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Step-by-Step Solution
Step 1: Identify the Relevant Gas Law
Recall the ideal gas law: $PV = nRT$. Here, $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the universal gas constant, and $T$ is the temperature.
Step 2: Express Pressure in Terms of Density and Molecular Weight
For a given gas, $n = \frac{\text{mass}}{M}$, where $M$ is the molecular weight. If $\rho$ is the density of the gas, then mass = $\rho \cdot V$. Substituting these into $PV = nRT$, we get:
$P = \frac{nRT}{V} = \frac{\bigl(\frac{\rho V}{M}\bigr)RT}{V} = \frac{\rho R T}{M}.$
Hence, for gas A and gas B, respectively:
$P_A = \frac{\rho_A R T}{M_A}, \quad P_B = \frac{\rho_B R T}{M_B}.$
Step 3: Form the Ratio of Pressures
According to the problem statement, the pressure of gas A ($P_A$) is twice that of gas B ($P_B$), and the density of A ($\rho_A$) is 1.5 times (or $\frac{3}{2}$ times) that of gas B ($\rho_B$). So we have:
$P_A = 2 P_B, \quad \rho_A = \frac{3}{2} \rho_B.$
Taking the ratio $\frac{P_A}{P_B}$ using the expressions from Step 2:
$\frac{P_A}{P_B} = \frac{\frac{\rho_A R T}{M_A}}{\frac{\rho_B R T}{M_B}}
= \frac{\rho_A}{\rho_B} \times \frac{M_B}{M_A} = 2.$
Step 4: Substitute the Ratio of Densities
We know $\frac{\rho_A}{\rho_B} = \frac{3}{2}$. So substitute this into the pressure ratio equation:
$\left(\frac{3}{2}\right) \times \frac{M_B}{M_A} = 2.
$
Step 5: Solve for the Molecular Weight Ratio
Rearrange to find $\frac{M_A}{M_B}$ (the ratio of the molecular weight of A to that of B):
$\frac{3}{2} \times \frac{M_B}{M_A} = 2
\quad \Longrightarrow \quad
\frac{M_B}{M_A} = \frac{2}{\frac{3}{2}} = \frac{4}{3}
\quad \Longrightarrow \quad
\frac{M_A}{M_B} = \frac{3}{4}.
$
Thus, the ratio of the molecular weights of gas A and gas B is $\frac{3}{4}.$
