If masses of all molecules of a gas are halved and the speed doubled. Then the ratio of initial and final pressure is :

Options

A. `2 : 1`
B. `1 : 2`
C. `1 : 4`
D. `4 : 1`

Answer

B

Explanation

The percentage exerted by a gas is given by `P=1/3rhoC^(2)`
But density `(rho)prop` mass (m)
`thereforeP_(2)/P_(1)=(1/3rho_(2)C_(2)^(2))/(1/3rho_(1)C_(1)^(2))=(2rho_(1)(C_(1)/C_(2))^(2))/(rho_(1)C_(1)^(2))`
`" "=2xx(1/2)^(2)=2/4=1/2`