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For a molecule of an ideal gas n=3xx10^(...

For a molecule of an ideal gas `n=3xx10^(8)cm^(-3)` and mean free path is `10^(-2)` cm. Calculate the diameter of the lomecule.

Text Solution

Verified by Experts

Mean free path is given by
`l= (1)/(sqrt(2) pi nd^2)`
`d^2 = (1)/(sqrt(2) pi n l )`
`d^2 = (1)/(sqrt(2) x 3.14 x 3 xx 10^(8) xx 10^(-2))= (10^(-6))/(1.414 xx 3.14 xx3)`
`d= sqrt(7.5 xx 10^(-8) ) = sqrt(7.5) xx 10^(-4)`
`d= 2.7 xx 10^(-4)` cm
Hence, the diameter of the molecule of the gas is `2.7 xx 10^(-4)` cm.
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