Gases. Chapter 4 E-mail: benzene4president@gmail.com Web-site: http://clas.sa.ucsb.edu/staff/terri/. Gases – Ch. 5. 1. Draw the following: a. Closed monometer attached to a flask filled with CO at 250 torr
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Gases Chapter 4 E-mail: benzene4president@gmail.com Web-site: http://clas.sa.ucsb.edu/staff/terri/
Gases – Ch. 5 1. Draw the following: a. Closed monometer attached to a flask filled with CO at 250 torr b. Open monometer at sea level attached to a flask filled with N2O at 600 torr
Gases – Ch. 5 2. Which of the following relationships is not true? a. PV = constant when temperature and moles of gas are held constant. b. V/T = constant when pressure and moles of gas are held constant. c. nT = constant when pressure and volume are held constant. d. P/n = constant when volume and temperature are held constant. e. All of these are true.
= n1T1 n2T2 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
Gases – Ch. 5 3. In an experiment 300 m3 of methane is collected at 785 torr and 165 °C. What is the volume (in m3) at STP?
Gases – Ch. 5 4. Body temperature is about 308 K. On a cold day, what volume of air at 273 K must a person with a lung capacity of 2.00 L breathe in to fill the lungs? a. 2.26 L b. 1.77 L c. 1.13 L d. 3.54 L e. none of these
Gases – Ch. 5 5. The volume of a helium balloon is 1.85 L at 24.0°C and 1.00 atm at sea level. The balloon is released and floats upward. At a certain altitude, the balloon has a volume of 2.14 L and the temperature is 15.2°C. What is the atmospheric pressure at this altitude? a. 0.538 atm b. 0.839 atm c. 0.891 atm d. 1.36 atm e. none of these
Gases – Ch. 5 6. 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. increases less than 10%. b. decreases less than 10%. c. increases more than 10%. d. decreases more than 10%. e. does not change.
Gases – Ch. 5 7. A gaseous mixture containing 1.5 mol Ar and 3.5 mol CO2 has a total pressure of 7.0 atm. What is the partial pressure of CO2? a. 1.8 atm b. 2.1 atm c. 3.5 atm d. 4.9 atm e. 2.4 atm
Gases – Ch. 5 8. The valve between a 3.25-L tank containing O2 (g) at 8.64 atm and a 2.48-L tank containing Ne (g) at 5.40 atm is opened. Calculate the ratio of partial pressures (O2:Ne) in the container. a. 1.31 b. 1.60 c. 2.10 d. 0.477 e. 0.615
Gases – Ch. 5 9. 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 b. 5.8 × 1022 c. 2.32 × 1024 d. 1.16 × 1022 e. 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 10. Potassium chlorate decomposes upon heating as follows: 2 KClO3 (s) → 2 KCl (s) + 3 O2 (g) A 2.36-g sample of KClO3 decomposes, and the oxygen at 23.8°C and 0.930 atm is collected. What volume of oxygen gas will be collected, assuming 100% yield? a. 0.504 L b. 1.93 L c. 0.756 L d. 0.0607 L e. 0.0404 L
Gases – Ch. 5 11. 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 12. 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 13. The density of an unknown gas at STP is 0.715 g/L. Identify the gas. a. NH3 b. Ne c. CH4 d. O2
Gases – Ch. 5 Molar mass (M) can be used to identify an unknown substance M = DRT/P or M = mRT/PV
Gases – Ch. 5 14. 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 c. 2.46 g/L d. 14.1 g/L e. none of these
Gases – Ch. 5 15. 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
Gases – Ch. 5 16. 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 17. 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 c. 49.5°C d. 25°C e. none of these
Gases – Ch. 5 18. 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 19. 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?
Gases – Ch. 5 Graham’s Law If two or more gases are effusing under the same conditions or
Gases – Ch. 5 20. 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 b. He c. Ar d. Kr e. Xe
Gases – Ch. 5 21. 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? c. Which flask has the greatest density?
Gases – Ch. 5 Useful equations KEavg = 3/2RT KE = 1/2mu2
Gases – Ch. 5 22. Under what conditions will a real gas behave like an ideal gas?
Gases – Ch. 5 23. Which of the following gases will have the lowest molar volume at STP? a. He b. CH2Cl2 c. CO2
Gases – Ch. 5 Van der Waal’s Equation a ⇒ compensates for the attractive forces between gas particles b ⇒ compensates for the volume of the gas particles
Answer key – Ch. 5 1. Draw the following: a. Closed monometer attached to a flask filled with CO at 250 torr b. Open monometer at sea level attached to a flask filled with N2O at 600 torr a. b.
Answer key – Ch. 5 2. Which of the following relationships is not true? a. PV = constant when temperature and moles of gas are held constant. b. V/T = constant when pressure and moles of gas are held constant. c. nT = constant when pressure and volume are held constant. d. P/n = constant when volume and temperature are held constant. e. All of these are true.
Answer key – Ch. 5 3. In an experiment 300 m3 of methane is collected at 785 torr and 165 °C. What is the volume (in m3) at STP? At STP the temperature and pressure are 273K and 760 torr respectively. P1V1/n1T1 = P2V2/n2T2 n is constant solving for V2⇒ V2 = P1V1T2/P2T1 V2 = (785 torr)(300 m3)(273 K)/(760 torr)(165 + 273K) V2 = 193 m3
Answer key – Ch. 5 4. Body temperature is about 308 K. On a cold day, what volume of air at 273 K must a person with a lung capacity of 2.00 L breathe in to fill the lungs? a. 2.26 L b. 1.77 L c. 1.13 L d. 3.54 L e. none of these P1V1/n1T1 = P2V2/n2T2 n and P is constant solving for V2⇒ V2 = V1T2/T1 V2 = (2 L)(273 K)/(308 K) V2 = 1.77 L
Answer key – Ch. 5 5. The volume of a helium balloon is 1.85 L at 24.0°C and 1.00 atm at sea level. The balloon is released and floats upward. At a certain altitude, the balloon has a volume of 2.14 L and the temperature is 15.2°C. What is the atmospheric pressure at this altitude? a. 0.538 atm b. 0.839 atm c. 0.891 atm d. 1.36 atm e. none of these P1V1/n1T1 = P2V2/n2T2 n is constant solving for P2⇒ P2 = P1V1T2/T1V2 P2 = (1 atm)(1.85 L)(15.2+273K)/(24+273K)(2.14 L) P2 = 0.839 atm
Answer key – Ch. 5 6. 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. increases less than 10%. b. decreases less than 10%. c. increases more than 10%. d. decreases more than 10%. e. does not change. Density = mass/volume the mass is staying constant so we need to determine what % the volume is changing P1V1/n1T1 = P2V2/n2T2 P and n is constant Solving for V2/V1⇒ V2/V1 = T2/T1 V2/V1 = (40+273K)/(20 + 273K) ⇒ V2/V1 = 1.068 ⇒ since the volume went up by 6.8% ⇒ the density decreased by 6.8 %
Answer key – Ch. 5 7. A gaseous mixture containing 1.5 mol Ar and 3.5 mol CO2 has a total pressure of 7.0 atm. What is the partial pressure of CO2? a. 1.8 atm b. 2.1 atm c. 3.5 atm d. 4.9 atm e. 2.4 atm P1V1/n1T1 = P2V2/n2T2 V and T is constant The combined gas law is not limited to just one gas – in this case we can make the 1st set of variables for CO2 and the 2nd set of variables for the gas mixture Solving for P1⇒ P1 = P2n1/n2 P1 = (7.0 atm)(3.5 mol)/(1.5 mol+3.5 mol) P1 = 4.9 atm
Answer key – Ch. 5 8. The valve between a 3.25-L tank containing O2 (g) at 8.64 atm and a 2.48-L tank containing Ne (g) at 5.40 atm is opened. Calculate the ratio of partial pressures (O2:Ne) in the container. a. 1.31 b. 1.60 c. 2.10 d. 0.477 e. 0.615 Each gas is affected by the valve opening P1V1/n1T1 = P2V2/n2T2 n and T is constant solving for P2⇒ P2 = P1V1/V2 for O2⇒ P2 = (8.64 atm)(3.25 L)/(3.25L+2.48L) P2 = 4.9 atm for Ne ⇒ P2 = (5.4 atm)(2.48L)/(3.25L+2.48L) = 2.3 atm The ratio of partial pressures = 4.9 atm/2.3 atm = 2.1
Answer key – Ch. 5 9. 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 b. 5.8 × 1022 c. 2.32 × 1024 d. 1.16 × 1022 e. none of these # of molecules can be derived from moles PV = nRT n = PV/RT n=(800torr)(4.5L)/(62.37torrL/molK)(300K) n = 0.192 mol 0.192 mol x 6.022x1023 molecules/mol = 1.16x1023 molecules
Answer key – Ch. 5 10. Potassium chlorate decomposes upon heating as follows: 2 KClO3 (s) → 2 KCl (s) + 3 O2 (g) A 2.36-g sample of KClO3 decomposes, and the oxygen at 23.8°C and 0.930 atm is collected. What volume of oxygen gas will be collected, assuming 100% yield? a. 0.504 L b. 1.93 L c. 0.756 L d. 0.0607 L e. 0.0404 L PV = nRT ⇒ V = nRT/P 2.36 g KClO3 x 1 mol KClO3 x 3mol O2 = 0.029 mol 2 mol KClO3 O2 122.55 g KClO3 V = (0.29mol)(0.08206atmL/molK)(296.8K)/(0.93atm) V = 0.756 L
Answer key – Ch. 5 11. 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 …continue to next slide
Answer key – Ch. 5 11. …continued 3.54 g Pb(NO3)2 x 1mol Pb(NO3)2 x (4 mol NO2+1mol O2) = 0.0267 mol 331 g Pb(NO3)2 2 mol Pb(NO3)2 Ptotal = (0.0267 mol)(0.08206atmL/molK)(300K)/(1.6L) Ptotal = 0.41 atm
Answer key – Ch. 5 12. 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 12. …continued CO ⇒ 3 mol C x 2 mol CO = 3 mol CO are formed O2⇒ 3 mol C x 1 mol O2 = 1.5 mol of O2 are consumed 2 mol C 2 mol C 2.5 mol – 1.5 mol = 1 mol O2 remains Ptotal = (3mol + 1mol)(0.08206atmL/molK)(296 K)/(3.5 L) Ptotal = 27.8 atm
Answer key – Ch. 5 13. The density of an unknown gas at STP is 0.715 g/L. Identify the gas. a. NH3 b. Ne c. CH4 d. O2 Molar mass can be useful to identify a substance M = DRT/P M = (0.715 g/L)(0.08206atmL/molK)(273K)/(1atm) M = 16 g/mol Unknown gas is CH4
Answer key – Ch. 5 14. 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 c. 2.46 g/L d. 14.1 g/L e. none of these Density is in the equation ⇒ M = DRT/P D = M P/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 15. 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. 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 Average Speed of A Average Speed of B
Answer key – Ch. 5 16. 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 ⇒ uavg = (8RT/π M)1/2 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 Average Speed of A Average Speed of B
Answer key – Ch. 5 17. 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 c. 49.5°C d. 25°C e. none of these uavg = (8RT/π M)1/2 uavg Ar = uavg Ne (8RTAr/π MAr)1/2 =(8RTNe/π MNe)1/2 8,R, πand ½ are constant TAr/MAr=TNe/MNe TAr = TNeMAr/MNe TAr = (298K)(39.95 g/mol)/(20.18 g/mol) T = 590 K or 317°C