Question

When $2g$ of a gas A is introduced into an evacuated flask kept at ${25}^{\circ }\text{C}$, the pressure is found to be $1atm$. If $3g$ of another gas B is then added to the same flask at the same temperature, the total pressure becomes $1.5atm$. Assuming ideal gas behaviour, the value of $\frac{M_B}{M_A}$to the nearest integer is :

Answer 1

From ideal gas equation,
$PV=nRT$
$PV=\left(\frac{m}{M}\right)RT\therefore M=\frac{mRT}{PV}$
Where m and M are given mass and the molar mass respectively.

After adding B the pressure in the flask increases by 0.5 atm.
On applying ideal gas equation for the gases A and B taking volume and temperature to be same for both,
$M_A=\frac{2\times RT}{1\times V};M_B=\frac{3\times RT}{0.5\times V}$

$\therefore \frac{M_A}{M_B}=\frac{2RT}{V}\times \frac{0.5V}{3RT}=\frac{2\times 0.5}{3}=\frac{1}{3}$
Therefore, the ratio $M_A:M_B=1:3$
$\frac{M_B}{M_A}=3$