When a flask of fixed volume is filled with `(x)/(2)` mol of an ideal gas A at a constant temperature, the pressure of the gas becomes 2 atom. Adint 2 y mol of another ideal gas B to the flask at the same temperature causes the pressure of the system to increase to 4.0 atm.

Explanation

For the gas, `A:P=2 atm and n=(x)/(2)` mol.
Hence, `PV=nRT or, 2xxV=(nRT)/(2)` . . [1]
For gas mixture: `P=4atm&n=(x)/(2)+2y=(1)/(2)(x+4y)mol`
`therefore`In case of the gas mixture, `4xxV=(1)/(2)(x+4y)RT` . . [2]
Deviding `[2]-:[1]` we get, `(4)/(2)=(x+4y)/(2)or,x=4y`