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?

A gas X,Y, at 35^@C has RMS speed 12ms^(-1). On heating the gas twice

1.	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?
Answer» SOLUTION :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)) andC_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)`