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(D) The pressure $P$ of an ideal gas is given by the kinetic theory formula: $P = \frac{1}{3} \frac{N m}{V} v_{rms}^2$,where $N$ is the number of mole

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$2P$

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(D) The pressure $P$ of an ideal gas is given by the kinetic theory formula: $P = \frac{1}{3} \frac{N m}{V} v_{rms}^2$,where $N$ is the number of molecules,$m$ is the mass of each molecule,$V$ is the volume,and $v_{rms}$ is the root-mean-square speed.Given that the mass of each molecule is halved $(m' = m/2)$ and the speed is doubled $(v' = 2v)$:The new pressure $P'$ is given by: $P' = \frac{1}{3} \frac{N m'}{V} (v')^2$.Taking the ratio of the new pressure to the initial pressure:$\frac{P'}{P} = \frac{m'}{m} 	imes \frac{(v')^2}{v^2} = \frac{m/2}{m} 	imes \frac{(2v)^2}{v^2} = \frac{1}{2} 	imes 4 = 2$.Therefore,$P' = 2P$.

If the mass of each molecule of a gas kept at a constant pressure $P$ in a container is halved and their speed is doubled,what will be the resulting pressure of the gas?

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