Question

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. 1/2
B. 2/3
C. 3/4
D. 2

Answer 1

Temperature of both gases are the same so you need to find the relationship between pressure, density and molecular mass when temperature is given (by using the ideal gas equation $PV = nRT$), in order to solve this question.

Complete step by step answer:
Let's find the relationship between density, pressure and molecular weight at a certain temperature. Let's say we have a sample of gas of molecular mass $M - gram$ and the sample weighs $m - gram$. Now no of moles of the sample will be $n = \dfrac{m}{M}$
Also density is ratio of mass and volume .So,
$density(\rho ) = \dfrac{{mass}}{{volume}} = \dfrac{m}{V}$
We know that ideal gas equation is $PV = nRT$
Putting value of V and n in the ideal gas equation we have,
$P(\dfrac{m}{\rho }) = \dfrac{m}{M}RT$
$\Rightarrow PM = \rho RT$

At a given temperature we have,
$\dfrac{{PM}}{\rho } = RT = {\rm{constant}}$
$\Rightarrow \dfrac{{{P_1}{M_1}}}{{{\rho _1}}} = \dfrac{{{P_2}{M_2}}}{{{\rho _2}}}\\\ \Rightarrow (\dfrac{{{P_1}}}{{{P_2}}})(\dfrac{{{\rho _2}}}{{{\rho _1}}}) = \dfrac{{{M_2}}}{{{M_1}}}$
Putting values of Pressure and density we have
$\dfrac{{{M_A}}}{{{M_B}}} = \dfrac{{{P_B}{\rho _A}}}{{{P_A}{\rho _B}}}\\\ \Rightarrow \dfrac{{{M_A}}}{{{M_B}}} = \dfrac{1}{2} \times 1.5 \\\ \therefore \dfrac{{{M_A}}}{{{M_B}}} = \dfrac{3}{4}$

Hence Option C is correct.

Note: Here all you need to do is find the relationship between pressure ,density and molecular weight using ideal gas equation of ideal gas and then compare the values to get the answer also if you can you should remember this new derived formula of ideal gas in terms of pressure, density, molecular weight and temperature it can save you some precious time in exam hall.