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(B) The pressure of an ideal gas is given by the kinetic theory formula: $P = \frac{1}{3} \frac{N m}{V} v_{rms}^2$.From this expression,we can see tha

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$2 P_0$

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(B) The pressure of an ideal gas is given by the kinetic theory formula: $P = \frac{1}{3} \frac{N m}{V} v_{rms}^2$.From this expression,we can see that the pressure $P$ is directly proportional to the mass of the molecules $m$ and the square of their root-mean-square velocity $v_{rms}^2$,assuming the number of molecules $N$ and volume $V$ remain constant: $P \propto m v_{rms}^2$.Let the initial state be $P_1 = P_0$,mass $m_1 = m$,and velocity $v_1 = v$.In the final state,the mass is halved: $m_2 = \frac{m}{2}$,and the velocity is doubled: $v_2 = 2v$.The ratio of the final pressure $P_2$ to the initial pressure $P_1$ is given by:$\frac{P_2}{P_1} = \frac{m_2 v_2^2}{m_1 v_1^2} = \frac{(\frac{m}{2}) (2v)^2}{m v^2}$.Simplifying the expression:$\frac{P_2}{P_1} = \frac{(\frac{m}{2}) (4v^2)}{m v^2} = \frac{2 m v^2}{m v^2} = 2$.Therefore,the final pressure is $P_2 = 2 P_1 = 2 P_0$.

$A$ gas at pressure $P_0$ is contained in a vessel. If the masses of all the molecules are halved and their velocities are doubled,the resulting pressure would be equal to

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