1 / 11

12.2 DALTON’S LAW OF PARTIAL PRESSURES

12.2 DALTON’S LAW OF PARTIAL PRESSURES. CONTEXT. We have looked at how P, T, V are connected We know the ideal gas law: PV=nRT We haven’t really discussed mixtures of gases If there is a mixture, how much of the total pressure is due to each gas?. In a mixture of non-reacting gases

isabella-paul
Télécharger la présentation

12.2 DALTON’S LAW OF PARTIAL PRESSURES

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 12.2 DALTON’S LAW OF PARTIAL PRESSURES

  2. CONTEXT • We have looked at how P, T, V are connected • We know the ideal gas law: PV=nRT • We haven’t really discussed mixtures of gases • If there is a mixture, how much of the total pressure is due to each gas?

  3. In a mixture of non-reacting gases • Each gas particle behaves independently • Particles spread themselves out to occupy the space available

  4. DALTON’S LAW OF PARTIAL PRESSURES The total pressure of a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases (the sum of the pressures that each gas would exert if it were alone). Ptot = P1 + P2 + P3 + …

  5. Partial Pressures of Atmospheric Gases • Air consists of a mixture of gases: N2 = 78.08% O2 = 20.95% Ar = 0.93 % CO2 = 0.03% If Standard atmospheric pressure is 101.325 kPa, determine the partial pressure of each gas. P N2 = 79.11 kPa P O2 = 21.23 kPa P Ar = 0.94 kPa P CO2 = 0.03 kPa PT = PN2 + PO2 + Par + PC02 PT≅ 101.325 kPa

  6. Gas Collection over Water • Gases can be collected by the downward displacement of water • The gas collected is a mixture of the desired gas and water. • This partial pressure of water vapour will be a component of the total pressure • To find the partial pressure of the desired gas, the vapour pressure of water must be subtracted. • The partial pressure of water vapour depends on the temperature of the water

  7. Gas Collection over Water Hydrogen gas from the reaction of Zinc with hydrochloric acid was collected by the displacement of water. Atmospheric conditions were 100.5 kPa and 20˚C. Determine the partial pressure of hydrogen gas. Ptot = Patm = PH2 + PH2O PH2 = Patm - PH2O = 100.5 kPa - 2.3 kPa = 98.2 kPa

  8. EXAMPLE • Oxygen gas is collected over water at 297 K and an atmospheric pressure of 1.0 atm. The volume of gas is 128 mL. Calculate the mass of O2 obtained. Hint: Water vapour at 297 K is 3 kPa. T = 297 K PT = 1.0 atm = 101.325 kPa PH2O = 3 kPa V = 128 mL = 0.128 L m = ? 1. PT = PO2 + PH2O PO2 = 101.325 kPa – 3 kPa PO2 = 98.325 kPa 2. PV =nRT n = PV / RT n = (98.325)(0.128)/(8.314)(297) n = 5.10 x 10-3 mol 3. m = n x M m = (5.10 x 10-3)(32.00) m = 0.16 g

  9. Homework • P. 460 # 22 -25 • Worksheet

More Related