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Kinetic Theory of Gases

A container c...

Question

A container contains $N_{2}$ gas at $T K$ . The number of moles of gas is $n_{0}$ . Consider it behaves like ideal gas and its $r m s$ speed is $v_{0}$ . Suppose the temperature of gas is tripled and $N_{2}$ molecules dissociate into atom then what will be the $r m s$ speed of atom?

v_{0} \sqrt{6}

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\sqrt{6 v_{0}}

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v_{0} \sqrt{3}

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\sqrt{3 v_{0}}

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Solution

The correct option is D $v_{0} \sqrt{6}$
Option A is correct.
$v_{0}$ = $\sqrt{3 R T / M}$
$v_{1}$ = $\sqrt{3 R T_{1} / M_{1}}$
But $T_{1} = 3 T$
and $M_{1} = M / 2$
By comparing both equation we get $v_{1} = v_{0} \sqrt{6}$

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