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

The single tool for this whole page is one boxed relation. Let us re-earn it from scratch so nothing is used blind.

Re-deriving the master ratio (the WHY behind the formula)

Look at the picture below. Two sealed rooms, each holding a gas at the same temperature $T$ . On the left are light molecules (blue); on the right, heavy ones (pink). Each has a tiny pinhole in the right wall.

Figure — Graham's law of effusion - diffusion (rate ∝ 1 - √M)

Figure 1 — Alt text: A dark chalkboard shows two boxes side by side. The left box holds many small blue circles (a light gas), each drawn with a long arrow meaning high speed. The right box holds fewer large pink circles (a heavy gas), each with a short arrow meaning low speed. Both walls have a tiny gap labelled "pinhole." A caption reads "same T ⇒ same average KE ⇒ v ∝ 1/√M," conveying that with equal energy the light molecules move faster and so reach and escape through the hole more often.

Follow the visual story in four moves — this is the derivation:

Same $T$ means same average kinetic energy. Temperature is the average translational kinetic energy of a molecule (this is the core claim of Kinetic Theory of Gases). So the average KE of a blue molecule equals that of a pink one: $\frac{1}{2} m \overline{v^{2}}$ is the same number on both sides (here $m$ is one molecule's mass). This is why in the figure the light blue balls and heavy pink balls were all "kicked with the same energy."
Same energy, but different mass, forces different speed. Since $\frac{1}{2} m \overline{v^{2}}$ is fixed, a bigger $m$ must come with a smaller $\overline{v^{2}}$ . Solving, the typical speed is $v_{r m s} = \frac{3 R T}{M}$ (see Root Mean Square Speed), where $M = N_{A} m$ is the molar mass. The heavy pink balls therefore crawl (short pink arrows in the figure); the light blue balls zoom (long blue arrows).
WHY rate tracks molecular speed — the flux argument. A molecule escapes only if it arrives at the hole moving outward. The number striking a patch of wall of area $A$ per second — the flux — is, from kinetic theory, $flux = \frac{1}{4} n^{*} ⟨ v ⟩ A$ Let us read each factor: $n^{*}$ is the number density (molecules per unit volume) — more molecules, more hits; $⟨ v ⟩$ is the average speed — faster molecules hit more often; and the $\frac{1}{4}$ is a fixed geometric factor coming from isotropy — molecules fly equally in all directions, so only the fraction heading toward the wall (and the cosine of their angle) counts, and averaging that over a hemisphere yields exactly $\frac{1}{4}$ . At the same $T$ and $P$ , $n^{*}$ is identical for both gases (equal $P, T$ ⇒ equal molecules per litre, from Ideal Gas Equation), and $\frac{1}{4} A$ is the same hole. So the only thing that differs between the two gases is $⟨ v ⟩$ . Hence $r \propto ⟨ v ⟩$ .
rms vs average speed — an honest footnote. Strictly, the flux uses the average speed $⟨ v ⟩$ , not the root-mean-square speed $v_{r m s}$ . For the Maxwell-Boltzmann Distribution these differ by a fixed constant: $⟨ v ⟩ = \frac{8 R T}{π M}$ while $v_{r m s} = \frac{3 R T}{M}$ , so $⟨ v ⟩ = \frac{8}{3 π} v_{r m s} \approx 0.921 v_{r m s}$ . Crucially, both are proportional to $1/ M$ . Since that shared $R T$ -and-constant factor cancels when we take a ratio of two gases, it makes no difference whether we track $⟨ v ⟩$ or $v_{r m s}$ — the $1/ M$ dependence is identical. We use $v_{r m s}$ below only because it is the speed the parent note built; the ratio is unchanged.

Putting it together, at fixed $T$ the factor $3 R T$ is a shared constant, so: $r \propto ⟨ v ⟩ \propto v_{r m s} = \frac{3 R T}{M} \propto \frac{1}{M}$

Now compare two gases by dividing their rates. The shared $3 R T$ (and the shared $\frac{1}{4} n^{*} A$ ) cancels top and bottom: $\frac{r _{1}}{r _{2}} = \frac{3 R T / M _{1}}{3 R T / M _{2}} = \frac{M _{2}}{M _{1}}$

The "faces of rate" — why $n$ , $V$ , $d$ , $ρ$ all obey the same ratio

The formula speaks about $r$ , but experiments measure other things. Here is why each measurable stands in for the rate — one justification each, no bundling:

Definition Translating rate into what you can actually measure

Moles, same time. Rate is moles-per-second: $r = n / t$ . If both gases effuse for the same time $t$ , then $r_{1} / r_{2} = (n_{1} / t) / (n_{2} / t) = n_{1} / n_{2}$ — the $t$ cancels. So amounts collected in equal time are in the ratio of rates.
Volume, same time. At the same $T, P$ , equal moles occupy equal volume (from Ideal Gas Equation, $V \propto n$ ). So $V_{1} / V_{2} = n_{1} / n_{2} = r_{1} / r_{2}$ . Volume is just a stand-in for moles.
Distance in a tube, same time. In a diffusion tube both fronts advance for the same elapsed time until they meet. Distance = speed × time, and with equal time the times cancel: $d_{1} / d_{2} = v_{1} / v_{2} = r_{1} / r_{2}$ . So how far each front travels is the rate ratio.
Density. At fixed $P, T$ , density $ρ \propto M$ (heavier molecules pack more mass per litre). Substituting $M \to ρ$ in the master ratio: $r_{1} / r_{2} = M_{2} / M_{1} = ρ_{2} / ρ_{1}$ .

Collecting these (each earned above): $\frac{r _{1}}{r _{2}} = \frac{n _{1}}{n _{2}} = \frac{V _{1}}{V _{2}} = \frac{d _{1}}{d _{2}} = \frac{M _{2}}{M _{1}} = \frac{ρ _{2}}{ρ _{1}}$

Time is the odd one out. Since $r = amount / t$ , for a fixed amount the time is the inverse of rate: $\frac{t _{1}}{t _{2}} = \frac{r _{2}}{r _{1}} = \frac{M _{1}}{M _{2}}$ . Slower gas ⇒ longer time.

Level 1 — Recognition

Recall Solution

Lighter = faster, so $C H_{4}$ is faster. The factor: $\frac{r _{C H_{4}}}{r _{S O_{2}}} = \frac{M _{S O_{2}}}{M _{C H_{4}}} = \frac{64}{16} = 4 = 2$ CH₄ effuses 2× faster. Sanity check: it is 4× lighter, and $4 = 2$ . ✓

Recall Solution

Larger rate ⇒ smaller mass. So Q has the larger molar mass. Rate and mass pull in opposite directions.

Recall Solution

$\frac{r _{1}}{r _{2}} = \frac{ρ _{2}}{ρ _{1}} = \frac{8}{2} = 4 = 2$ The lighter (less dense) gas 1 effuses twice as fast. ✓

Level 2 — Application

Recall Solution

$\frac{r _{u nk}}{r _{H_{2}}} = \frac{1}{3} = \frac{M _{H_{2}}}{M _{u nk}} = \frac{2}{M _{u nk}}$ Square both sides (to undo the root): $\frac{1}{9} = \frac{2}{M _{u nk}} ⟹ M_{u nk} = 2 \times 9 = 18 g/mol$ A gas 3× slower is $3^{2} = 9$ × heavier: $2 \times 9 = 18$ . ✓ (This is water vapour, $H_{2} O$ .)

Recall Solution

Same volume, so $r = V / t \Rightarrow r \propto 1/ t$ . The unknown takes twice as long, so it is half as fast: $\frac{r _{u nk}}{r _{N e}} = \frac{t _{N e}}{t _{u nk}} = \frac{30}{60} = \frac{1}{2} = \frac{M _{N e}}{M _{u nk}}$ $\frac{1}{4} = \frac{20}{M _{u nk}} ⟹ M_{u nk} = 80 g/mol$ Half the speed ⇒ 4× the mass ⇒ $20 \times 4 = 80$ . ✓

Recall Solution

Same time ⇒ volumes are in the ratio of rates: $\frac{V _{O_{2}}}{V _{H e}} = \frac{r _{O_{2}}}{r _{H e}} = \frac{M _{H e}}{M _{O_{2}}} = \frac{4}{32} = \frac{1}{8} = \frac{1}{2 2}$ $V_{O_{2}} = 40 \times \frac{1}{2 2} = \frac{40}{2.828} \approx 14.14 mL$ Heavier O₂ is slower, so less volume gets out. ✓

Level 3 — Analysis

Recall Solution

Study the tube figure below as you read. Both fronts advance for the same time until they meet, so (from the "distance in a tube" face of rate) distances are in the ratio of speeds: $\frac{d _{N H_{3}}}{d _{H C l}} = \frac{M _{H C l}}{M _{N H_{3}}} = \frac{36.5}{17} = 2.147 \approx 1.465$ The distances add to the tube length: $d_{N H_{3}} + d_{H C l} = 100$ . Let $d_{H C l} = x$ , then $d_{N H_{3}} = 1.465 x$ : $1.465 x + x = 100 ⟹ 2.465 x = 100 ⟹ x \approx 40.6 cm$ So $d_{N H_{3}} = 100 - 40.6 = 59.4 cm$ . In the figure, the long blue arrow (NH₃) covers most of the tube while the short pink arrow (HCl) barely moves; the white ring sits where they meet. The ring forms about $59.4 cm$ from the NH₃ end (i.e. closer to the HCl end, as expected — lighter NH₃ travels farther).

Figure 2 — Alt text: A dark chalkboard shows a long horizontal tube of length 100 cm. A blue arrow starts at the left "NH₃ end" and travels far to the right, about 59.4 cm, labelled fast. A pink arrow starts at the right "HCl end" and travels only about 40.6 cm to the left, labelled slow. Where the two arrows meet, a thick white vertical bar marks the white ring of NH₄Cl. The picture shows that the lighter NH₃ covers more of the tube, so the ring forms nearer the heavier HCl end.

Recall Solution

Each gas effuses at a rate proportional to its own $v_{r m s}$ and to how much of it is present. With equal starting moles, the escaped amounts are simply in the ratio of rates: $\frac{n _{A}^{esc}}{n _{B}^{esc}} = \frac{r _{A}}{r _{B}} = \frac{M _{B}}{M _{A}} = \frac{16}{4} = 4 = 2$ So the escaped gas is enriched to a mole ratio $A : B = 2 : 1$ — that is $\frac{2}{3} \approx 66.7%$ A. The light gas is preferentially skimmed off. This is the principle behind Isotope Separation (Uranium Hexafluoride).

Level 4 — Synthesis

Recall Solution

(a) $\frac{r _{235}}{r _{238}} = \frac{352}{349} = 1.008596 \approx 1.00429$ A separation of only $\approx 0.43%$ per stage — the masses are almost equal. (b) Two stages multiply: $(1.00429)^{2} \approx 1.00860$ , i.e. $\approx 0.86%$ . This is why real plants use thousands of cascaded stages. Small advantage, compounded relentlessly.

Recall Solution

(a) $\frac{r _{X}}{r _{C O_{2}}} = 1.5 = \frac{M _{C O_{2}}}{M _{X}} = \frac{44}{M _{X}}$ $2.25 = \frac{44}{M _{X}} ⟹ M_{X} = \frac{44}{2.25} \approx 19.56 g/mol$ (b) At the same $P, T$ , density $ρ \propto M$ , so: $\frac{ρ _{X}}{ρ _{C O_{2}}} = \frac{M _{X}}{M _{C O_{2}}} = \frac{19.56}{44} \approx 0.444$ Faster gas ⇒ lighter ⇒ less dense. ✓

Level 5 — Mastery

Recall Solution

Qualitative: He (lighter) escapes faster, so it leaves the vessel preferentially. The gas left behind becomes progressively richer in the heavier Ne — the remaining mixture's average molar mass rises over time. Quantitative (initial instant, equal moles present): $\frac{n _{H e}^{esc}}{n _{N e}^{esc}} = \frac{r _{H e}}{r _{N e}} = \frac{M _{N e}}{M _{H e}} = \frac{20}{4} = 5 \approx 2.236$ So for every $1$ mole of Ne that escapes, about $2.24$ moles of He escape at the start. As He depletes, this lead shrinks (fewer He molecules to present at the hole).

Recall Solution

$\frac{r _{o x}}{r _{N_{2}}} = 0.7906 = \frac{28}{M _{o x}}$ $0.790 6^{2} = 0.625 = \frac{28}{M _{o x}} ⟹ M_{o x} = \frac{28}{0.625} = 44.8 g/mol$ On the rounding: the arithmetic gives $44.8 g/mol$ , not exactly $44$ . The $\approx 0.8$ excess comes from the rate ratio $0.7906$ being a rounded input — a truly $44.0$ g/mol gas would give $28/44 = 0.7977$ , close but not identical. Molar masses of real molecules are quantised (you can only build them from whole atoms), so we snap $44.8$ to the nearest chemically possible value: $44 g/mol$ . Identifying the oxide: we need a nitrogen oxide $N_{a} O_{b}$ with $14 a + 16 b = 44$ . Try $a = 2, b = 1$ : $2 (14) + 16 = 44$ . ✓ That is $N_{2} O$ (dinitrogen monoxide, "laughing gas"). Other candidates fail: $NO$ is $30$ , $N O_{2}$ is $46$ , so $N_{2} O$ is the unique fit. The $\sim 2%$ gap between $44.8$ and $44$ is measurement/rounding noise, not a fifth atom. ✓

Recall Solution

Rate $\propto v_{r m s} = 3 R T / M$ . The constant $3 R$ cancels in the ratio: $\frac{r _{A}}{r _{B}} = \frac{T _{A} / M _{A}}{T _{B} / M _{B}} = \frac{T _{A}}{M _{A}} \cdot \frac{M _{B}}{T _{B}} = \frac{300}{4} \cdot \frac{16}{1200} = \frac{300 \times 16}{4 \times 1200}$ $= \frac{4800}{4800} = 1 = 1$ They effuse at equal rates! B is 4× heavier (which alone would make it $4 = 2$ × slower) but 4× hotter (which alone makes it $4 = 2$ × faster). The two effects cancel exactly. This is why "same $T$ " is a required condition for the simple law.

Recall wrap-up

Recall Which face of "rate" for which clue?

Same time, compare amounts ::: use $r_{1} / r_{2} = V_{1} / V_{2} = n_{1} / n_{2}$ Same amount, compare times ::: use $t_{1} / t_{2} = r_{2} / r_{1} = M_{1} / M_{2}$ (inverse!) Diffusion tube, meeting point ::: use $d_{1} / d_{2} = M_{2} / M_{1}$ AND $d_{1} + d_{2} = L$ Densities given ::: use $r_{1} / r_{2} = ρ_{2} / ρ_{1}$ Different temperatures ::: abandon simple form, use $v_{r m s} = 3 R T / M$ each Distinguish $m$ from $M$ ::: $m$ = one molecule's mass; $M = N_{A} m$ = molar mass (per mole)

Connections

Parent topic note — the full derivation these exercises drill.
Root Mean Square Speed — the $v_{r m s} = 3 R T / M$ used in the re-derivation and L5·Q3.
Kinetic Theory of Gases — why temperature is shared kinetic energy, and the flux $\frac{1}{4} n^{*} ⟨ v ⟩ A$ .
Maxwell-Boltzmann Distribution — the spread of speeds and the $⟨ v ⟩$ vs $v_{r m s}$ distinction.
Ideal Gas Equation — supplies $V \propto n$ and $ρ \propto M$ at fixed $P, T$ .
Isotope Separation (Uranium Hexafluoride) — the real payoff of L4·Q1.