Question

9) If you mix 1.00 moles of N2, 0.20 moles of O2 and 0.05 moles of H2O at 20.00°C in a 25.00L vessel. What is the pressure? How much pressure does each gas individually exert? How do the pressures of each gas, or partial pressures, relate to the total pressure? (Remember that you are assuming the gases are ideal particles)

Answer 1

Total number of mole s= 1.00 mol N2 +0.20 mol O2 + 0.05 mol H2O Partial pressure of H2O = mole fraction * P total = 1.25 Mole =0.04 *1.20 atm Temperature = 20.00 C or 20+273 = 293 K = 0.048 atm Volume =25.00 L The ideal gas law is following: PV = nRT Here, R; (0.0821 atm.L mol-!.K) is the universal gas constant. Dalton's law: According to Dalton's law the total pressure of gases mixture (gases of mixture do not react chemically) is equal to the sum of all the partial pressure of the component gases of mixture at constant temperature and volume. Pressure is calculates as follows: PTotal = P + P + P +... P=nRT/V = 1.25 *0.0821 *293/25 Here, Pro is the total pressure of mixture of gases, P. is the partial pressure of gas 1, P, is the partial pressure of gas 2 and P, is the partial pressure of gas 3. =1.20 atm Mole fraction of N2=1.00/1.25= 0.8 Mole fraction of O2=0.20/1.25=0.16 Partial pressure : The pressure exerted by an individual gas in a mixture is known as its partial pressure. Mole fraction of H2O=0.05/1.25=0.04 Mole fraction: Partial pressure of N2= mole fraction * P total Mole fraction is a ratio of mole of solute or solvent to the total number of moles (mole of solute + moles of solvent) or number of mole of solution. = 0.8 *1.20 atm = 0.96 atm Partial pressure of O2= mole fraction * P total =0.16 *1.20 atm =0.192 atm Raoult's law: According to the Raoult's law in the mixture the partial pressure of a gas is equal to the product of its mole fraction and the total pressure of gases: PA = X. Protal Partial pressure of H20 = mole fraction * P total