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Q.
The root mean square velocity of hydrogen molecules at $300\, K$ is $1930$ metre / sec. Then the r.m.s velocity of oxygen molecules at $1200 \,K$ will be
Kinetic Theory
Solution:
Root-mean square-velocity is given by
$v _{ rms }=\sqrt{\frac{3 R T}{M}}$
i.e., $v _{ rms } \propto \sqrt{\left(\frac{T}{M}\right)}$
$\therefore \frac{\left( v _{ rms }\right) O _2}{\left( v _{ rms }\right) H _2}=\sqrt{\left[\frac{ T _{ O _2}}{ T _{ H _2}} \times \frac{ M _{ H _2}}{ M _{ O _2}}\right]}=\frac{1}{2}$
$\therefore\left( v _{ rms }\right) O _2=\left( v _{ rms }\right) H _2 \times \frac{1}{2}=\frac{1930}{2}=965\, m / s$