A person evacuated a cylinder and filled in $4~\mathrm{g}$ of gas A at $25~\mathrm{^\circ C}$. Pressure was found to be $1~\mathrm{atm}$. Another person filled another $8~\mathrm{g}$ of gas B in the cylinder at the same temperature. The final pressure was found to be $1.5~\mathrm{atm}$. What is the ratio of molecular masses of A and B (assuming ideal gas behaviour)?
I tried to do it this way:
Taking the molecular mass of A as $x\ \mathrm{g}$ and that of B as $y\ \mathrm{g}$, I formed the respective ideal gas equations but the required terms ($x$ and $y$) just cancelled out.
$$N(a) = 4/x ;\quad N (b) = 8/y$$
So$$V (a) = \frac{4 \times 22.4}{x};\quad V (b) = \frac{8 \times 22.4}{y}$$
So, ideal gas equation for A:
$$1 X \frac{4 \times 22.4}{x} = \frac {4 \times R \times 25} x$$
Here $x$ cancels out. Similarly $y$ cancels out in its equation.
This is the problem I am facing.
Another thing I want to know is that the total pressure (i.e $1.5~\mathrm{atm}$), will that be the sum of the individual pressures of gases A and B?