WS 4.1 Key All gas molecules or atoms are always in constant
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WS 4.1 Key 1. a) All gas molecules or atoms are always in constant, random motion. Collisions between molecules produce pressure without a loss in total energy, (elastic collisions), KE is proportional to Temp. b) Ideal gas molecules have negligible volume and gas molecules don’t attract to each other. NOTE: these postulates are for Ideal Gases only. π 2. a) π²π¬πππ = πΉπ» ππΉπ» π b) ππππ = √ 3. 4. 5. 6. πππππ πππππ = √ππ πππππ πππππ = √ππ (32π ) 4.67 8. 20π .48 π ) (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 molβK ππππ = √ ππππ ππ π―π ππππ ππ πΆπ π J π ππ.ππ H2 effuses 4.67 times faster than CO2 = √ π.ππ = 6.85π πππππ πππππ 7. ( π π π²π¬πππ = (8.31 π π ππππ ππ π΅π―π ππππ ππ π―π = √ππ π π.πππ NH3 compared to He: 0.48 times (slower of course) = √ππ.πππ ) = ππ πππ πππππ―π πππππππππππ ππ π.ππ =√ π π.ππ ππ π.ππ =√ = 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 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 the greater number of molecules exerts twice the pressure. b) The piston would move upward to double the volume because the faster moving molecules exert twice the pressure. c) The piston would move downward to halve the volume so that molecules are more crowded in order to produce enough collisions to match the 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 10.4 atm 585 mL 0.250 mol 295 K 14. a) ππ = ππ π b) 0.382ππππ2 × 15. π·π π½π = ππ π»π π2 = 16. π π ππ (94.6πππ)(10.0πΏ) = (8.31)(298πΎ) π π 32.0 π π2 = ππ. π π πΆπ 1 πππ π= π·π π½π (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 =π= 17. π = πββ³ π βπ = = = π. πππ πππ (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 18. Total Pressure Partial Pressure of He Density Average Kinetic energy per molecule I and III are the same, then II I, II, III III, I, II since T is same avg. KE is same for all 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) Ne n = 0.043 mol n = PV/RT = (532/760)(0.5)/(0.0821)(298) H2 n = 0.014 mol b. the total pressure after all stopcocks 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 N2 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) π = b) π = c) πππππ πππππ ππ π π ππ π π = = (0.9489ππ‘π).090πΏ) (.0821)(298πΎ) (0.0313ππ‘π).090πΏ) (.0821)(298πΎ) ππ.ππ π/πππ =√ π.ππ π/πππ = = π. ππ × ππ−π πππ H2 = π. ππ × ππ−π πππ then convert to molecules: 6.92 x 1019 molecules H2O H2 is 2.99 times faster d) H2O shows hydrogen bonding which will causes it to deviate from ideal behavior, particularly at low temperatures and high pressures. 22. a) PV=nRT π· = ππΉπ» b) ππ = ππ β ππ‘ππ‘ππ π½ = (.πππππ)(.ππππ)(ππππ²) (πππ³) (ππππ πππππ‘πππ) ππ = (. 33) β 74.9πππ = ππππ·π = π. ππππππ ππ ππ. πππ·π ππ = ππ ππ‘ππ‘ππ = .20πππ = .33 .60πππ 23. ππππ ππππ‘πππ ππ π»π ππ .978 πππ ππππ πππππ‘πππ ππ π2 ππ 0.0218 (π +π )π π ππ‘ππ‘ππ = π»π π2 = 12.099 atm π PHe= XHe Ptotal = 0.978 x 12.099atm = 11.8 atm PO2= XO2 Ptotal = 0.0218 x 12.099atm = 0.264 atm 24. a) All have the same avg. KE because they are at the same temperature and temperature is a measure of KE. b) He because they have the lowest mass. c) He because partial pressure is proportional to the number of moles of gas and the number of moles increases with decreasing molar mass. d) SO2 will deviate because they are the largest, have the greatest LDF interactions and they are also polar molecules so they will have the strongest attractive forces. e) SO2 because its polarity provides higher attractive forces. f) Decreasing the temperature and increasing the pressure.
