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(B) The pressure exerted by an ideal gas is given by the formula $P = \frac{1}{3} \rho v_{rms}^2$,where $\rho = \frac{M}{V}$ is the density and $v_{rm

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$2{P_0}$

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(B) The pressure exerted by an ideal gas is given by the formula $P = \frac{1}{3} \rho v_{rms}^2$,where $\rho = \frac{M}{V}$ is the density and $v_{rms}$ is the root-mean-square speed.Since $\rho = \frac{N \cdot m}{V}$ (where $N$ is the number of molecules and $m$ is the mass of one molecule),we have $P = \frac{1}{3} \frac{N \cdot m}{V} v_{rms}^2$.This implies $P \propto m \cdot v_{rms}^2$.Let the initial mass be $m_1 = m$ and initial speed be $v_1 = v$. The initial pressure is $P_0 \propto m \cdot v^2$.After the changes,the new mass is $m_2 = \frac{m}{2}$ and the new speed is $v_2 = 2v$.The new pressure $P$ is proportional to $m_2 \cdot v_2^2 = (\frac{m}{2}) \cdot (2v)^2 = (\frac{m}{2}) \cdot (4v^2) = 2mv^2$.Comparing the new pressure $P$ with the initial pressure $P_0$,we get $P = 2P_0$.

$A$ gas at a pressure ${P_0}$ is contained in a vessel. If the masses of all the molecules are halved and their speeds are doubled,the resulting pressure $P$ will be equal to:

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