1. Chapter 5 Gases

2. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–2 QUESTION Four bicycle tires are inflated to the following pressures. Which one has the highest pressure? Tire A 3.42 atm; Tire B 48 lbs/sq in; Tire C 305 kPa; Tire D 1520 mmHg. (Recall; 1.00 atm = 760 mmHg = 14.7 lb/sq in = 101.3 kPa) 1.Tire A 2.Tire B 3.Tire C 4.Tire D

3. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–3 ANSWER Choice 1 Even though it has the smallest number, it represents the highest pressure of the four. When all four are changed to a common label (use conversion factors found on page 181 and dimensional analysis) 3.42 atm is a higher pressure than the others. Section 5.1: Pressure

4. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–4 QUESTION Why is it critical that the temperature be held constant when applying Boyle’s law to changing the pressure of a trapped gas? 1.Gas molecules may expand at higher temperatures; this would change the volume. 2.Changing the temperature causes the gas to behave in non-ideal fashion. 3.Changing the temperature affects the average particle speed, which could affect the pressure. 4.Allowing the temperature to drop below 0°C would cause the trapped gas to no longer follow Boyle’s Law.

5. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–5 ANSWER Choice 3 provides the connection between temperature and pressure that would introduce another variable when studying the pressure- volume relationship. Boyle’s law shows that if the molecules of a trapped gas maintain the same average velocity when their volume is changed, the pressure will be inversely related to the volume change. Section 5.2: The Gas Laws of Boyle, Charles, and Avogadro

6. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–6 QUESTION After examining this figure, which of the following statements offers accurate, consistent statements about the gases shown at constant pressure?

7. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–7 QUESTION (continued) 1.At –273°C all gases occupy nearly the same volume; the different slopes are because of differences in molar masses. 2.At zero Celsius the gases have different volumes because the larger the molecule, the larger the volume. 3.Since the pressure is constant, the only difference that could cause the different slopes is in the number of moles of gas present in each sample. For example, the N 2 O sample must have the fewest moles. 4.Although it does appear to do so, the volumes do not reach zero but if the graph used K instead of °C the volume would reach zero for all the gases.

8. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–8 ANSWER Choice 3 is the correct conclusion. If all the gases had the same number of moles in their samples, the volume should be the same for all four at any temperature (with constant pressure). This is reflected in Avogadro’s law. Section 5.2: The Gas Laws of Boyle, Charles, and Avogadro

9. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–9 QUESTION Each of the balloons hold 1.0 L of different gases. All four are at 25°C and each contains the same number of molecules. Of the following which would also have to be the same for each balloon? (obviously not their color) 1.Their density 2.Their mass 3.Their atomic numbers 4.Their pressure

10. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–10 ANSWER Choice 4 is consistent with Avogadro’s Law. The temperature, pressure, number of moles, and volume are related for a sample of trapped gas. If two samples have three variables out of the four the same, the fourth variable must be the same as well. Section 5.2: The Gas Laws of Boyle, Charles, and Avogadro

11. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–11 QUESTION If a 10.0 L sample of a gas at 25°C suddenly had its volume doubled, without changing its temperature what would happen to its pressure? What could be done to keep the pressure constant without changing the temperature? 1.The pressure would double; nothing else could be done to prevent this. 2.The pressure would double; the moles of gas could be doubled. 3.The pressure would decrease by a factor of two; the moles of gas could be halved. 4.The pressure would decrease by a factor of two; the moles could be doubled.

12. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–12 ANSWER Choice 4 describes two opposing changes. When the volume increases, the pressure of a trapped gas will decrease (at constant temperature and constant moles of gas). However, if the pressure drops, more collisions could be restored by adding more particles of gas in the same ratio as the pressure decline. Section 5.2: The Gas Laws of Boyle, Charles, and Avogadro

13. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–13 QUESTION The typical capacity for human lungs is approximately 5,800 mL. At a temperature of 37°C (average body temperature) and pressure of 0.98 atm, how many moles of air do we carry inside our lungs? 1.1.9 mol 2.0.22 mol 3.230 mol 4.No one wants moles in their lungs.

14. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–14 ANSWER Choice 2 is correct. The units for temperature must be in K, pressure in atm, and volume in L. Then using the universal constant 0.08206 L atm/ K mol in the PV = nRT equation moles may be solved. Section 5.3: The Ideal Gas Law

15. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–15 QUESTION The aroma of fresh raspberries can be attributed, at least in part, to 3-(para-hydroxyphenyl)-2-butanone. What is the molar mass of this pleasant smelling compound if at 1.00 atmosphere of pressure and 25.0°C, 0.0820 grams has a volume of 12.2 mL? 1.13.8 g/mol 2.164 g/mol 3.40.9 g/mol 4.224 g/mol

16. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–16 ANSWER Choice 2 is correct. Using PV = nRT and properly substituting 0.0122 L for V, 298 K for T, using 0.08206 for R and solving for n the number of moles represented by 0.0820 grams can be obtained. Then the grams in one mole can be found. Section 5.3: The Ideal Gas Law

17. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–17 QUESTION If a person exhaled 125 mL of CO 2 gas at 37.0°C and 0.950 atm of pressure, what would this volume be at a colder temperature of 10.0°C and 0.900 atm of pressure? 1.3.12 mL 2.0.130 L 3.0.120 L 4.22.4 L

18. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–18 ANSWER Choice 3 correctly accounts for the changing conditions. The combination of Charles and Boyle’s laws provides the mathematical basis for the calculation. After converting the temperature to K units, the equation P 1 V 1 /T 1 = P 2 V 2 /T 2 may be used by solving for V 2. Section 5.3: The Ideal Gas Law

19. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–19 QUESTION Under STP conditions what is the density of O 2 gas? 1.Not enough information is given to solve this. 2.1.31 g/L 3.1.43 g/L 4.0.999 g/L

20. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–20 ANSWER Choice 3 is the correct density for O 2 at STP. At STP conditions the volume of one mole of a gas is approximately 22.4 L. One mole of oxygen = 32.0 grams. The density is calculated by dividing mass by volume. Section 5.4: Gas Stoichiometry

21. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–21 QUESTION The primary way your body produces exhaled CO 2 is by the combustion of glucose, C 6 H 12 O 6 (molar mass = 180. g/mol.). The balanced equation is shown here: C 6 H 12 O 6 (aq) + 6 O 2 (g)  6 CO 2 (g) + 6 H 2 O (l) If you oxidized 5.42 grams of C 6 H 12 O 6 while tying your boots to climb Mt. Everest, how many liters of O 2 did you use (assume STP conditions)? 1.0.737 L 2.0.672 L 3.4.05 L 4.22.4 L

22. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–22 ANSWER Choice 3, assuming you were not holding your breath, is correct. The number of moles of glucose must first be determined (5.42/180. = 0.0301 moles), then this is multiplied by 6 to account for the stoichiometric ratio between glucose and oxygen. From this, PV = nRT may be used with the appropriate substitutions. Section 5.4: Gas Stoichiometry

23. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–23 QUESTION Freon-12 had been widely used as a refrigerant in air conditioning systems. However, it has been shown to be related to destroying Earth’s important ozone layer. What is the molar mass of Freon-12 if 9.27 grams was collected by water displacement, in a 2.00 liter volume at 30.0°C and 764 mmHg. Water’s vapor pressure at this temperature is approximately 31.8 mmHg. 1.120. g/mol 2.12.0 g/mol 3.115 g/mol 4.92.7 g/mol

24. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–24 ANSWER Choice 1 is the molar mass of Freon-12. The pressure must be corrected for the presence of water by subtracting 31.8 mmHg from the total pressure. This should also be converted to atm. The temperature must be converted to K. Then PV = nRT can be used if n is written as g/mm and solved for mm. Section 5.5: Dalton’s Law of Partial Pressures

25. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–25 QUESTION Radon (element 86) is a radioactive noble gas that can be found in some homes. In those situations radon can diffuse in the air and end up inhaled by the occupants. What is the ratio of the velocity of radon gas compared with O 2 in the same room at the same conditions? 1.2.6 times faster 2.0.38 times as fast 3.0.14 times as fast 4.6.9 times faster

26. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–26 ANSWER Choice 2 shows the approximate ratio for the more massive radon velocity compared to the lighter oxygen molecules. This is an application of Graham’s law. Velocity of Rn/Velocity of O 2 = (mm O 2 ).5 /(Rn).5 Section 5.7: Effusion and Diffusion

27. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–27 QUESTION If the mole fraction of O 2 in our atmosphere at standard conditions is approximately 0.209, what is the partial pressure of the oxygen in every breath you take? 1.1.00 atm 2.4.78 atm 3.159 torr 4.3640 mmHg

28. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–28 ANSWER Choice 3 shows the correct pressure. This calculation is based on the application of Dalton’s law of partial pressures. For one atmosphere of pressure 0.209 is caused by oxygen so 760 .209 = 159 torr. Section 5.5: Dalton’s Law of Partial Pressures

29. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–29 QUESTION Kinetic molecular theory helps explain the definition of temperature based on molecular motion. Which statement describes an important aspect of this connection? 1.Temperature is inversely related to the kinetic energy of the gas particles. 2.At the same temperature, more massive gas particles will be moving faster than less massive gas particles. 3.As the temperature of a gas sample increases, the average velocity of the gas particles increases. 4.Kinetic energy is directly related to temperature. This is valid for any units of temperature.

30. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–30 ANSWER Choice 3 states a correct connection between temperature and velocity. The important relationship is summarized in the following equation: ½  mass  velocity 2 = 3/2 RT (T must be in K). Section 5.6: The Kinetic Molecular Theory of Gases

31. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–31 QUESTION As the temperature of a gas increases, which statement best correlates to information about molecular velocity? 1.The average molecular velocity will increase, but the distribution of molecular velocities will stay the same. 2.The average molecular velocity will stay the same, but the molecular velocity distribution will spread. 3.The average molecular velocity will increase, and the distribution of the molecular velocities will spread. 4.The average molecular velocity will stay the same, and the distribution of the molecular velocities will stay the same.

32. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–32 ANSWER Choice 3 accurately reflects the connection between molecular velocity and an increase in temperature. Figure 5.21 on page 206 visually shows the relationship. Kelvin temperature is directly related to molecular velocity - with greater temperature the distribution of available velocities also increases. Section 5.6: The Kinetic Molecular Theory of Gases

33. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–33 QUESTION Real gases exhibit their most “ideal” behavior at which relative conditions? 1.Low temperatures and low pressures 2.High temperatures and high pressures 3.High temperatures and low pressures 4.Low temperatures and high pressures

34. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–34 ANSWER Choice 3 provides the correct choice. At those conditions gas molecules are farthest apart and exert their least influence on each other thereby permitting their behavior to follow mathematical predictions. Section 5.8: Real Gases