2 g of a gas A is introduced into an evacuated flask kept at ${25}^{\circ }$C. The pressure is found to be 1 atm. If 3g of another gas B is added to the same flask the total pressure becomes 1.5atm, assuming ideal gas behavior. The ratio of molecular weight of gas A to gas B $\left(M_A:M_B\right)$ is

According to Ideal gas law.

PV = nRT

$$\begin{gathered} n = \frac{m}{M} \\ \Rightarrow PV\hspace{0.17em}= \frac{m}{M}\times RT \end{gathered}$$

$$\begin{gathered} \Rightarrow PM = \frac{mRT}{V} \\ or M = \frac{mRT}{PV} ....\left(i\right) \\ {m}_1\left(A\right)=2g \\ {P}_1=1atm \\ and \\ m_2\left(B\right)=3g \\ {P}_2=1.5atm \\ M_A = \frac{m_{A}RT}{P_{A}V} \\ and \\ {M}_B = \frac{m_{B}RT}{P_{B}V} \\ So, \\ \frac{M_A}{M_B} =\frac{\frac{m_{A}RT}{P_{A}V}}{\frac{m_{B}RT}{P_{B}V}} \\ \Rightarrow \frac{M_A}{M_B}=\frac{m_A\times P_B}{P_A\times m_B} \left[T=cons\tan t\right] \\ =\frac{2\times 0.5}{1\times 3}=\frac{1}{3} \left[increase in pressure=0.5atm\right] \end{gathered}$$

M_A : M_B = 1:3