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(C) Given: $m_A = \frac{1}{2}m_B$ and $u_{rms,A} = 2 	imes u_{rms,B}$.We know that pressure $P = \frac{1}{3} \frac{N}{V} m u_{rms}^2$,where $m$ is th

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$2$

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(C) Given: $m_A = \frac{1}{2}m_B$ and $u_{rms,A} = 2 	imes u_{rms,B}$.We know that pressure $P = \frac{1}{3} \frac{N}{V} m u_{rms}^2$,where $m$ is the mass of one molecule and $N$ is the number of molecules.Thus,$P_A = \frac{1}{3} \frac{N_A}{V_A} m_A u_{rms,A}^2$ and $P_B = \frac{1}{3} \frac{N_B}{V_B} m_B u_{rms,B}^2$.Since $V_A = V_B$ and $N_A = N_B$,the ratio is $\frac{P_A}{P_B} = \frac{m_A u_{rms,A}^2}{m_B u_{rms,B}^2}$.Substituting the given values: $\frac{P_A}{P_B} = \frac{(\frac{1}{2}m_B) (2 u_{rms,B})^2}{m_B u_{rms,B}^2} = \frac{\frac{1}{2} m_B 	imes 4 u_{rms,B}^2}{m_B u_{rms,B}^2} = 2$.Therefore,the ratio of the pressures is $2$.

The mass of molecule $A$ is half the mass of molecule $B$. The $rms$ speed of $A$ is twice the $rms$ speed of $B$. If the number of molecules in samples of $A$ and $B$ are equal,what is the ratio of the pressures of samples $A$ and $B$ in separate containers of equal volume?

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