The temperature of an ideal gas is increased from `140 K` to `560 K`. If a `140 K` the root mean square velocity of the gas molecule is `V`, at `560 K` it becomes

Options

A. `5 V`
B. `2 V`
C. `V//2`
D. `V//4`

Answer

B

Explanation

The `V_(rms)` at `140 K` is `V`
`:. V=sqrt((3Rxx140)/(M))`
At `540K=V' = sqrt((3Rxx560)/(M))`
`= sqrt((3R xx 140 xx 4)/(M)) = 2sqrt((3R xx 140)/(M)) = 2V`