The root mean square velocity of a gas molecule at 100 K and 0.5 atm pressure is `106.4 m s^(-1)`. If the temperature is rasied to 400 K and the pressure is raised to 2 atm, then root mean square velocity becomes

Options

A. `106.4 ms^(-1)`
B. `425.6 ms^(-1)`
C. `212.8 ms^(-1)`
D. `851.2ms^(-1)`

Answer

C

Explanation

According to Kinetic Gas equation
`u = sqrt((3RT)/(M)) " or " (u_(1))/(u_(2)) = sqrt((T_(1))/(T_(2)))`
Hence, `(106.4)/(u_(2)) sqrt((100)/(400)) " or " u_(2) = 212.8 ms^(-1)`