Graham's law of effusion - diffusion (rate ∝ 1 - √M): Question bank

Reminder of the anchors we lean on:

Effusion = escape through a tiny pinhole into vacuum.
Diffusion = spreading/mixing of gases by random motion.
The equal thing at same $T$ is average kinetic energy, $\overline{K E} = \frac{3}{2} k_{B} T$ — not speed, not momentum.
The law: $\frac{r _{1}}{r _{2}} = \frac{M _{2}}{M _{1}}$ (the other gas's mass on top).

True or false — justify

Two gases at the same temperature have the same average speed.

False. They share the same average kinetic energy (

\frac{1}{2} m \overline{v^{2}}

), and since KE ties mass to speed squared, the lighter gas is faster. Equal energy ≠ equal speed.

Two gases at the same temperature have the same average kinetic energy.

True.

\overline{K E} = \frac{3}{2} k_{B} T

depends only on

T

— this is the whole engine of Graham's law and is independent of the molecule's mass or identity.

Doubling the temperature doubles the effusion rate.

False. Rate

\propto v_{r m s} \propto T

, so doubling

T

multiplies the rate by

2 \approx 1.41

, not by 2. The square root guards this too.

A heavier gas always effuses more slowly than a lighter one at the same

T

and

P

True. Since

r \propto 1/ M

and

M

grows with mass, larger

M

gives smaller rate — no exceptions among ideal gases at matched

T, P

If you raise the pressure of only one gas, Graham's simple ratio still holds.

False. The clean

M_{2} / M_{1}

form assumes equal

P

as well as

T

. More molecules per volume means more hits on the hole, so pressure changes the rate independently of mass.

Graham's law applies equally to effusion and diffusion ratios.

True — both scale as

1/ M

. But diffusion's actual speed is throttled by collisions, so only the ratio transfers cleanly, not the absolute travel time.

At the same

T

, the gas with higher density effuses faster.

False. At same

T, P

, density

ρ \propto M

, so higher density means heavier molecules, which effuse slower:

r \propto 1/ ρ

CO and N₂ (both

M = 28

) effuse at identical rates.

True. Graham's law sees only molar mass; identical

M

gives identical rates regardless of the gases being chemically different.

Spot the error

"H₂ is 16 times lighter than O₂, so H₂ effuses 16 times faster."

Error: forgot the square root. Mass links to speed squared, so the speed (and rate) ratio is

16 = 4

, not 16. H₂ effuses 4× faster.

\frac{r _{1}}{r _{2}} = \frac{M _{1}}{M _{2}}

— keep matching subscripts."

Error: masses must swap. Rate is inverse to

M

, so the heavier gas's mass belongs on top:

\frac{r _{1}}{r _{2}} = \frac{M _{2}}{M _{1}}

. Sanity check: lighter gas must come out faster.

"The lighter gas escapes faster because it has more kinetic energy."

Error: KE is equal, not larger. Both gases have the same

\overline{K E} = \frac{3}{2} k_{B} T

. The lighter one is faster because its mass is small, not because its energy is big.

"In the NH₃–HCl tube, the white ring forms in the middle since they meet halfway."

Error: they don't move at equal speeds. NH₃ (

M = 17

) is lighter and faster than HCl (

M = 36.5

), so it travels farther; the ring forms nearer the HCl end.

"Rate ∝

1/ M

, so a gas 4× slower is 2× heavier."

Error: undo the root correctly. Slower by 4× means

1/ M

shrank 4×, so

M

grew 4×, so

M

grew

4^{2} = 16

×. The gas is 16× heavier.

"We can apply Graham's law to compare a hot gas with a cold gas by just using their masses."

Error: $T$ must match. The

3 R T

factor only cancels when temperatures are equal. Different

T

requires the full

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

for each gas.

"Because effusion is molecules leaking out, doubling the hole size doubles the mass ratio effect."

Error: hole size cancels. Both gases share the same hole, so its area drops out of the ratio. Graham's law is about the relative rates, which depend only on mass at fixed

T, P

Why questions

Why is it

1/ M

and not

1/ M

Because kinetic energy

\frac{1}{2} m v^{2}

(not momentum

m v

) is the equal quantity, so mass appears with

v^{2}

. Solving for

v

and taking the root gives

v \propto 1/ M

, and rate follows speed.

Why does the other gas's molar mass go on top of the ratio?

Rate is inversely proportional to

M

, so when you divide two rates the masses flip: the gas you're comparing against ends up in the numerator.

Why must temperature be equal for the simple form?

The full speed is

3 R T / M

; only when

T_{1} = T_{2}

does

3 R T

cancel between the two gases, leaving a pure mass ratio.

Why does effusion happen one molecule at a time without collisions in the hole?

The pinhole is smaller than the average distance between collisions (the mean free path), so molecules pass through individually rather than as a colliding stream — that's what makes rate track

v_{r m s}

directly.

Why can density replace mass in Graham's law?

At the same

T

and

P

, equal volumes hold equal molecule counts (ideal-gas behaviour), so density is proportional to molar mass:

ρ \propto M

. Swapping

M \to ρ

leaves the ratio unchanged.

Why is UF₆ used to separate uranium isotopes despite tiny mass differences?

The masses of

^{235}

UF₆ and

^{238}

UF₆ differ by only ~1%, giving a rate ratio near

352/349 \approx 1.004

; because the effect is so small, thousands of effusion stages are cascaded to enrich the lighter isotope (see Isotope Separation (Uranium Hexafluoride)).

Why does real diffusion in air look slow even for light gases?

Molecules collide constantly and zig-zag, so despite huge instantaneous speeds their net progress across a room is slow — the collisions cancel much of the motion, though the ratio of two gases' diffusion still obeys

1/ M

Edge cases

Two gases with exactly equal molar masses — what is their rate ratio?

Exactly

1

M_{2} / M_{1} = 1 = 1

. Graham's law can't distinguish them, so they effuse at identical rates.

What happens to the rate as molar mass approaches zero (hypothetical)?

r \propto 1/ M \to \infty

M \to 0

— the rate grows without bound, matching the picture that ever-lighter molecules move ever faster (a limiting behaviour, not a real gas).

What happens to

v_{r m s}

as temperature approaches absolute zero?

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

T \to 0

, so effusion stops. All molecular motion ceases at

0 K

, consistent with Kinetic Theory of Gases.

If one side of an effusion setup is not a vacuum, does Graham's law still hold?

The pure form assumes escape into vacuum; a back-pressure lets molecules re-enter, so the net rate falls and the simple ratio only approximates reality until pressures are balanced.

For a mixture of two gases effusing through the same hole, which enriches the escaping stream?

The lighter component, since it effuses faster; the gas leaving is momentarily richer in the low-

M

species — the principle behind isotope enrichment.

Does Graham's law hold for a single gas comparing two temperatures?

Only via the full

v_{r m s}

: with

M

fixed,

r \propto T

, so the same gas effuses faster when hotter. The mass-ratio shortcut doesn't apply — there's no second mass to compare.

If the two gases are at the same

T

but you measure distance travelled instead of moles, does the ratio change?

No — distance, moles, and volume per unit time are all proportional to rate, so

\frac{d _{1}}{d _{2}} = \frac{n _{1}}{n _{2}} = \frac{V _{1}}{V _{2}} = \frac{M _{2}}{M _{1}}

all give the same ratio.

Connections

Parent topic — Graham's law
Root Mean Square Speed — the speed these traps hinge on.
Kinetic Theory of Gases — source of the equal-KE principle.
Maxwell-Boltzmann Distribution — why "average" speed is a distribution, not a single value.
Ideal Gas Equation — justifies $ρ \propto M$ at fixed $T, P$ .
Isotope Separation (Uranium Hexafluoride) — the edge-case application.