1.

The r.m.s. speed of oxygen is v at a particular temperature. If the temperature is doubled and oxygen molecule dissociate into oxygen atoms, the r.m.s. speed becomes

Answer»

v
`sqrt(2)v`
`2v`
`4v`

Solution :As, `v_(rms)=sqrt((3RT)/(M))`
For OXYGEN molecules at temperature T,
`v_(rms)=v=sqrt((3RT)/(M))""...(i)`
Now, temperature is doubled and oxygen molecules dissociate into oxygen atom (molar mass BECOMES `M//2`) then rms speed will be
`v._(rms)=sqrt((3R(2T))/((M//2)))=2sqrt((3RT)/(M))=2v` [USING eqn(i)]


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