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    				| 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 | |