In a vessel,an ideal gas is at a pressure P. | vedclass

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

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$2 P$

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(B) The pressure $P$ of an ideal gas is given by the kinetic theory formula: $P = \frac{1}{3} \frac{N m}{V} v_{rms}^2$ where $N$ is the number of molecules,$m$ is the mass of each molecule,$V$ is the volume,and $v_{rms}$ is the root-mean-square speed. From this relation,we can see that $P \propto m \cdot v_{rms}^2$. Let the initial mass be $m_1 = m$ and initial speed be $v_1 = v$. The initial pressure is $P_1 = P$. Given the new conditions: $m_2 = \frac{m}{2}$ $v_2 = 2v$ Now,calculating the new pressure $P_2$: $\frac{P_2}{P_1} = \frac{m_2}{m_1} 	imes \left( \frac{v_2}{v_1} ight)^2$ $\frac{P_2}{P} = \left( \frac{m/2}{m} ight) 	imes \left( \frac{2v}{v} ight)^2$ $\frac{P_2}{P} = \left( \frac{1}{2} ight) 	imes (2)^2$ $\frac{P_2}{P} = \frac{1}{2} 	imes 4 = 2$ Therefore,$P_2 = 2P$.

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

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