Rate of Effusion Calculator
Calculate gas effusion rates using Graham's Law. Compare effusion speeds of two gases based on molar mass with step-by-step solutions and RMS velocity
Rate of Effusion Calculator
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Formula: rate1/rate2 = sqrt(M2/M1)
Worked example โ Hydrogen effuses 3.98 times faster than oxygen
Formula
rate1/rate2 = sqrt(M2/M1)
Graham's Law of Effusion states the rate of effusion is inversely proportional to the square root of molar mass. Lighter gases effuse faster. The ratio of rates equals the square root of the inverse ratio of molar masses.
Worked Examples
Example 1: Hydrogen vs Oxygen
Problem:Compare the effusion rates of hydrogen (M=2.016 g/mol) and oxygen (M=32.00 g/mol).
Solution:rate(H2)/rate(O2) = sqrt(M(O2)/M(H2)) = sqrt(32.00/2.016) = sqrt(15.873) = 3.984
Result:Hydrogen effuses 3.98 times faster than oxygen
Example 2: Unknown Gas Identification
Problem:A gas effuses 0.354 times as fast as helium (M=4.003). Find the molar mass of the unknown gas.
Solution:rate(X)/rate(He) = sqrt(M(He)/M(X)) 0.354 = sqrt(4.003/M(X)) 0.354^2 = 4.003/M(X) M(X) = 4.003/0.1253 = 31.95 g/mol (likely O2)
Result:Unknown gas molar mass = 31.95 g/mol (oxygen)
Frequently Asked Questions
What is Graham's Law of Effusion?
Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, rate1/rate2 = sqrt(M2/M1), where M1 and M2 are the molar masses of the two gases. This means lighter gases effuse faster than heavier ones. For example, hydrogen (M=2) effuses 4 times faster than oxygen (M=32) because sqrt(32/2) = 4. The law was discovered by Scottish chemist Thomas Graham in 1848.
What is the difference between effusion and diffusion?
Effusion is the escape of gas molecules through a tiny hole (smaller than the mean free path) into a vacuum, while diffusion is the spreading of gas molecules through another gas or medium. Both processes follow Graham's Law, but effusion is simpler because molecules pass through one at a time without collisions. Diffusion is slower due to molecular collisions and is described by Fick's laws. In practice, effusion is used in isotope separation and leak detection, while diffusion governs mixing and transport in bulk systems.
How is effusion used in isotope separation?
Gaseous diffusion was historically used to enrich uranium-235 from uranium-238 by converting uranium to UF6 gas and passing it through porous barriers. Since U-235-F6 (M=349) is slightly lighter than U-238-F6 (M=352), it effuses slightly faster with a separation factor of sqrt(352/349) = 1.0043. Because this factor is so close to 1, thousands of stages are needed for significant enrichment. This process was used in the Manhattan Project and remained in use until centrifuge technology largely replaced it.
What is the RMS speed of a gas molecule?
The root-mean-square (RMS) speed is the square root of the average squared velocity of gas molecules: vrms = sqrt(3RT/M), where R is the gas constant (8.314 J/mol/K), T is temperature in Kelvin, and M is molar mass in kg/mol. At room temperature (298 K), nitrogen molecules have an RMS speed of about 515 m/s, while helium molecules travel at about 1360 m/s. The RMS speed is directly related to effusion rates and is slightly higher than the average molecular speed.
How do you identify an unknown gas from its effusion rate?
Rearrange Graham's Law to solve for the unknown molar mass: M_unknown = M_known x (rate_known / rate_unknown) squared. If a gas effuses at half the rate of helium, whose molar mass is 4.00 g/mol, then M = 4.00 x (1/0.5) squared = 4.00 x 4 = 16.0 g/mol, which points to methane. Because the relationship is a square, a small error in the measured rate ratio doubles into the molar mass, so timings should be repeated.
How does temperature change the rate of effusion?
Molecular speeds scale as the square root of absolute temperature, so the effusion rate of a single gas is proportional to the square root of T divided by molar mass. Doubling the absolute temperature raises the effusion rate by a factor of the square root of 2, about 1.41. Crucially, the ratio between two gases is unaffected, because the temperature factor is the same for both. That is why Graham's Law compares gases only when they are at the same temperature and pressure.
What is the relationship between effusion time and effusion rate?
Rate and time are reciprocals for a fixed amount of gas escaping through the same aperture, so if the rate goes as the inverse square root of molar mass, the time goes as the square root: t1/t2 = square root of (M1/M2). A heavier gas therefore takes longer. If oxygen at 32.00 g/mol takes 60 seconds to escape, helium at 4.00 g/mol takes 60 divided by the square root of 8, about 21 seconds.
Why does a helium balloon deflate faster than an air-filled one?
Two effects work in the same direction. Helium's molar mass of 4.00 g/mol against about 29 g/mol for air means helium atoms move roughly the square root of 29/4, about 2.7 times faster, so they find and pass through pores far more often. Helium atoms are also very small, so they permeate the latex membrane itself rather than merely escaping at the neck. Mylar balloons hold helium longer because the metallised film is much less permeable.
What conditions must hold for Graham's Law to be valid?
The aperture must be smaller than the mean free path of the molecules, so that molecules arrive at the hole individually rather than as a flowing stream. That means low pressure, a genuinely pinhole opening, and gases behaving close to ideally. Both sides must be at the same temperature, and the comparison assumes the same pressure driving each gas. If the hole is large or the pressure high, flow becomes hydrodynamic and the dependence on molar mass largely disappears.
Why does effusion depend on molar mass rather than molecular size?
Kinetic theory gives the effusive flux as one quarter of the number density times the mean molecular speed, multiplied by the hole area. Molecular diameter never enters that expression, because each molecule crosses the aperture independently. Speed does enter, and it scales as the square root of T/M, which is where the inverse square-root dependence on molar mass comes from. Size matters for diffusion through a bulk gas and for permeation through a membrane, but not for effusion through a pinhole.
References
Background & Theory
History
Reviewed for accuracy by Manoj Kumar, Mathematics Educator ยท Editorial policy
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