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Gases Chapter 5 E-mail: [email protected] Web-site: http://clas.sa.ucsb.edu/staff/terri/ Gases – Ch. 5 1. Draw the following: a. A closed monometer attached to a flask filled with CO at 250 torr b. An open monometer at sea level attached to a flask filled with N2O at 600 torr Gases – Ch. 5 Gases – Ch. 5 2. Determine if the following are directly or inversely proportional – assume all other variables are constant a. Pressure and volume b. Pressure and temperature Gases – Ch. 5 3. The valve between two tanks is opened. See below. Calculate the ratio of partial pressures (O2:Ne) in the container after the valve is opened. a. b. c. d. e. 1.31 1.60 2.10 0.477 0.615 3.25-L 8.64 atm O2 2.48-L 5.40 atm Ne Gases – Ch. 5 Combined Gas Law If one variable (P, V, n or T) of a gas changes at least one other variable must change P1V1 = P2V2 n1T1 n2T2 Gases – Ch. 5 4. In an experiment 300 m3 of methane is collected over water at 785 torr and 65 °C. What is the volume of the dry gas (in m3) at STP? The vapor pressure of water at 65 °C is 188 torr. Gases – Ch. 5 5. Consider a sample of neon gas in a container fitted with a movable piston (assume the piston is mass-less and frictionless). The temperature of the gas is increased from 20.0°C to 40.0°C. The density of neon a. b. c. d. e. increases less than 10%. decreases less than 10%. increases more than 10%. decreases more than 10%. does not change. Gases – Ch. 5 6. A gaseous mixture containing 1.5 mol Ar, 6 mol He and 3.5 mol Ne has a total pressure of 7.0 atm. What is the partial pressure of Ne? a. b. c. d. e. 1.4 atm 2.2 atm 3.8 atm 4.6 atm 2.7 atm Gases – Ch. 5 7. A mixture of oxygen and helium is 92.3% by mass oxygen. What is the partial pressure of oxygen if atmospheric pressure is 745 Torr? Gases – Ch. 5 8. A sample of oxygen gas has a volume of 4.50 L at 27°C and 800.0 torr. How many oxygen molecules are in the sample? a. b. c. d. e. 1.16 × 1023 5.8 × 1022 2.32 × 1024 1.16 × 1022 none of these Gases – Ch. 5 Ideal gas law Considering one set of variables for a gas under “ideal” conditions PV = nRT R ⇒ Universal gas constant R = 0.08206 atmL/molK R = 62.37 torrL/molK R = 8.314 KPaL/molK or J/molK Gases – Ch. 5 9. Consider the combustion of liquid hexane: 2 C6H14 (l) + 19 O2 (g) → 12 CO2 (g) + 14 H2O (l) 1.52-g of hexane is combined with 2.95 L of oxygen at 312K and 890 torr. The carbon dioxide gas is collected and isolated at 297 K and 0.930 atm. What volume of carbon dioxide gas will be collected, assuming 100% yield? a. b. c. d. e. 0.504 L 1.93 L 2.23 L 0.607 L 4.04 L Gases – Ch. 5 10. A 3.54-g sample of lead(II) nitrate (molar mass = 331 g/mol) is heated in an evacuated cylinder with a volume of 1.60 L. The salt decomposes when heated, according to the following equation: 2 Pb(NO3)2 (s) 2 PbO (s) + 4 NO2 (g) + O2 (g) Assuming complete decomposition, what is the pressure (in atm) in the cylinder after decomposition and cooling to a temperature of 300. K? Assume the PbO(s) takes up negligible volume. Gases – Ch. 5 11. 2.5 mol of O2 gas and 3.0 mol of solid carbon, C (s) are put into a 3.50-liter container at 23°C. If the carbon and oxygen react completely to form CO (g), what will be the final pressure (in atm) in the container at 23°C? Gases – Ch. 5 12. The density of an unknown gas at STP is 0.715 g/L. Identify the gas. a. NO b. Ne c. CH4 d. O2 Gases – Ch. 5 Molar mass (M) can be used to identify an unknown substance M = DRT P or mRT M= PV Gases – Ch. 5 13. Air is 79% N2 and 21% O2 by volume. Calculate the density of air at 1.0 atm, 25°C. a. b. c. d. e. 0.590 g/L 1.18 g/L 2.46 g/L 14.1 g/L none of these Gases – Ch. 5 14. These plots represent the speed distribution for 1.0 L of oxygen at 300 K and 1000 K. Identify which temperature goes with each plot. Gases – Ch. 5 Average Speed = 8 Gases – Ch. 5 15. These plots represent the speed distribution for 1.0 L of He at 300 K and 1.0 L of Ar at 300 K. Identify which gas goes with each plot. Gases – Ch. 5 16. Calculate the temperature at which the average velocity of Ar (g) equals the average velocity of Ne (g) at 25°C. a. b. c. d. e. 317°C 151°C 49.5°C 25°C none of these Gases – Ch. 5 17. Order the following according to increasing rate of effusion if all gases are at the same T and P. F2, Cl2, NO, NO2, CH4 Gases – Ch. 5 18. It takes 12 seconds for 8 mL of hydrogen gas to effuse through a porous barrier at STP. How long will it take for the same volume of carbon dioxide to effuse at STP? Gases – Ch. 5 Graham’s Law If two or more gases are effusing under the same conditions 1 2 = 1 2 1 2 2 1 Gases – Ch. 5 19. The diffusion rate of H2 gas is 6.45 times faster than that of a certain noble gas (both gases are at the same temperature). What is the noble gas? a. b. c. d. e. Ne He Ar Kr Xe Gases – Ch. 5 20. Consider two 5 L flasks filled with different gases. Flask A has carbon monoxide at 250 torr and 0 °C while flask B has nitrogen at 500 torr and 0 °C. a. Which flask has the molecules with the greatest average kinetic energy? b. Which flask has the greatest collisions per second? Gases – Ch. 5 Useful equations KEavg = 3/2RT KE = 1/2mu2 ⇒ = 4 2 Gases – Ch. 5 21. Under what conditions will a gas behave the most like an ideal gas? Gases – Ch. 5 Gases – Ch. 5 22. Which of the following gases will have the lowest molar volume at STP? a. He b. CH2Cl2 c. CO2 Gases – Ch. 5 The molar volume can be derived from the ideal gas law: = At STP the molar volume of an ideal gas is 22.41 L/mol As a gas deviates from ideal behavior the molar volume decreases Gases – Ch. 5 [ + a 2 ](V – nb) = nRT a ⇒ compensates for the attractive forces between gas particles b ⇒ compensates for the volume of the gas particles Gases – Ch. 5 Gas a (atmL2/mol2) b (L/mol) He 0.034 0.0237 Ne 0.211 0.0171 Ar 1.35 0.0322 Kr 2.32 0.0398 Xe 4.19 0.0511 N2 1.39 0.0391 CO2 3.59 0.0427 CH4 2.25 0.0428 NH3 4.17 0.0371 H2O 5.46 0.0305 Gases – Ch. 5 You have completed ch. 5 Answer key – Ch. 5 1. Draw the following: a. A closed monometer attached to a flask filled with CO at 250 torr b. An open monometer at sea level attached to a flask filled with N2O at 600 torr a. b. Answer key – Ch. 5 2. Determine if the following are directly or inversely proportional – assume all other variables are constant a. Pressure and volume inversely b. Pressure and temperature directly Answer key – Ch. 5 3. The valve between two tanks is opened. See below. Calculate the ratio of partial pressures (O2:Ne) in the container after the valve is opened. a. 1.31 Each gas is affected by the valve opening b. 1.60 P1V1 = P2V2 where n and T are constant c. 2.10 n1T1 n2T2 d. 0.477 P1V1 solving for P ⇒ P = 2 2 e. 0.615 V2 for O2 ⇒ P2 = (8.64 atm)(3.25 L) (3.25L+2.48L) P2 = 4.9 atm for Ne ⇒ P2 = (5.4 atm)(2.48 L) = 2.3 atm (3.25L+2.48L) The ratio of partial pressures ⇒ 2 = 4.9 atm= 2.1 2.3 atm Answer key – Ch. 5 4. In an experiment 300 m3 of methane is collected over water at 785 torr and 65 °C. What is the volume of the dry gas (in m3) at STP? The vapor pressure of water at 65 °C is 188 torr. When a gas is collected over water there will be water vapor in the collection chamber. A dry gas implies that the water vapor has been removed. The PCH4 = Ptotal – PH2O = 785 torr – 188 torr = 597 torr At STP the temperature and pressure are 273K and 760 torr respectively. Using the combined gas law where n is constant P1V1 = P2V2 n1T1 n2T2 3)(273 K) (597 torr)(300 m P V T solving for V2 ⇒ V2 = 1 1 2 ⇒ V2 = P2 T 1 (760 torr)(65 + 273 K) V2 = 190 m3 Answer key – Ch. 5 5. Consider a sample of neon gas in a container fitted with a movable piston (assume the piston is mass-less and frictionless). The temperature of the gas is increased from 20.0°C to 40.0°C. The density of neon a. b. c. d. e. increases less than 10%. Density = D2 = (m2/V2) where mass is constant decreases less than 10%. D1 (m1/V1) increases more than 10%. D2 = V1 decreases more than 10%. D1 V2 does not change. Using P1V1 = P2V2 where P and n are constant n1T1 n2T2 V Solving for V1 ⇒ 1 = T1 V2 V2 T2 So D2 = T1 = (20+273K) ⇒ D2 = 0.936 or 93.6% ⇒ D1 T2 (40 + 273K) D1 the density decreased by 6.4 % Answer key – Ch. 5 6. A gaseous mixture containing 1.5 mol Ar, 6 mol He and 3.5 mol Ne has a total pressure of 7.0 atm. What is the partial pressure of Ne? a. 0.95 atm b. 2.2 atm PNe = XNePtotal c. 3.8 atm XNe = nnNe d. 2.7 atm total e. 4.8 atm X = 3.5mol = 0.318 Ne 11mol PNe = (0.318)(7.0 atm) = 2.2 atm Answer key – Ch. 5 7. A mixture of oxygen and helium is 92.3% by mass oxygen. What is the partial pressure of oxygen if atmospheric pressure is 745 Torr? If you had a 100 g sample ⇒ 92.3 g of O2 and 7.7 g of He 92.3 g O2 = 2.88 mol O 2 31.998 g/mol 7.7 g of He = 1.92 mol He 4.80 total moles 4.0026 g/mol PO2 = XO2Ptotal = 2.88 mol O2 745 torr = 447 torr 4.80 total mol Answer key – Ch. 5 8. A sample of oxygen gas has a volume of 4.50 L at 27°C and 800.0 torr. How many oxygen molecules does it contain? a. 1.16 × 1023 # of molecules can be derived from moles b. 5.8 × 1022 PV = nRT c. 2.32 × 1024 n = 22 d. 1.16 × 10 (800torr)(4.5L) e. none of these n= (62.37torrL/molK)(300K) n = 0.192 mol 0.192 mol x 6.022x1023 molecules/mol = 1.16x1023 molecules Answer key – Ch. 5 9. Consider the combustion of liquid hexane: 2 C6H14 (l) + 19 O2 (g) → 12 CO2 (g) + 14 H2O (l) 1.52-g of hexane is combined with 2.95 L of oxygen at 312K and 890 torr. The carbon dioxide gas is collected and isolated at 297 K and 0.930 atm. What volume of carbon dioxide gas will be collected, assuming 100% yield? a. 0.504 L Need to determine the limiting reagent b. 1.93 L 1.52 g = 0.0177 mol vs. n = C H 6 14 c. 2.23 L 86.07 g/ d. 0.607 L nO2 = PV (890 torr/760 torr/atm)(2.95L) = 0.135 mol RT (0.08206 atmL/molK)(312K) e. 4.04 L 0.135/19 < 0.0177/2 so O2 is the LR 0.135 mol O2 (12 mol CO2) = 0.0853 mol CO2 (19 molO2) V = nRT = (0.0853 mol)(0.08206 atmL/molK)(297K) = 2.23 L P (0.93atm) Answer key – Ch. 5 10. A 3.54-g sample of lead(II) nitrate (molar mass = 331 g/mol) is heated in an evacuated cylinder with a volume of 1.60 L. The salt decomposes when heated, according to the following equation: 2 Pb(NO3)2 (s) 2 PbO (s) + 4 NO2 (g) + O2 (g) Assuming complete decomposition, what is the pressure (in atm) in the cylinder after decomposition and cooling to a temperature of 300. K? Assume the PbO(s) takes up negligible volume. The reaction produces 2 gases so the pressure in the container is the total pressure ⇒ Ptotal = ntotalRT V 3.54 g Pb(NO3)2 (1mol Pb(NO3)2 ) (4 mol NO2+1mol O2) = 0.0276 mol gas (331 g Pb(NO3)2) 2 mol Pb(NO3)2 Ptotal = (0.0267 mol)(0.08206atmL/molK)(300K) = 0.41 atm (1.6L) Answer key – Ch. 5 11. 2.5 mol of O2 gas and 3.0 mol of solid carbon, C (s) are put into a 3.50-liter container at 23°C. If the carbon and oxygen react completely to form CO (g), what will be the final pressure (in atm) in the container at 23°C? 2 C (s) + O2 (g) 2 CO (g) Determine limiting reagent C ⇒ 3 mol C/2 = 1.5 O2 ⇒ 2.5 mol/1 = 2.5 ⇒ C is the LR Since C is the LR ⇒ in addition to the CO formed there will be excess O2 in the container so the pressure will be the total pressure ⇒ Ptotal = ntotalRT/V …continue to next slide Answer key – Ch. 5 11. …continued 2 C (s) O2 (g) 3 mol 2.5 mol -3 -3(2) +3(2) 0 1 mol 3 mol 1 2 CO (g) 0 mol 2 Since there is 4 mol of gas in the container Ptotal = (4 )( 0.08206 )(23+273) 3.5 Ptotal = 27.8 atm Answer key – Ch. 5 12. The density of an unknown gas at STP is 0.715 g/L. Identify the gas. a. NO Molar mass can be useful to identify a substance b. Ne M = DRT P c. CH4 (0.715 g/L)(0.08206atmL/molK)(273K) M = d. O2 (1atm) M = 16 g/mol Unknown gas is CH4 Answer key – Ch. 5 13. Air is 79% N2 and 21% O2 by volume. Calculate the density of air at 1.0 atm, 25°C. a. 0.590 g/L b. 1.18 g/L Density is in the equation ⇒ c. 2.46 g/L M = DRT P d. 14.1 g/L MP D = e. none of these RT Since we have 2 gases ⇒ D = (MN2 PN2 )+(MO2 PO2) RT D= (28g/mol)(0.79atm)+(32g/mol)(0.21 atm) (0.08206 atmL/molK)(298K) D = 1.18 g/L Answer key – Ch. 5 14. These plots represent the speed distribution for 1.0 L of oxygen at 300 K and 1000 K. Identify which temperature goes with each plot. Average Speed of A Average Speed of B According to the average speed equation ⇒ uavg = (8RT/π M)1/2 we can see the relationship between average speed and temperature as T ↑ uavg ↑ since uavg B > uavg A ⇒ TB>TA Plot A ⇒ 300K Plot B ⇒ 1000K Answer key – Ch. 5 15. These plots represent the speed distribution for 1.0 L of He at 300 K and 1.0 L of Ar at 300 K. Identify which gas goes with each plot. According to the average speed equation ⇒ = Average Speed of A Average Speed of B 8 we can see the relationship between average speed and molar mass as molar mass ↑ uavg ↓ since uavg B > uavg A ⇒ MB < MA Plot A ⇒ Ar Plot B ⇒ He Answer key – Ch. 5 16. Calculate the temperature at which the average velocity of Ar (g) equals the average velocity of Ne (g) at 25°C. a. 317°C b. 151°C = 8 c. 49.5°C uave Ar = uave Ne d. 25°C 8 8 = e. none of these 8,R, and π constant TAr = TNe ⇒ T = MArTNe Ar MNe MAr MNe TAr = (298K)(39.95 g/mol) (20.18 g/mol) T = 590 K or 317°C Answer key – Ch. 5 17. Order the following according to increasing rate of effusion: F2, Cl2, NO, NO2, CH4 As molar mass ↑ average speed ↓ rate of effusion ↓ Since the relative molar masses are Cl2 (70.9 g/mol) > NO2 (46.01 g/mol) > F2 (38 g/mol) > NO (30.01 g/mol) > CH4 (16.042 g/mol) Therefore the relative rates of effusion are Cl2 < NO2 < F2 < NO < CH4 Answer key – Ch. 5 18. It takes 12 seconds for a given volume of hydrogen gas to effuse through a porous barrier. How long will it take for the same volume of carbon dioxide? timeCO2 = MCO2 timeH2 MH2 timeCO2 = timeH2 MCO2 MH2 44.01g/mol 2.016g/mol timeCO2 = 56 s timeCO2 = (12 s) Answer key – Ch. 5 19. The diffusion rate of H2 gas is 6.45 times as great as that of a certain noble gas (both gases are at the same temperature). What is the noble gas? a. Ne Molar mass can be used to identify b. He rateH2 Munk c. Ar = rateunk MH2 d. Kr rateH2 e. Xe Munk =MH2 rateunk Munk = 2.016 g/mol 6.45 Munk = 83.87 g/mol Unknown gas is Kr Answer key – Ch. 5 20. Consider two 5 L flasks filled with different gases. Flask A has carbon monoxide at 250 torr and 0 °C while flask B has nitrogen at 500 torr and 0 °C. a. Which flask has the molecules with the greatest average kinetic energy? According to KEavg = 3RT ⇒ we see the relationship ⇒ as T 2 ↑ KEavg↑ ⇒ since both flasks are at the same T they will have the same KEavg b. Which flask has the greatest collisions per second? According to = 4 2 we see three relationships ⇒ as T↑ Z↑ or as molar mass↑ Z↓ or as N/V (or P)↑ Z↑ ⇒ so since both flasks have the same T and molar mass but the PB > PA ⇒ ZB > ZA Answer key – Ch. 5 21. Under what conditions will a real gas behave like an ideal gas? An “ideal” gas is one that in reality adheres to the ideal gas law ⇒ meaning experimental values agree with calculated values using PV = nRT Gases are more likely to behave “idealy” when the pressure is low and/or the temperature is high Deviations from the ideal gas law is due to the attractive forces between the gas particles and the volume of the gas particles relative to the volume of the container Answer key – Ch. 5 22. Which of the following gases will have the lowest molar volume at STP? a. He b. CH2Cl2 The molar volume of an “ideal” gas is 22.4 L/mol c. CO2 as the attractive forces of the gas particles ↑ the molar volume ↓ – later (in ch 16) we will learn the specifics of attractive forces however for now we can use the relationship that as molar mass ↑ attractive forces ↑ (an exception is water – although water is rather on the light side it has quite strong attractive forces called H-Bonds which we’ll see further in ch 16) Therefore since CH2Cl2 has the highest molar mass it has the strongest attractive forces and the lowest molar volume