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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 $P \propto m \cdot v_{rms}^2$,where $m$ is the mass of a molecule and $v_{rms}$ is the root-mean-square speed.Let the initial state be $P_1 = P_0$,$m_1 = m$,and $v_{rms,1} = v$. In the final state,the mass is halved $(m_2 = m/2)$ and the $rms$ speed is doubled $(v_{rms,2} = 2v)$.Taking the ratio of the pressures:$\frac{P_2}{P_1} = \frac{m_2}{m_1} 	imes \left( \frac{v_{rms,2}}{v_{rms,1}} ight)^2$Substituting the values:$\frac{P_2}{P_0} = \left( \frac{m/2}{m} ight) 	imes \left( \frac{2v}{v} ight)^2 = \frac{1}{2} 	imes (2)^2 = \frac{1}{2} 	imes 4 = 2$.Therefore,the new pressure is $P_2 = 2P_0$.

$A$ gas is filled in a container at pressure ${P_0}$. If the mass of all molecules is halved and their $rms$ speed is doubled,what will be the new pressure?

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