The quantity of gas in a closed vessel is halved and the velocities of its molecules are doubled. The final pressure of the gas will be

Solution :
$${v_{rms}} = \sqrt {\frac{{3P}}{\rho }} = \sqrt {\frac{{3PV}}{m}} \Rightarrow P = \frac{{m{{\left( {{v_{rms}}} \right)}^2}}}{{3V}}$$
If $${P'}$$ be the final pressure then
$$\frac{{P'}}{P} = \left[ {\frac{{\frac{{\frac{m}{2}{{\left( {2{v_{rms}}} \right)}^2}}}{{3V}}}}{{\frac{{m{{\left( {{v_{rms}}} \right)}^2}}}{{3V}}}}} \right] = 2 \Rightarrow P' = 2P.$$