Rate of Effusion Calculator
Compare how quickly two gases diffuse or effuse with Graham's law. This calculator can solve for either gas rate or for an unknown molar mass when the other values are known.
Result
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Quick Answer: Graham's law says r1 / r2 = sqrt(M2 / M1). A lighter gas diffuses or effuses faster because its molar mass appears in the denominator under the square root.
How to Use Graham's Law
- Choose diffusion or effusion mode: The same mass relationship applies, but the wording of the result changes with the process you choose.
- Pick the unknown: You can solve for a gas rate or for one of the molar masses.
- Enter the remaining values: Keep both rates on the same basis, such as relative speed, time inverse, or a shared measurement scale.
- Interpret the ratio: The lower-molar-mass gas should come out with the higher rate.
Rate of Effusion Calculator Formula
r1 / r2 = sqrt(M2 / M1)
| Variable | Meaning | Unit |
|---|---|---|
| r1 | Rate of gas 1 | same arbitrary rate basis as r2 |
| r2 | Rate of gas 2 | same arbitrary rate basis as r1 |
| M1 | Molar mass of gas 1 | g/mol |
| M2 | Molar mass of gas 2 | g/mol |
Worked Examples
Example 1 - Helium versus oxygen
- Gas 1 molar mass: 4 g/mol
- Gas 2 molar mass: 32 g/mol
- Gas 2 rate: 1.00
Result: Gas 1 rate = 2.83
Helium is much lighter than oxygen, so it diffuses and effuses about 2.8 times faster.
Example 2 - Find an unknown molar mass
- Gas 1 rate: 3.0
- Gas 2 rate: 1.5
- Gas 1 molar mass: 16 g/mol
Result: Gas 2 molar mass = 64 g/mol
When gas 1 is twice as fast, gas 2 must be four times heavier under Graham's law.
Example 3 - Nitrogen versus hydrogen
- Gas 1 molar mass: 28 g/mol
- Gas 2 molar mass: 2 g/mol
- Gas 1 rate: 1.0
Result: Gas 2 rate = 3.74
Hydrogen moves much faster because its molar mass is far smaller than nitrogen's.
Graham's Law Reference Table
| Range | Meaning | Action |
|---|---|---|
| Rate ratio greater than 1 | Gas 1 is faster than gas 2. | Gas 1 should also be the lighter gas if the inputs are consistent. |
| Rate ratio equal to 1 | Both gases move at the same rate in the model. | Their molar masses should match. |
| Rate ratio less than 1 | Gas 2 is faster than gas 1. | Check that gas 2 also has the lower molar mass. |
Frequently Asked Questions
It is the relationship stating that a gas rate is inversely proportional to the square root of its molar mass when conditions are otherwise comparable.
Diffusion describes mixing through another gas, while effusion describes a gas escaping through a tiny opening. Graham's mass relationship is commonly used for both.
The lighter gas diffuses faster because the rate varies with the inverse square root of molar mass.
Yes. If you know both gas rates and one molar mass, the calculator can rearrange Graham's law to find the other.
They only need to be on the same basis. The law uses a ratio, so both rates must be comparable measurements.
Note: This calculator uses the standard idealized Graham's-law relationship. Real gas behavior can deviate under some conditions.
References
Last reviewed: March 14, 2026