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Grahams Law of Effusion

General Chemistry • Gases

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Graham’s Law of Effusion — Ratios, Unknown Mass, & Predictions

For gases at the same \(T\) and \(P\), Graham’s law states that an effusion-related quantity \(Q\) (molecular speed, effusion rate, distance traveled, or amount effused in a fixed time) is inversely proportional to the square root of the molar mass: \( Q \propto \dfrac{1}{\sqrt{M}} \). Consequently, \( \dfrac{Q_A}{Q_B}= \sqrt{\dfrac{M_B}{M_A}} \), whereas time to effuse a fixed amount satisfies \( \dfrac{t_A}{t_B}= \sqrt{\dfrac{M_A}{M_B}} \).

Assumes same \(T\) and \(P\) for A and B.

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Frequently Asked Questions

What is Graham's law of effusion?

Graham's law states that an effusion-related quantity Q is inversely proportional to the square root of molar mass M at the same temperature and pressure. It is commonly written as Q_A/Q_B = sqrt(M_B/M_A).

How do you calculate the ratio of effusion rates for two gases?

Use the rate form Q_A/Q_B = sqrt(M_B/M_A), where Q can be an effusion rate, molecular speed, distance traveled in a fixed time, or amount effused in a fixed time. Lighter gases (smaller M) give larger Q values.

How do you find an unknown molar mass using an observed ratio?

Rearrange Graham's law. For a rate-type ratio, M_A = M_B / (Q_A/Q_B)^2; for a time ratio, M_A = (t_A/t_B)^2 x M_B.

When should you use time ratio instead of rate ratio?

Use the time ratio t_A/t_B when the problem compares how long each gas takes to effuse the same amount. Use the rate ratio when comparing how fast each gas effuses or how much passes in a fixed time.

What assumptions are required for Graham's law to be accurate?

The gases should be at the same temperature and pressure, and effusion should occur through a tiny orifice where molecules pass without significant collisions. Bulk diffusion with many collisions can deviate from the ideal relationship.