# Graham's Law

on . Posted in Classical Mechanics

Graham's law, also called Graham's law of effusion, relates to the diffusion of gases.  It states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass, assuming other conditions such as temperature and pressure are constant.  Or you could say the process in which one gas scatters itself within another.  In simpler terms, lighter gases diffuse or effuse faster than heavier gases under the same conditions.  This means that gases with lower molar masses will spread out more quickly and travel greater distances in a given amount of time compared to gases with higher molar masses.

This concept is particularly useful in understanding the behavior of gases and predicting their relative rates of diffusion or effusion.  It has applications in various fields, including chemistry, physics, and atmospheric science.

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### Graham's Law formula

$$R_1\;/\;R_2 = \sqrt { M_2 \;/\; M_1 }$$     (Graham's Law0

$$R_1 = R_2 \; \sqrt{ M_2 } \;/\; M_1$$

$$R_2 = R_1 \; M_1 \;/\; \sqrt{ M_2 }$$

$$M_2 = R_1^2 \; M_1^2 \;/\; R_2^2$$

$$M_1 = R_2 \; \sqrt{ M_2 } \;/\; R_1$$

Symbol English Metric
$$R_1$$ = the rate of effusion of the first gas (volume or number of moles per unit time) $$mol\;/\;sec$$ $$mol\;/\;s$$
$$R_2$$ = the rate of effusion for the second gas $$mol\;/\;sec$$ $$mol\;/\;s$$
$$M_2$$ = molar mass of gas two $$lbm\;/\;mol$$ $$kg\;/\;mol$$
$$M_1$$ = molar mass of gas one $$lbm\;/\;mol$$ $$kg\;/\;mol$$