AP Chemistry WS 4.1 Key All gas molecules or atoms are always in
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AP Chemistry WS 4.1 Key 1. a) All gas molecules or atoms are always in constant, random motion. KE is proportional to Temp. Note: assumptions for the KMT also apply to real gases under all but exceptional conditions of low temperatures and high pressures. b) Ideal gas molecules have negligible volume in comparison to their container and gas ideal gas molecules don’t attract or repel each other. NOTE: these postulates are for Ideal Gases only, but again they apply to real gases under most conditions. π 2. a) π²π¬πππ = πΉπ» ππΉπ» π b) ππππ = √ 3. 4. 5. 6. πππππ πππππ = √ππ πππππ πππππ = √ππ (32π ) 4.67 π π molβK ) (274πΎ) = 3415.41 π½/πππ with Sigfigs. 3420 J/mol (π)(π.ππ π±/πππ π²)(ππππ²) .πππππ ππ/πππ = 581.80 π/π with SigFigs 582 m/s ππ = √π.π He effuses 2.8 times faster than O2 ππππ ππ π―π ππππ ππ πͺπΆπ π J π ππππ = √ ππππ ππ π―π ππππ ππ πΆπ π π π²π¬πππ = (8.31 π ππ.ππ H2 effuses 4.67 times faster than CO2 = √ π.ππ = 6.85π πππππ πππππ π ππππ ππ π―π ππππ ππ π΅π―π = √ππ π ππ.ππ = √π.πππ He diffuses 2.06 times faster than NH3 ; therefore, the reciprocal will be the diffusion rate of NH3 compared to He: 0.49 times 7. ( 8. 20π .49 ) = ππ πππ πππππ―π πππππππππππ π = √π.πππ π π.ππ π = √π.πππ = ππ π/πππ ππΉπ» π 9. a) A is O2 and B is He. O2 is heavier and would have slower speeds compared to He. ππππ = √ ππΉπ» π b) B is higher temperature. Hotter gas has greater KE, therefore greater speed than cooler gas. ππππ = √ 10. 1.22 ππ‘π × 1.22ππ‘π × 1.22ππ‘π × 760πππ»π 1ππ‘π 101.325πππ 1ππ‘π 760ππππ 1ππ‘π = 927.2 πππ»π (927π πππππ) = 123.62 πππ (124π πππππ) = 927.2 ππππ (927π πππππ) 11. 520πππ»π 749πππ»π − 670πππ»π = 79πππ»π 749πππ»π + 103πππ»π = 852πππ»π 12. a) The piston would move upward to double the volume because doubling the number of molecules exerts twice πΉ the pressure. More collisons ∴ more force and more pressure. π = π b) The piston would move upward to double the volume because the faster moving molecules exert twice the π pressure. π²π¬πππ = πΉπ» π c) Double pressure ∴ volume decreases by ½ . Molecules are more crowded, increased # of collisions, greater force, greater pressure. 13. P V n T 2.00 atm 1.00 L 1.500 mol 16.2 K 30.3 kPa 1.250 L 0.0152 mol 27oC 650 torr 11.2 L 0.333 mol 350 K 585 mL 0.250 mol 295 K 10.4 atm 14. a) ππ = ππ π π= b) 0.382ππππ2 × 15. π·π π½π = ππ π»π π2 = 16. π π = 32.0 π π2 1 πππ (94.6πππ)(10.0πΏ) (8.31)(298πΎ) (πππππππ)(πππππ³) ππ π»π ππ (ππππ²) (750π‘πππ)(450ππΏ)(288πΎ)(π1 ) (750π‘πππ)(350ππΏ)(π2 ) πββ³ given π βπ π πββ³ = π βπ π πββ³ π βπ 18. = = π. πππ πππ = ππ. π π πΆπ π·π π½π =π= 17. π = ππ π π = (πππππππ)(πππππ³) ππ π»π rearrange and solve for T2 = πππ. π π² This is really just Charles Law the information, rearrange solving for molar mass(M) ππ π π =β³ (1ππ‘π)(17.04π/πππ) (.0821)(273) = (4.93π)(.0821)(400πΎ) 1.05ππ‘π = πππ. ππ/πππ = .76 π/πΏ Total Pressure Partial Pressure of He Density Average Kinetic energy per molecule 1 1 2 1 19. a. the number of moles of each gas. n = PV/RT = (265/760)(1.0)/(0.0821)(298) n = 0.014 mol n = PV/RT = (800/760)(1.0)/(0.0821)(298) n = 0.043 mol n = PV/RT = (532/760)(0.5)/(0.0821)(298) n = 0.014 mol N2 Ne H2 b. the total pressure after all valves are opened. P = nRT/V = (0.071)(0.0821)(298)/(2.5) P = 0.695 atm (528 torr) c. the partial pressure of each gas. PN2 = XN2Ptot = (0.014/0.071)(528 torr) N2 PN2 = 104 torr PNe = XNePtot = (0.043/0.071)(528 torr) Ne PNe = 320 torr PH2 = XH2Ptot = (0.014/0.071)(528 torr) H2 PH2 = 104 torr 740 = PH2 + 32torr ο PH2 = 708 torr 20. a) Ptot = PH2 + PH2O b) ππ = ππ π π = ππ π π = (0.932ππ‘π)(2.0πΏ) (.0821)(303πΎ) = π. π × ππ−π πππ 21. a) ππ»2 = 745 πππ»π − 23.8πππ»π = πππ. π ππ π―π π= ππ π π = (.949ππ‘π)(.090 πΏ) π (298πΎ) = π. ππ × ππ−π πππππ π―π 3 2 3 1 1 3 1 1 Both I and III have the same # of molecules, therefore same pressure b) First convert 23.8 mm Hg to atm ππ (0.0313ππ‘π)(.090 πΏ) 6.02 × 1023 π= = = π. ππ × ππ−π πππππ π―π πΆ × = π. ππ × ππππ πππππππππ π―π πΆ 1 ππππ π π π (298πΎ) c) πππ‘π ππ π»2 πππ‘π ππ π»2π = √18.02 √2.02 = π. ππ πππππ d) H2O because it is a polar molecule and in fact attracts to other water molecules due to hydrogen IMF bonding. The stronger the IMF, the more the molecule or atom will deviate from ideal gas behavior. At low T and high P conditions molecules with stronger IMF will attract more to each other and collide less often with the container walls. a) π·π» = 22. b) ππ π π = (0.60πππ)π (300πΎ) 20.0 πΏ π·πΆπ = ππΆπ β π·π» = ( π.ππππ 1πππ 23. 24.2 g He × 4.00π = 6.05 πππ π»π ππ = (6.185πππ)π (298πΎ) 1.5 πΏ = π. ππππππ ππ ππ. ππ ππ·π π.ππππ = 12.1ππ‘π ) β ππ. ππππ·π = ππ. ππππ·π 4.32 g π2 × ππ»π = ( 1πππ 32.00π = 0.135 πππ π2 6.05 πππ 6.185 πππ ) β 12.1ππ‘π = ππ. ππππ ππ2 = ( 0.135 πππ 6.185 πππ ) β 12.1ππ‘π = π. ππ πππ 24. a) b) All gases have the same KE avg. because temperature is the same and temperature is a measure of KE. He because it has the lowest MM. At the same temperature, the square of the velocity of any gas varies inversely with its molar ππΉπ» mass. ππππ = √ c) d) e) f) π He because partial pressure is proportional to the number of moles of gas present and the # of moles increases with decreasing molar mass. 3.00 g of each gas, so if you convert to moles of each, you will see that the lower MM will have the most moles. SO2 is the largest, most complex and they are the most polar. SO2 uses dipole-dipole bonding which is the strongest of the IMF’s, the other gases only use LDF. SO2 because again it uses dipole-dipole intermolecular bonding which is stronger than the LDF, therefore higher BP. decreasing T and increasing P.
