At room temperature `(300K)`, the rms speed of the molecules of a certain diatomic gas is found to be `1930 m//s`. Can you gusess name of the gas? Find the temperature at which the rms speed is double of the speed in part one `(R = 25//3 J// mol -K)`

Answer

`H_(2), 1200K`

Explanation

`V_(rms) = sqrt((3RT)/(M)) rArr 1930`
`M = (25 xx3)/(193 xx193) kg//mol = (25 xx 3)/(193xx193) xx 1000g//mol`
`= 2.01348 g//mol ~~ 2g//mol`
Hence, gas is `H_(2)`.
Let `T_(1)` be temperature at which `V_(rms)` is doubled.
`2V_(rms) = sqrt((3RT_(1))/(M))`
from, equation (i) and (ii), `sqrt((T_(1))/(T)) = 2`
`rArr T_(1) = 4T =4 (300)K, T_(1) = 1200K`
`V_(rms) =sqrt((3RT)/(M)) rArr 1930 = sqrt((3((25)/(3))300)/(M))`
`M = (25 xx3)/(193 xx 193) kg//mol`
`=(25 xx3)/(193xx193) xx 100 g//mol, =2.01348 g//mol`
`~~ 2g//mol, 2V_(rms) = sqrt((3RT_(1))/(M)) sqrt((T_(1))/(T)) =2`
`rArr T_(1) = 4T = 4 (300)K T_(1) = 1200K`