Pressure of 1 ~g ideal gas X at 300 ~K is 2 ~atm . When 2 ~g of another gas Y is introduced in the same vessel at same temperature, the pressure become l atm. The correct relationship between molar masses of X and Y is

Step-by-step solution

Given, Initially, array ll p=2 ~atm , & T=300 ~K V=v ~L , & m_x=1 ~g array Finally, array rlrl p & =2+1 & T & =300 ~K & =3 ~atm & m_y & =2 ~g array V=v ~L Let molar mass of gas X and Y be M_x and M_y respectively. Applying ideal gas equation to initial and final conditions. Initially, 2 V= 1 M_x R T ...(i) Finally, 3 V= ( 1 M_x + 2 M_y ) R T ....(ii) Dividing (i) by (ii), we get M_y=4 M_x .