1 (304)

Prévia do material em texto

10 Gases Solutions to Exercises Additional Exercises 10.101 P₁V₁ = P₂V₂; V₂ = P₁ V₂ = 3.0atm 730 torr 1.0mm³ 760 1 atm torr = 10.102 PV = nRT, n = PV/RT. Since RT is constant, n is proportional to PV. Total available n = (15.0 L 1.00 10² atm) - (15.0 L 1.00 atm) = 1485 = 1.49 10³ L-atm Each balloon holds 2.00 L 1.00 atm = 2.00 L-atm 1485 L - atm available = 742.5 = 742 balloons 2.00 L- - atm/balloon (Only 742 balloons can be filled completely, with a bit of He left over.) 10.103 P = nRT V ; n = 1.4 10⁻⁵ mol, V = 0.600 L, T = 23°C = 296 K P = 1.4 10⁻⁵ mol 0.08206 mol L K atm 0.600 296K L = 5.7 X 10⁻⁴ atm = 0.43 mm Hg 10.104 (a) Change mass CO₂ to mol P = 1.00 atm, T = 27°C = 300 K. 10⁶ tons CO₂ 2000 ton lb 453.6 lb X 44.01 1 mol = 1.237 X 10¹¹ = 1 10¹¹ mol V = nRT P = 1.237 10" mol 1.00 atm 300 K X mol K atm = 3.045 X 10¹² = 3 10¹² L (b) 1.237 10¹¹ mol CO₂ 44.01 mol 1 1.2 cm³ 1000 1L = 4.536 X = 5x L (c) n = 1.237 10¹¹ mol, P = 90 atm, T = 36°C = 309 K V = nRT P = 1.237 10" mol 90 309 atm K 0.08206L mol K atm = 3.485 10¹⁰ = 3 10¹⁰ L 10.105 (a) n = PV RT = 3.00 atm 0.08206 mol K - atm 300 K = 13.4 mol C₃H₈(g) (b) 0.590 g (1) 110 10³ mL 1 mol = 1.47 X 10³ mol C₃H₈ (1) 1 mL 44.094 g (c) Using C₃H₈ in a 110 L container as an example, the ratio of moles liquid to moles gas that can be stored in a certain volume is 1.47 X 10³ mol liquid = 110. 13.4 mol gas A container with a fixed volume holds many more moles (molecules) of because in the liquid phase the molecules are touching. In the gas phase, the molecules are far apart (statement 2, section 10.7), and many fewer molecules will fit in the container. 298