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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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$2P_0$

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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 $P \propto m v_{rms}^2$.Let the initial mass be $m_1 = m$ and initial speed be $v_1 = v$.The final mass is $m_2 = \frac{m}{2}$ and the final speed is $v_2 = 2v$.Taking the ratio of the final pressure $P_2$ to the initial pressure $P_1 = P_0$:$\frac{P_2}{P_1} = \frac{m_2}{m_1} 	imes \left( \frac{v_2}{v_1} ight)^2 = \left( \frac{m/2}{m} ight) 	imes \left( \frac{2v}{v} ight)^2$.$\frac{P_2}{P_0} = \frac{1}{2} 	imes (2)^2 = \frac{1}{2} 	imes 4 = 2$.Therefore,the resultant pressure is $P_2 = 2P_0$.

In a vessel,the gas is at pressure $P_0$. If the mass of all the molecules is halved and their speed is doubled,then the resultant pressure will be

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