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If the rms velocity of the molecules of ...

If the rms velocity of the molecules of a gas in a container be doubled then the pressure of the gas will.

A

becomes 4 times of the previous value

B

becomes 2 times of its previous value

C

remains same

D

becomes `1/4` of its previous value

Text Solution

Verified by Experts

`V_(rms) = sqrt((3PV)/(m)) rArr V_(rms)^(2) = (3PV)/(m) rArr P = (V_(rms)^(2)m)/(3V)`
Here `V_(rms)` is doubled, then pressure of a gas will be
`P. = ((2V_(rms))^(2)m)/(3V) = 4((V_(rms)^(2)m)/(3V))` , P. = 4P
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