When the temperature of a gas is raised from `27^(@)C` to `90^(@)C`, the percentage increase in the rms velocity of the molecules will be

Options

A. `10%`
B. `15%`
C. `20%`
D. `17.5%`

Answer

A

Explanation

`v_(rms) = sqrt((3RT)/(M))`
`implies (v_2)/(v_2) = sqrt((T_2)/(T_1)) = sqrt((273+90)/(273+30)) = 1.1`
`%` increase = `((v_2)/(v_1)-1)xx100=0.1xx100=10%`.