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      

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