Ch 21 Solutions - Stress can be defined as any type of change that causes physical, emotional, PDF

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Stress can be defined as any type of change that causes physical, emotional, or psychological strain. Stress is your body's response to anything that requires attention or action. Everyone experiences stress to some degree...

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Chapter 21 Physical properties of gases Q1. Use the kinetic molecular theory to explain the following observed properties of gases: a Gases occupy all the available space in a container. b Gases can be easily compressed compared with their corresponding liquid forms. c A given volume of a gaseous substance weighs less than the same volume of the substance in the liquid state. d Gases will mix together readily. e The total pressure of a mixture of gases is equal to the sum of the pressures exerted by each of the gases in the mixture. A1. a

b c

d e

Molecules of gases are in constant, rapid, random motion and the forces between molecules are negligible. They continue to move outwards until stopped by the walls of the container, filling all the space available. Most of the volume occupied by a gas is space, so compression can be achieved by reducing the space between the particles. The molecules in a gas are spread much further apart than those of a liquid. A given mass of gas would occupy a much greater volume than the same mass of the liquid phase. Therefore, the density of the gas is less. Gases mix easily together because of the large amount of space between the molecules. The pressure exerted by a gas depends on the number of collisions of gas particles and the wall of the container. The pressure is independent of the type of gas involved. The total pressure exerted by a mixture of gases will depend on the total number of collisions each gas has with the container.

Q2. Use the ideas of the kinetic molecular theory of gases to explain the following observations: a Tyre manufacturers recommend a maximum pressure for tyres. b The pressure in a car’s tyres will increase if a long distance is travelled on a hot day. c You can smell dinner cooking as you enter the house. d A balloon will burst if you blow it up too much. A2. a

b

Tyres have a recommended maximum pressure to give a comfortable ride as well as good traction on the road. If the pressure in a tyre is too high, the gas inside cannot be compressed as easily and passengers will be more aware of bumps on the road. During a long journey on a hot day, the air in a tyre warms up. This means the air molecules have increased kinetic energy, and collisions with the walls of the tyres will increase, and so the pressure will increase.

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c

d

Particles from the cooking food escape the pot and move randomly through the house. If the food has an odour, and if there are enough particles in the air, you will detect the odour as you enter the house. As air is pumped into a balloon, air molecules collide with the rubber of the balloon, forcing it to expand. If too much air is pumped in, the balloon reaches a stage where it cannot stretch any further. If the number of collisions by molecules per given surface area is increased still further, the rubber will break.

Q3. Horse-drawn coaches had ‘tyres’ made from a solid strip of rubber attached to a wooden wheel. Why would these give a much bumpier ride than the modern pneumatic (air-filled) tyre? A3. When an air-filled tyre passes over a bump in the road, the air molecules inside the tyre move and so absorb some of the shock. A solid rubber tyre can distort and absorb some shock, but is not as effective as an air-filled tyre. Q4. The graph in Figure 21.3 (page 358) shows the distribution of molecular kinetic energies for oxygen at 20°C, 30°C and 40°C. Predict the shape of the graph of the distribution of molecular kinetic energies of the same sample of oxygen gas at –20°C. A4. The peak of the graph would be higher and closer to the Y-axis. Q5. Convert each of the following pressures to the units specified: a 1400 mmHg to atm, Pa and bars b 80 000 Pa to atm, mmHg and bars c 4.24 atm to mmHg and Pa d 120 kPa to mmHg, atm and bars e 140 kPa to Pa f 92 000 Pa to kPa A5. These conversions should be applied as needed: 1.00 atm = 760 mmHg = 101.3 kPa = 1.013 × 105 Pa = 1.013 bar a 760 mmHg = 1.00 atm 1400 So 1400 mmHg = 1.00 × 760 atm = 1.84 atm 760 mmHg = 1.013 × 105 Pa 1400 5 So 1400 mmHg = 1.013 × 10 × 760 Pa = 1.87 × 105 Pa 760 mmHg = 1.013 bar 1400 So 1400 mmHg = 1.013 × 760 bar = 1.87 bar b 1.013 × 105 Pa = 1.00 atm

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80 000 5 So 80 000 Pa = 1.00 × 1.013 × 10 atm = 0.790 atm 1.013 × 105 Pa = 760 mm Hg

80 000 5 So 80 000 Pa = 760 × 1.013 × 10 mm Hg = 600 mmHg 1.013 × 105 Pa = 1.013 bar

80 000 5 So 80 000 Pa = 1.013 × 1.013 × 10 bar

c

d

= 0.800 bar 1.00 atm = 760 mmHg So 4.24 atm = 760 × 4.24 mm Hg = 3220 mm Hg 1.00 atm = 1.013 × 105 Pa So 4.24 atm = 1.013 × 105 × 4.24 Pa = 4.30 × 105 Pa 101.3 kPa = 760 mmHg

760 So 120 kPa = 101.3

× 120 kPa = 900 mm Hg

101.3 kPa = 1.00 atm

1.00 So 120 kPa = 101.3

× 120 atm = 1.18 atm

101.3 kPa = 1.013 bar

1.013 So 120 kPa = 101.3 e

× 120 atm = 1.20 bars 140 kPa = 140 × 1000 Pa = 1.40 × 105 Pa

f

92 000 92 000 Pa = 1000

kPa = 92 kPa

Q6. Convert the following volumes to the unit specified: a 2 L to mL b 4.5 dm3 to mL c 2250 mL to L d l20 mL to L e 5.6 mL to L f 3.7 dm3 to m3 g 285 mL to m3 h 4.70 × 10–3 m3 to dm3 and cm3 A6. These conversions should be applied as needed: 1 mL = 1 cm3, 1 L = 1 dm3, 1 L = 1000 mL = 1000 cm3, 1 m3 = 103 dm3 = 106 cm3 a 2 L = 2 × 1000 mL = 2 × 103 mL b 4.5 dm3 = 4.5 L = 4.5 × 1000 mL = 4.5 × 103 mL 2250 c 2250 mL = 1000 L = 2.25 L 120 d 120 mL = 1000 L = 0.12 L

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e f g h

56 56 mL = 1000 L = 5.6 × 10–2 L 3.7 3.7 dm3 = 1000 m3 = 3.7 × 10–3 m3

285 285 mL = 285 cm = 106

m3 = 2.85 × 10–4 m3 = 4.70 × 10–3 × 1000 dm3 = 4.70 dm3 = 4.70 × 1000 cm3 = 4.70 × 103 cm3 3

4.70 × 10–3 m3

Q7. In the kinetic molecular theory, pressure is described as the force per unit area of surface. Explain what happens to the pressure in each of the following situations: a The temperature of a filled aerosol can is increased. b A gas in a syringe is compressed. A7. a

b

As temperature increases, the average kinetic energy of gas molecules in the can will increase. This will lead to an increase in the frequency and force of collisions of gas molecules with the inside walls of the aerosol cans. This will cause an increase in pressure. As the syringe is compressed, the inside surface area of the syringe will decrease. The number of collisions of molecules per unit area per second with the inside walls of the syringe will increase. This will cause a pressure increase.

Q8. Copy and complete the following table, which refers to a given mass of gas kept at a constant temperature. Initial conditions Final conditions Pressure Volume Pressure Volume a 2 atm 50 L 4 atm b 732 mmHg 20 L 500 mmHg 3 c 800 mmHg 350 cm 120 mmHg d 105 kPa 650 cm3 800 cm3 A8. Remember to use Boyle’s law: P1V1 = P2V2. Each of the pressure and volume units needs to be the same. These conversions need to be applied as needed: 1.00 atm = 760 mmHg = 101.325 kPa = 101 325 Pa = 1.01 bar 1 mL = 1 cm3, 1 L = 1 dm3, 1 L = 1000 mL = 1000 cm3 2 50 a V2 = 4 = 25 L 73220 b V2 = 500 = 29 L 800350 120 c V2 = = 2.3 × 103 cm3 or 2.3 L

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d

105650 800 P2 = = 85.3 kPa

Q9. A sample of air in a syringe has a volume of 120 mL when the pressure is 100 000 Pa at room temperature. If the temperature of the gas remains constant, what volume would the air occupy if the pressure was changed to: a 200 000 Pa? b 45 000 Pa? c 64 kPa? A9. Remember to use Boyle’s law: P1V1 = P2V2. Each of the pressure and volume units needs to be the same. 100 000120 200 000 = 60 mL a V = 2

b

c

100 000120 45 000 V2 = = 267 mL This conversion needs to be applied: 101.325 × 103 kPa = 101 325 Pa. 100 000 120 V = 64 1000 = 188 mL 2

Q10. The pressure on a 9.0 mL sample of carbon dioxide is increased from 110 kPa to 280 kPa. Calculate the new volume occupied by the gas, assuming the temperature does not change. A10. Remember to use Boyle’s law: P1V1 = P2V2. Each of the pressure and volume units needs to be the same. 110 9 V2 = 280 = 3.5 mL Q11. A sample of nitrogen occupies 100 mL at 40°C and a pressure of 80 kPa. What pressure would be needed to reduce the volume to 25 mL at 40°C? A11. Remember to use Boyle’s law: P1V1 = P2V2. Each of the pressure and volume units needs to be the same. 80100 25 P2 = = 320 Pa Q12. Convert the following Celsius temperatures to absolute temperatures: a 100°C b 175°C c –145°C Heinemann Chemistry 1 (4th edition)  Reed International Books Australia Pty Ltd

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A12. Remember to convert temperatures by using: T = t + 273, where T is the absolute temperature (in kelvin), and t is the temperature on the Celsius scale. It is also conventional not to use the degree symbol when writing the absolute temperature. For example, 25C would be written as 298 K. a T =100 + 273 = 373 K b T = 175 + 273 = 448 K c T = –145 + 273 = 128 K Q13. A syringe contains 90 mL of gas measured at 1 atm pressure at 25°C. a What volume would the gas occupy at: i 1 atm and 120°C? ii 1 atm and –90°C? b To what temperature would the gas need to be changed for the volume to be: i 30 mL? ii 120 mL? (The pressure remains at 1 atm.) A13. V1 V2 Remember to use Charles’ Law: T1 = T 2 . The temperature must be in kelvin, the volume units must be the same, and the pressure is constant. 90 (120  273) 25  273 a i V2 = = 120 mL 90 ( 90  273) 25  273 ii V2 = = 55 mL V2T1 b Charles’ law needs to rearranged to give: T = V1 . 2

i ii

30 ( 25  273) 90 T2 = = 99 K or –174C 120( 25  273) 90 T2 = = 397 K or 124C

Q14. A cylinder, volume 20 000 L, contains methane. A second cylinder, with volume 500 L, contains 40 mol of methane. Both gas samples are at the same temperature and pressure. Calculate: a the amount of methane in the first cylinder b the mass of methane in the first cylinder

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A14. V1 V2 This law can be used: n1 = n2 . The temperature and pressure must remain constant. 20 00040 500 a n(methane) = = 1600 mol m b Since n = M and the molar mass of methane is 16 g mol–1. m(methane) = 1600 × 16 g = 25.6 kg Q15. A balloon contains 0.35 mol of helium and has a volume of 5.3 L at a certain temperature and pressure. A further 0.12 mol of helium is added, keeping the temperature and pressure constant. Calculate the new volume of the balloon. A15. V1 V2 This law can be used: n1 = n2 . The temperature and pressure must remain constant. V1 n2 n The law must be rearranged to give: V2 = 1 5.3 (0.35  0.12) 0.35 V2 = = 7.1 L Q16. Calculate the volume of the following gases at SLC. a 1.4 mol of chlorine (Cl2) b 1.0 × 10–3 mol of hydrogen (H2) c 1.4 g of nitrogen (N2) A16. Remember that, under standard laboratory conditions (SLC), 1 mol of any gas has a V volume of 24.5 L. Use the formula: n = Vm , where n is the amount in mol, V is the volume in L and Vm is the molar volume in L mol–1. To calculate V the formula is rearranged to V = n × Vm. a V(Cl2) = 24.5 × 1.4 = 34 L b V(H2) = 24.5 × 1.0 × 10–3 L = 2.5 × 10–2 L = 25 mL m c Since n = M and the molar mass of nitrogen is 28.0 g mol–1. 24.5 1.4 V(N2) = 28.0 = 1.2 L

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Q17. Calculate the mass of the following gas samples. All volumes are measured at SLC. a 2.8 L of neon (Ne) b 50 L of oxygen (O2) c 140 mL of carbon dioxide (CO2) A17. Remember, that under standard laboratory conditions (SLC), 1 mol of any gas has a volume of 24.5 L. These questions involve three steps: Step 1: Calculate the amount (in mol) of gas at SLC. Step 2: Find the molar mass (M) of the gas. m Step 3: Use n = M to find the mass by reorganising as m = nM. 2.8 a n(Ne) = 24.5 = 0.114 mol m(Ne) = 0.114 × 20.1 = 2.3 g 50 b n(O2) = 24.5 = 2.04 mol m(O2) = 2.04 × 32.0 = 65 g 140 10  3 c n(CO2) = 24.5 = 0.0057 mol m(CO2) = 0.0057 × 44.0 = 0.251 g Q18. A fixed amount of gas occupies a volume of 30 L at 20°C and a pressure of 100 kPa. What volume would it occupy at: a 50°C and a pressure of 250 kPa? b –45°C and a pressure of 70 000 Pa? c 80°C and a pressure of 1 atm? A18. P1V1

P2V 2

The combined gas equation needs to be used: T1 = T2 . Temperature should be in kelvin, and each of the pressure and volume units needs to be the same. P1V 1T 2 a V = T1P2 2

b

V2

10030 (50  273) = (20  273) 250 = 13.2 L P1V 1T 2 = T1P2

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V2

c

9

100 30 (  45  273) (20 273) 70 = = 33.3 L P1V 1T 2 TP = 1 2 10030 (80  273) = (20  273) 101 = 35.8 L

Q19. A balloon has a volume of 150 L at a pressure of 101 kPa and a temperature of 27°C. It rises to an altitude of 15 km, where the temperature is –30°C and the pressure is 12 kPa. What is the volume of the balloon at this altitude? A19. P1V1 T1

P2V 2 T2

The combined gas equation needs to be used: = . Temperature should be in kelvin, and each of the pressure and volume units needs to be the same. P1V1T2 V = T1 P2 2

101150(  30  273) (27  273) 12 = = 1023 L Q20. A volume of 32.0 mL of hydrogen at 25°C and 75 kPa is compressed to 16 mL and heated to 50°C. What will be the new pressure exerted by the hydrogen? A20. P1V1 P2V 2 The combined gas equation needs to be used: T1 = T2 . Temperature should be in kelvin, and each of the pressure and volume units needs to be the same. P1 V 1 T 2 P = T1V 2 2

7532.0(50  273) (25  273) 16 = = 160 kPa Q21. A 4.20 L sample of gas at 23°C and 0.25 atm is transferred to a 6.50 L vessel. To what temperature must the gas be heated so that its pressure increases to 0.60 atm?

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A21. P1V1 P2V 2 The combined gas equation needs to be used: T1 = T2 . Temperature should be in kelvin, and each of the pressure and volume units needs to be the same. P2V 2T1 PV T2 = 1 1 0.60 6.5 ( 23  273) 0.25 4.20 = = 1099 K or 826C = 830C (to 2 significant figures) Q22. 0.25 mol of nitrogen is placed in a flask of volume 5.0 L at a temperature of 5°C. What is the pressure in the flask? A22. The general gas equation needs to be used: PV = nRT. Temperature should be in kelvin, pressure in kPa and volume in L. R = 8.31 J K–1 mol–1. nRT P = V 0.25 8.31(5  273) 5.0 = = 116 kPa Q23. Calculate the mass of helium in a balloon if the volume is 100 L at a pressure of 95 000 Pa and a temperature of 0°C. A23. The general gas equation needs to be used: PV = nRT. Temperature should be in kelvin, pressure in kPa and volume in L. R = 8.31 J K–1 mol–1. m The second step involves use of n = M . The molar mass of helium is 4.0 g mol–1. PV n(He) = RT 95 100 = 8.31(273 0) = 4.19 mol m(He) = 4.19 × 4.0 = 16.8 g Q24. What volume of gas, in litres, is occupied by: a 0.20 mol of hydrogen at 115 kPa and 40°C? b 12.5 mol of carbon dioxide at 5 atm and 150°C? Heinemann Chemistry 1 (4th edition)  Reed International Books Australia Pty Ltd

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c

11

8.50 g of hydrogen sulfide (H2S) at 100 kPa and 27°C?

A24. The general gas equation needs to be used: PV = nRT. Temperature should be in kelvin, pressure in kPa and volume in L. R = 8.31 J K–1 mol–1. nRT a V = P 0.20 8.31(40  273) 115 = = 4.5 L nRT b V = P 12.5 8.31 (150  273) 5 101.325 = = 87 L m c Since n = M and the molar mass of hydrogen sulfide is 34.1 g mol–1 nRT 8.50 8.31( 27  273) 100 V = P = 34.1 × = 6.23 L Q25. At a given temperature, a sample of nitrogen, of mass 11.3 g, exerts a pressure of 102 kPa in a gas cylinder of volume 10.0 L. Calculate the temperature of the gas. A25. The general gas equation needs to be used: PV = nRT. Temperature should be in kelvin, pressure in kPa and volume in L. R = 8.31 J K–1 mol–1. m Since n = M and the molar mass of nitrogen is 28.0 mol–1, PV T = nR

102 × 10.0 × 28.0 = 11.3 × 8.31 = 304 K = 31.1C Q26. Which sample of gas will contain the greater amount (mol) of gas: 3.2 L of nitrogen at 25°C and a pressure of 1.2 bar or 2.5 L of helium at 23°C and a pressure of 1.2 atm? A26. The general gas equation needs to be used: PV = nRT. Temperature should be in kelvin, pressure in kPa and volume in L. R = 8.31 J K–1 mol–1. These conversions need to be applied as needed: 1.00 atm = 101.325 kPa = 1.01 bar.

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PV n(N2) = RT 1203.2 = 8.31 ( 25  273) = 0.155 mol PV n(He) = RT (1.2 101) 2.5 = 8.31 (23  273) = 0.123 mol  there is a greater amount of nitrogen. Q27. Magnesium reacts with hydrochloric acid according to the equation: Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g) Calculate the volume of hydrogen produced at SLC when the following masses of magnesium react with excess hydrochloric acid: a 2.4 g b 0.54 g c 0.0018 g d 1.6 g A27. The balanced equation shows that 1 mol of magnesium produces 1 mol of hydrogen gas. Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g) m Remember that n = M . (The molar mass of magnesium is 24.3 g mol–1.) Remember that under SLC, 1 mol of any gas has a volume of 24.5 L.

V V Use the formula n = m rearranged as V = n × Vm. a

b

c

d

2.4 n(Mg) = 24.3 = 0.10 mol 1 n(H2)/n(Mg) = 1  V(H2) = 24.5 × 0.10 L = 2.45 L 0.54 n(Mg) = 24.3 = 0.022 mol  V(H2) = 24.5 × 0.022 L = 0.54 L 0.0018 n(Mg) = 24.3 = 0.000 074 mol  V(H2) = 24.5 × 0.000 074 L = 0.0018 L or 1.8 mL 1.6 n(Mg) = 24.3 = 0.0658 mol  V(H2) = 24.5 × 0.0658 L = 1.61 L

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Q28. Propane (C3H8) burns in oxygen according to the equation: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) Calculate: i the volume of oxygen at SLC used ii the volume of carbon dioxide at SLC produced when the following masses of propane react completely with excess oxygen: a 22 g b 5.0 g c 0.145 g d 16.5 g e 3.4 kg A28. The balanced equation shows that 1 mol of propane reacts with 5 mol of oxygen, and produces 3 mol of carbon dioxide. C3H8(g) + 5O2(g)  3CO2 (g) + 4H2O(g) m M Remember that n = . (The molar mass of propane is 44 g mol–1.) Remember that under SLC, 1 mol of any gas has a volume of 24.5 L.

V V Use the formula n = m rearranged as V = n × Vm. a

i

ii

b

i ii

c

i ii

d

i ii

e

i

22 n(C3H8) = 44 = 0.50 mol n ( O2 ) 5 = n ( C3 H8) 1  V(O2) = 5 × 0.50 × 24.5 = 61 L n ( CO2 ) 3 = n ( C3 H8) 1  V(CO2) = 3 × 0.50 × 24.5 = 37 L 5.0 n(C3H8) = 44 = 0.114 ...