Question

5. One liter of gas A at 2 atm pressure and two liters of gas B at 3 atm are mixed in a 4 liter flask. What will be the final pressure if the gases initially and finally are at the same temperature? What will be the average molecular weight of the mixture if the weight of A + B is 24 g and the temperature is 60?

Answer 1

11 Volume of the gas i is = 1 L Pressure of the gas 'À is PA = 2 atm l Volume of the gas B is V₂ = 2L Pressure of the gas B is pp = 3 atm for Agar Before mixing, the ideal gas law is PV = nRT - A AT and for B gas is pp = n RT. 0 Let as Pressure suppose of a after mixing the gas is pl and for B partial gas is "final volume of Flask = 4 L=v gas equation After arising for A gas the ideal 18 px V = nRT - CB for B gas is PP x V = n RT ... (1)

from equation ② and i we get PAVA = Px v or, P = PANA from equation ④ and 1h we get- PBY = P x v oh, Pe= PBVB Here in the gas mixture and lp are partial pressure. Hence applying Dalton's partial pressure law we get-t it Total pressure = (Pat Peliste n Total pressure = | PAYA & PY) So, final pressure = (2x1) + (3X2) atm = = 2 atm final pressure = 2 atm. Therefore

Mu weight of (A + B) = 24 g Let us suppose average molecular weight = M g/mol, So, 24 g of (A+b) gas = 24 mae of gas Hence, no. of mal of gas mixture = (24) mal Temperature is 60c = (273+60) K=333K Final Pressure (P) = 2 alm. Volume of the gas = 4L We apply ideal gas equation - Pv=nRT or. 2x4 = (2 ) x 0.082*333 ai 8 = 655.344 ori m = 6551344 = 81.918 g/mol M Average molecular weight of the mixture = 81.918 g/mal.