A gas X,Y, at `35^@C` has RMS speed `12ms^(-1)`. On heating the gas twice to the original absolute temperature, the dimer totally dissociated to give monomer. What is the RMS speed of `XY_2` molecules at the given elevated temperature?

Explanation

RMS velocity of (C) is given as
`C= sqrt((3RT)/(M))`
Given temperature `T_1 = 308K `
Elevated temperature `T_2 = 616K`
` C_1 = sqrt((3RT_1)/(M_1)) and C_2 = sqrt((3RT_2)/(M_2))`
The ratio of RMS velocities
` (C_2)/(C_1) = sqrt((T_2)/(T_1) (M_1)/(M_2))= sqrt((161 )/( 308 ) xx ( 2M_2)/(M_2))=2`
RMS speed of `XY_2` molecules = `C_2`
`2 xx C_1 = 2 xx 12 = 24ms^(-1)`