MCQ
Two vessels separately contain two ideal gases $A$ and $B$ at the same temperature, the pressure of $A$ being twice that of $B.$ Under such conditions, the density of $A$ is found to be $1.5$ times the density of $B.$ The ratio of molecular weight of $A$ and $B$ is
  • A
    $0.5$
  • B
    $0.67$
  • $0.75$
  • D
    $2$

Answer

Correct option: C.
$0.75$
c
According to an ideal gas equation, the molecular weight of an ideal gas is

$M=\frac{\rho R T}{P} \quad\left(\text { as } P=\frac{\rho R T}{M}\right)$

where $P, T$ and $\rho$ are the pressure, temperature and density of the gas respectively and $R$ is the universal gas constant.

$\therefore$ The molecular weight of $A$ is

$M_{A}=\frac{\rho_{A} R T_{A}}{P_{A}}$ and that of $B$ is $M_{B}=\frac{\rho_{B} R T_{B}}{P_{B}}$

Hence, their corresponding ratio is

$\frac{M_{A}}{M_{B}}=\left(\frac{\rho_{A}}{\rho_{B}}\right)\left(\frac{T_{A}}{T_{B}}\right)\left(\frac{P_{B}}{P_{A}}\right)$

Here, $\frac{\rho_{A}}{\rho_{B}}=1.5=\frac{3}{2}, \frac{T_{A}}{T_{B}}=1$ and $\frac{P_{A}}{P_{B}}=2$

$\therefore \quad \frac{M_{A}}{M_{B}}=\left(\frac{3}{2}\right)(1)\left(\frac{1}{2}\right)=\frac{3}{4}$

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