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A gas X,Y, at 35^@C has RMS speed 12ms^(...

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?

Text Solution

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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)))`
Substitution the values, `sqrt((616)/(308) xx (2M_(2))/(M_(2))) = 2`
RMS speed of `XY_(2)` molecules ` = C_(2), 2 xx C_(1) = 2 xx 12 = 24ms^(-1)`
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