Given below are two statements: Statement I : The average momentum...

Momentum is a vector quantity and average momentum of molecules of a sample is zero due to random motion.

${P}_{avg} = 0$

RMS speed of a molecule is given by

${v}_{rms} = \sqrt{\dfrac{3RT}{M}}$

Now the temperature is doubled ${T}_{new} = 2T$ and due to dissociation ${M}_{new} = \dfrac{M}{2}$.

$⇒ \dfrac{{v}_{new}}{v} = \dfrac{\sqrt{\dfrac{2T}{\dfrac{M}{2}}}}{\sqrt{\dfrac{T}{M}}}$

$⇒ {v}_{new} = 2v$