Suppose that a 1.0 liter vessel contains n2o gas at 1.0 atm pressure. suppose that a second vessel, 6.0 l in volume contains oxygen at 6.0 atm pressure. now suppose that the two vessels are connected by a pipe of negligible volume and the two gases mix and react to produce as much nitrogen dioxide as possible. assume that the temperature remains constant. what is the pressure in the apparatus at the end of the reaction?

P = 5.868 atm

Explanation:

assume T = 25°C = 298.15 K

∴ PN2O = 1 atm

∴ XN2O = nN2O/nt

∴ n = PV/RTideal gas

⇒ nN2O = ((1atm)(1L))/((0.082atm.L/K.mol)(298.15K)) = 0.041 mol N2O

∴ PO2 = 6 atm

∴ XO2 = nO2/nt

∴ nO2 = ((6atm)(6L))/((0.082atm.L/K.mol)(298.15K)) = 1.472 mol O2

⇒ nt = nN2O + nO2 = 0.041 + 1.472 = 1.513 mol

⇒ XN2O = 0.041/1.513 = 0.03

⇒ XO2 = 1.472/1.513 = 0.973

⇒ Pt = ((1atm)(0.03)) + ((6atm)(0.973))

⇒ Pt = 5.868 atm

pv=nrt

explanation: