Question

When 2 grams of a gas $A$ are introduced into an evacuated flask kept at ${25}^{o}C$ the pressure is found to be one atmosphere, if 3 grams of gas ${}^{\prime }{B}^{\prime }$ are then added to the same flask the total pressure becomes 1.5 $atm$. The ratio of molecular weights of the gases ($M_A:M_B$ ) is [Assume ideal gas behavior] :

• A. 2 : 3
• B. 3 : 2
• C. 1 : 3
• D. 4 : 3

Answer 1

The correct option is C 1 : 3

$PV=nRT$

For gas $A$, $P_{A}V=n_{A}RT⟶\left(1\right)$

For gas $B,$ $P_{B}V=n_{B}RT⟶\left(2\right)$

Number of moles of gas $A,$

$n_A=\frac{2}{M_A},$ ($M_A=$ molar mass of gas $A$).

Number of moles of gas $B,$

$n_A=\frac{3}{M_B},$ ($M_B=$ molar mass of gas $B$).

Pressure of gas $A,$ $P_A=1$ $atm$.

Total pressure, $P_T=P_A+P_B=1.5$ $atm$.

So, pressure of gas $B,P_B=P_T-P_A=1.5-1=0.5$ $atm$.

Hence from equation $\left(1\right)$ & $\left(2\right),$ $V,R$ & $T$ are same for both the gasses.

$\frac{P_A}{P_B}=\frac{n_A}{n_B}=\frac{2\times M_B}{M_A\times 3}\therefore \frac{M_B}{M_A}=\frac{3P_A}{2P_B}=\frac{3\times 1}{2\times 0.5}=\frac{3}{1}$

$\therefore$ Ratio of molecular weights of the gasses $=M_A:M_B=1:3$