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(A) The pressure of an ideal gas is given by the kinetic theory formula:$p = \frac{1}{3} \frac{M}{V} v_{rms}^2 = \frac{1}{3} \frac{N m}{V} v^2$where $

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$2: 1$

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(A) The pressure of an ideal gas is given by the kinetic theory formula:$p = \frac{1}{3} \frac{M}{V} v_{rms}^2 = \frac{1}{3} \frac{N m}{V} v^2$where $m$ is the mass of a molecule, $N$ is the number of molecules, $V$ is the volume, and $v$ is the root-mean-square speed.Let the initial pressure be $p = \frac{1}{3} \frac{N m}{V} v^2$.When the mass of each molecule is doubled $(m' = 2m)$ and the speed is halved $(v' = v/2)$, the new pressure $p'$ is:$p' = \frac{1}{3} \frac{N (2m)}{V} \left(\frac{v}{2}\right)^2$$p' = \frac{1}{3} \frac{N (2m)}{V} \left(\frac{v^2}{4}\right) = \frac{1}{2} \left( \frac{1}{3} \frac{N m}{V} v^2 \right) = \frac{1}{2} p$Therefore, the ratio of initial pressure to final pressure is:$\frac{p}{p'} = \frac{p}{p/2} = \frac{2}{1}$Thus, the ratio is $2: 1$.

In an ideal gas, if the masses of all molecules are doubled and their speeds are halved, then the ratio of initial and final pressures of the gas is

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