Q.Equal moles of hydrogen and oxygen gases are placed in a container, with a pin-hole through which both can escape. What fraction of oxygen escapes in the time required for one-half of the hydrogen to escape? (NEET phase I)
Graham's Law of Diffusion – From Intuition to Precision
Imagine you're in a room where someone opens a bottle of perfume at one end. You don't smell it instantly — it takes time for the perfume molecules to wander across the room. Now imagine the same experiment with a bottle of ammonia. You'd smell the ammonia much faster. Why? The ammonia molecules are lighter.
That's the core physical idea: lighter gas molecules move faster, on average, than heavier ones at the same temperature. Since diffusion and effusion are processes driven by molecular motion, a lighter gas will spread out (diffuse) or escape through a tiny hole (effuse) more quickly than a heavier gas.
Note
Diffusion is the mixing of gases due to random molecular motion. Effusion is the escape of a gas through a tiny hole into a vacuum. Graham's Law applies to both.
The Precise Statement
Graham's Law of Diffusion/Effusion states:
At constant temperature and pressure, the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass (or density).
Mathematically, for two gases A and B:
rBrA=MAMB=ρAρB
where:
r = rate of diffusion/effusion (volume or moles per unit time)
M = molar mass
ρ = density (at same T and P)
r2r1=M1M2
Why the Square Root? (The Physics)
The reason comes from kinetic molecular theory. At a given temperature, the average kinetic energy of gas molecules is the same for all gases:
21mv2=constant
Here m is the mass of one molecule and v is its speed. Rearranging:
v∝m1
Since molar mass M is proportional to molecular mass m, the average molecular speed is inversely proportional to M. And since the rate of diffusion/effusion is directly proportional to this average speed, you get Graham's Law.
Watch out
A common mistake is to invert the ratio. If gas A is lighter (MA<MB), then rA>rB. Check: MB/MA>1, so rA/rB>1 — correct. Always put the lighter gas in the numerator if you want a ratio > 1.
Worked Example
Problem: Hydrogen (M=2g/mol) and oxygen (M=32g/mol) are allowed to effuse through identical pinholes. How much faster does hydrogen effuse?
Step 1. By Graham's law, rO2rH2=MH2MO2=232=16=4 -- hydrogen effuses 4 times faster than oxygen under the same conditions.
Step 2. In a given time t, the amount of gas that effuses is proportional to its rate. If half (0.5 mol out of 1 mol, taking equal starting moles) of the hydrogen escapes in time t, this corresponds to an "amount effused" of 0.5 mol for H2 in that time. …