Question

Name Dalton's Law of Partial Pressures - Many gas samples are a mixture of gases. For example, air is a mixture of gases. The gases dissolved in blood make up a mixture as well. In a gas mixture, each gas exerts its partial pressure (the specific pressure contribution of the gas to the total pressure of the gas mixture). Dalton's law states that the total pressure of a gas mixture (Pita) is the sum of the partial pressures of the gases in the mixture. Protal = P + P + P + P + ..... ) The partial pressure of each gas can be calculated by multiplying the total pressure (P by the gas percentage (%). P, (%)(Total) (Note: the percentage must be expressed as the decimal equivalent) The table below lists the partial pressures for the gases in air on a typical day. GOO P Gas Nitrogen, N Oxygen, O Carbon Dioxide, CO Water Vapor, H0 Other trace gases Total air P Partial Pressure (mm Hg) IPN2 = 594 594 Pas 160 2 Co2 = 0.25 P2 = 5.7 Pote=0. 05 0 P Total = 760 Percentage (%) 78 1 0.033 10.75 .22 100 • Use the percentages in the table above to prove that oxygen in the atmosphere has a partial pressure of 160.0 mm Hg. Problem solving - Show your work and box your answer for each problem 1.00 atm = 760 mm Hg = 760 torr = 14.7 psi = 29.9 in. Hg = 101 325 Pa 1. What is the volume in liters of 4.0 moles of methane gas, CH., at 18 °C and 1.40 atm?

Answer 1

I deal Cras Law PV = nRT Where p - Pressure = 1.4o, catm V = Volume in Litou na mole of gaas = 4.0 mole (My R= Gres constant = 0.082 ltr. catm mole .k T= Temp in kelvin - 118 + 273115) k T= 29115k V= HRT P . 4 (more) x 0.082 atm ) x 291-15 (6) - 1.40 (cetm). Moleik V = 68.2122, Litore od volume of Chy gas . At moshenic pressure (Plath = I atm = 760 mmily Partial Puresswn of O₂ in cutmosphere is : Poz = (% & O2 in Admosphere ) x (P) atm - 21x760 = 159.6 mm Hg ñ 160 mmHg 100