question_answer

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 [NEET (Re) 2015]

A)  \[\frac{2}{3}\]             

B)       \[\frac{3}{4}\]

C)  2                    

D)       \[\frac{1}{2}\]

Correct Answer: B

Solution :

\[{{\rho }_{A}}=1.5{{\rho }_{B}}\]       \[{{\rho }_{B}}\]

\[{{\rho }_{A}}=2{{\rho }_{B}}\]          \[{{p}_{B}}\]

According to ideal gas equation, we have

Pressure, \[p=\frac{\rho RT}{M},\] where M is molecular weight of ideal gas.

Such that, \[\frac{p}{\rho }=\frac{RT}{M}\Rightarrow M=\frac{\rho RT}{p}\]

where, R and T are constant.

So,       \[M\propto \frac{\rho }{p}\]

\[\Rightarrow \]   \[\frac{{{M}_{A}}}{{{M}_{B}}}=\frac{{{\rho }_{A}}}{{{\rho }_{B}}}\times \frac{{{p}_{B}}}{{{p}_{A}}}=1.5\times \frac{1}{2}=0.75=\frac{3}{4}\]