The mass of molecule A is half the mass of mo | vedclass

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(C) The pressure of an ideal gas is given by the kinetic theory formula: $P = \frac{1}{3} \frac{N}{V} m v_{rms}^2$. Given: $m_A = \frac{1}{2} m_B$,$v_

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$2$

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(C) The pressure of an ideal gas is given by the kinetic theory formula: $P = \frac{1}{3} \frac{N}{V} m v_{rms}^2$. Given: $m_A = \frac{1}{2} m_B$,$v_{rms,A} = 2 v_{rms,B}$,$N_A = N_B$,and $V_A = V_B$. Taking the ratio of pressures: $\frac{P_A}{P_B} = \frac{\frac{1}{3} \frac{N_A}{V_A} m_A v_{rms,A}^2}{\frac{1}{3} \frac{N_B}{V_B} m_B v_{rms,B}^2}$. Since $N_A = N_B$ and $V_A = V_B$,the expression simplifies to: $\frac{P_A}{P_B} = \frac{m_A}{m_B} 	imes \left( \frac{v_{rms,A}}{v_{rms,B}} ight)^2$. Substituting the given values: $\frac{P_A}{P_B} = \frac{1}{2} 	imes (2)^2 = \frac{1}{2} 	imes 4 = 2$.

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