At room temperature, the rms speed of the molecules of a certain diatomic gas is found to be `1930 m//s`. The gas is

Options

A. `H_(2)`
B. `F_(2)`
C. `O_(2)`
D. `CI_(2)`

Answer

A

Explanation

`v_(rms) = sqrt((3 RT)/(M_(0)))`
`M_(0) = (3 RT)/(v_(rms)^(2)) = (3 xx 8.3 xx 300)/((1930)^(2)) = 2 xx 10^(-3) kg = 2g`
Molecular mass `= 2 g` i.e. gas is `H_(2)`