the walls; inter-molecular collisions merely redistribute speeds and preserve isotropy, so the pressure result is unchanged. (See Kinetic Theory of Gases.)

P V = n R T

holds for a real gas at very high pressure.

False — at high pressure the molecules' own volume and their mutual attractions matter, so the ideal assumptions break. You then need the Real Gases and Van der Waals Equation.

The constant

R

is more fundamental than

k_{B}

False —

k_{B}

is the fundamental per-molecule constant;

R = N_{A} k_{B}

is just

k_{B}

scaled up to one mole. Nature counts molecules, not moles.

Spot the error

Each statement contains one flaw. Name it.

"When a molecule hits the

x

-wall, its momentum change is

- m v_{x}

Error: it should be

- 2 m v_{x}

. The velocity reverses from

+ v_{x}

- v_{x}

, so the change is

- m v_{x} - (+ m v_{x}) = - 2 m v_{x}

, not just losing

m v_{x}

"Force on the wall uses the full speed

v

because that's how fast the molecule travels."

Error: only the perpendicular component

v_{x}

reverses and delivers momentum; the parallel components (

v_{y}, v_{z}

) just slide along the wall and give no push to that wall.

"Time between hits on the same wall is

L / v_{x}

Error: the molecule must travel to the far wall and back, a distance

2 L

, so

Δ t = 2 L / v_{x}

, not

L / v_{x}

"We can set

⟨ v_{x}^{2} ⟩ = ⟨ v_{y}^{2} ⟩ = ⟨ v_{z}^{2} ⟩

for a gas flowing through a pipe."

Error: isotropy fails when there's a bulk flow — a preferred direction exists, so the three mean squares differ. The equality only holds for a gas with no net drift.

"Substituting

⟨ E_{k} ⟩ = \frac{3}{2} k_{B} T

is a mathematical identity we derived earlier."

Error: it is a physical definition/input, coming from the Equipartition Theorem, not something the mechanics gave us. It is the bridge that turns Newton's laws into thermodynamics.

"Using

T = 25

for room temperature in

P V = n R T

is fine as long as you're consistent."

Error:

T

must be absolute (Kelvin) because it is proportional to kinetic energy;

25^{\circ} C = 298.15

K. At

0^{\circ} C

molecules still move, so Celsius would wrongly predict zero energy.

⟨ v^{2} ⟩ = ⟨ v ⟩^{2}

, so we can just square the average speed."

Error: the mean of the squares is not the square of the mean. Because of the spread in speeds,

⟨ v^{2} ⟩ > ⟨ v ⟩^{2}

; only

⟨ v^{2} ⟩

(the mean-square) enters the pressure. See Root Mean Square Speed.

Why questions

Force yourself to explain the reason, not restate the fact.

Why does the wall feel a steady pressure and not a jittery series of taps?

Because ~

1 0^{23}

molecules strike every second; individual taps blur into a smooth average push, just as raindrops on a roof sound continuous. Pressure is that time-average.

Why does pressure depend on

⟨ v^{2} ⟩

rather than on the average speed

⟨ v ⟩

Because both the momentum per hit (

\propto v_{x}

) and the hit-frequency (

\propto v_{x}

) scale with velocity, so their product carries

v_{x}^{2}

. Faster molecules count twice over.

Why can we replace the sum

\sum v_{x, i}^{2}

N ⟨ v_{x}^{2} ⟩

Because

⟨ v_{x}^{2} ⟩

is defined as

\frac{1}{N} \sum v_{x, i}^{2}

; multiplying the average by

N

just undoes the average. No physics is assumed there — it's bookkeeping.

Why must the collisions be elastic for the derivation to give a constant pressure?

Elastic means kinetic energy is conserved, so

⟨ v^{2} ⟩

(and hence

T

) stays fixed. If energy leaked into the walls, the gas would cool and

P

would drift down.

Why does

P V = n R T

contain no reference to the molecule's mass, yet the derivation is full of

m

The mass cancels when we insert the temperature bridge:

N m ⟨ v^{2} ⟩ = N \cdot 3 k_{B} T

, so

m

disappears into

T

. Temperature already encodes the mass-speed combination.

Why is isotropy (

⟨ v_{x}^{2} ⟩ = ⟨ v_{y}^{2} ⟩ = ⟨ v_{z}^{2} ⟩

) a physical assumption and not automatic?

It states the gas has no preferred direction — true only in equilibrium with no flow, gravity gradient, or field. It's an assumption about the state, which is why a moving or stratified gas can violate it.

Why does squeezing the box (smaller

V

) raise the pressure even if

T

is unchanged?

With a shorter path

2 L

between walls, each molecule hits more often per second, so more momentum is dumped per second on the same area. This is the

1/ V

P = N k_{B} T / V

Edge cases

The scenarios the tidy final formula quietly assumes away.

What does the kinetic model predict for pressure as

T \to 0

⟨ v^{2} ⟩ \to 0

, so

P \to 0

— molecules stop moving and stop kicking the walls. This defines absolute zero as the point of vanishing translational motion.

P V = n R T

valid for a single molecule (

N = 1

Only in a time-averaged, statistical sense — one molecule gives wildly fluctuating, spiky "pressure". The law is meaningful because

N

is astronomically large, smoothing the fluctuations.

What happens to the ideal-gas prediction as density

N m / V

becomes very large?

It fails, because molecular volume and attractions can no longer be ignored; real molecules aren't points. This is precisely the regime of the Real Gases and Van der Waals Equation.

Does the derivation care about the shape of the container (we used a cube)?

No — the cube is just convenient. Summed over all walls and orientations, only the total volume

V

survives, so any shape gives the same

P V = n R T

For a gas of two different molecular masses mixed together, what determines each species' partial pressure at temperature

T

Each species independently satisfies

P_{i} V = N_{i} k_{B} T

; the total is the sum (Dalton's law). Mass drops out because both share the same

T

, hence the same average kinetic energy.

If a container wall is moving (like a piston being pushed in), is the collision still "elastic" in our sense?

No — a moving wall changes the rebound speed, doing work on the gas and heating it. The static derivation assumes rigid, stationary walls; moving walls belong to the study of work and internal energy.

Does a monatomic and a diatomic gas at the same

T

N

V

have the same pressure?

Yes — pressure depends only on the translational kinetic energy

\frac{3}{2} k_{B} T

, which is identical for both. The extra rotational modes of the diatomic store energy but don't push on the walls (see Equipartition Theorem).

Recall One-line summary to lock in

Pressure = momentum bookkeeping; the $\frac{1}{3}$ is geometry (isotropy); the $T$ is a definition ( $⟨ E_{k} ⟩ = \frac{3}{2} k_{B} T$ ); and the law breaks exactly when the "point particles, no forces, elastic, isotropic" assumptions break.