1.

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.

Answer» Given, `n=3xx10^(8)cm^(-1),lambda=10^(-2)` cm
Mean free path is given by
`lambda=1/(sqrt2pind_(2))`
`rArr" "d^(2)=1/(sqrt2pinlambda`
`rArr" " d^(2)=1/(sqrt2xx3.14xx3xx10^(8)xx10^(-2))`
`=10^(-6)/(1.414xx3.14xx3)`
`rArr" "d=sqrt(7.5xx10^(-6))=sqrt7.5xx10^(-4)`
`rArr" " d=2.7xx10^(-4)`cm
Hence, the diameter of the molecule of the gas is `2.7xx10^(-4)`cm


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