The molecules of a given mass of gas have root mean square speeds of `100 ms^(-1)` at `27^(@)C` and `1.00` atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at `127^(@)C` and `2.0` atmospheric pressure?

Options

A. `150/(sqrt(3)) m//s`
B. `125/(sqrt(3)) m//s`
C. `200/(sqrt(3)) m//s`
D. `100 (sqrt(3)) m//s`

Answer

C

Explanation

We know that for a given mass of gas
`v_(rms) = sqrt((3RT)/(M))`
where `R` is gas constant , `T` is temperature in kelvin and `M` is molar mass of the gas.
Clearly for a given gas `v_(rms) prop sqrt(T)`, as `R, M` are constant
Hence `((v_(rms))_1)/((v_(rms))_2) = sqrt((T_1)/(T_2))`
given, `(v_(rms))_(1) = 100m//s`
`T_(1) = 27^(@)C = 27+273 = 300K`
`T_(2) = 127^(@)C = 127+273 = 400K`
`:.` From eq, (i), `(100)/((v_(rms))_(2))= sqrt((300)/(400)) = (sqrt(3))/(2)`
`implies (v_(rms))_(2) = (2 xx 100)/(sqrt(3)) = (200)/(sqrt(3)) m//s`.