Graham’s law application: Helium has four times the atomic weight of hydrogen. Relative to hydrogen, helium will diffuse at what multiple of the rate?

Introduction / Context:Graham’s law states that the rate of diffusion (or effusion) of a gas is inversely proportional to the square root of its molar mass. This principle is widely used for order-of-magnitude estimates in mass transfer and gas separation contexts.

Given Data / Assumptions:

• Comparison is qualitative using Graham’s law: r ∝ 1/√M.
• Statement references atomic weight ratio He:H = 4:1, treating hydrogen on a per-atom basis in the intended textbook-style simplification.
• Isothermal, same pressure conditions.

Concept / Approach:If helium’s mass is four times hydrogen’s reference mass, then the rate ratio is r_He / r_H = √(M_H / M_He) = √(1/4) = 1/2. Many pedagogy problems use this simplified mass ratio to reinforce the inverse square-root dependence without delving into diatomic versus monatomic complications.

Step-by-Step Solution:Write Graham’s law: r_A / r_B = √(M_B / M_A).Set M_He / M_H = 4. Then r_He / r_H = √(1/4) = 1/2.Therefore helium diffuses at half the rate of hydrogen (per the problem’s stated ratio).

Verification / Alternative check:Using molecular gases (He = 4, H2 = 2) would give r_He / r_H2 = √(2/4) = 1/√2. However, the stem explicitly frames a 4:1 ratio, yielding 1/2 per the simplified approach.

Why Other Options Are Wrong:Options 4 and 2 invert Graham’s dependence; 1/4 corresponds to mass ratio 16:1; 1/√2 applies if comparing He to H2, which is not how the stem is framed.

Common Pitfalls:Confusing atomic hydrogen with molecular hydrogen; always match the mass basis used in the question.

Final Answer:1/2