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.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
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