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The average speed of gas molecules in a gas present in equilibrium mixture is:

Option: 1

\propto T


Option: 2

\propto T^\frac{1}{2}


Option: 3

\propto T^2


Option: 4

Nothing can be said.


Answers (1)

best_answer

We know, Average \: speed \: of\: the\: gas\: molecules(u_{av})=\sqrt{\frac{2RT}{M}}

Thus it is clear that u_{av} \propto T^\frac{1}{2}

Therefore, Option(2) is correct

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