PROBLEM: What is the total pressure in atmospheres of a gas mixture that contains 1.0 g H2 and 8.0 g of Ar in a 3.0-L container at 27°C? What are the partial pressures of the two gases? Step 1: Convert g into moles for each gas 1.0 g H2 x 1 mol / 2.02 g = 0.495 mol H2 8.0 g Ar x 1 mol / 39.95 g = 0.200 mol Ar Step 2: Use PV = nRT to solve for pressure of each gas PV = nRT – – > PH2 = nRT/V – – > PH2 = 0.495 mol(0.0821 atm•L•mol–1•L–1) 300.K 3.0 L PH2 = 4.06395 – – > 4.1 atm (sig. digits) PV = nRT – – > PAr = nRT/V – – > PAr = 0.200 mol(0.0821 atm•L•mol–1•L–1) 300.K PAr = 1.642 – – > 1.6 atm (sig. digits) Step 3: Add the two partial pressures together to get the total pressure 4.1 atm + 1.6 atm = 5.7 atm