Pressure of 1g of an ideal gas A at `27^@C` is found to be 2 bar. When 2g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Find the relationship between their molecular masses.

Explanation

Ideal gas equation , `pV=nRT or pV=W/M RT`
For two gases `p_1V_1=W_1/M_1RT_1(or)V_1W_1/(p_1W_1)RT_1`
`p_2V_2=W_2/M_2RT_2(or)V_2W_2/(p_2W_2)RT_2`
Since `V_1=V_2` also T and R are constants
`W_1/(p_1M_1)=W_2/(p_2M_2)`
Pressure of gas B= 1 bar
`1/(2M_A)=2/(1M_B)`
`thereforeM_B=4M_A`