chapter 5 gases. copyright © houghton mifflin company. all rights reserved.crs question, 5–2...

Chapter 5

GasesGases

Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 5–2

QUESTIONFour 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 A2. Tire B3. Tire C4. Tire D


ANSWERChoice 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


QUESTIONWhy 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 wouldchange the volume.

2. Changing the temperature causes the gas to behave in non-idealfashion.

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 thetrapped gas to no longer follow Boyle’s Law.


ANSWERChoice 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


QUESTIONAfter examining this figure, which of the following statements offers accurate, consistent statements about the gases shown at constant pressure?


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 N2O 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.


ANSWERChoice 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.



QUESTIONEach 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 density2. Their mass3. Their atomic numbers4. Their pressure


ANSWERChoice 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.



QUESTIONIf 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 toprevent 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.


ANSWERChoice 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.



QUESTIONThe 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 mol2. 0.22 mol3. 230 mol4. No one wants moles in their lungs.


ANSWERChoice 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


QUESTIONThe 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/mol2. 164 g/mol3. 40.9 g/mol4. 224 g/mol


ANSWERChoice 2 is correct. Using PV = nRT and properly substituting0.0122 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.



QUESTIONIf a person exhaled 125 mL of CO2 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 mL2. 0.130 L3. 0.120 L4. 22.4 L


ANSWERChoice 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 P1V1/T1 = P2V2/T2 may be used by solving for V2.



QUESTIONUnder STP conditions what is the density of O2 gas?

1. Not enough information is given to solve this.2. 1.31 g/L3. 1.43 g/L4. 0.999 g/L


ANSWERChoice 3 is the correct density for O2 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


QUESTIONThe primary way your body produces exhaled CO2 is by the

combustion of glucose, C6H12O6 (molar mass = 180. g/mol.). The

balanced equation is shown here:

C6H12O6 (aq) + 6 O2 (g) 6 CO2 (g) + 6 H2O (l)

If you oxidized 5.42 grams of C6H12O6 while tying your boots to

climb Mt. Everest, how many liters of O2 did you use (assume STP

conditions)?

1. 0.737 L2. 0.672 L3. 4.05 L4. 22.4 L


ANSWERChoice 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


QUESTIONFreon-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/mol2. 12.0 g/mol3. 115 g/mol4. 92.7 g/mol


ANSWERChoice 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


QUESTIONRadon (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 O2 in the same room at the same

conditions?

1. 2.6 times faster2. 0.38 times as fast3. 0.14 times as fast4. 6.9 times faster


ANSWERChoice 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 O2 = (mm O2).5/(Rn).5

Section 5.7: Effusion and Diffusion


QUESTIONIf the mole fraction of O2 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 atm2. 4.78 atm3. 159 torr4. 3640 mmHg


ANSWERChoice 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


QUESTIONKinetic 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 thegas particles.

2. At the same temperature, more massive gas particles will bemoving 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 validfor any units of temperature.


ANSWERChoice 3 states a correct connection between temperature and velocity. The important relationship is summarized in the following equation: ½ mass velocity2 = 3/2 RT (T must be in K).

Section 5.6: The Kinetic Molecular Theory of Gases


QUESTIONAs the temperature of a gas increases, which statement best correlates to information about molecular velocity?

1. The average molecular velocity will increase, but thedistribution of molecular velocities will stay the same.

2. The average molecular velocity will stay the same, but themolecular velocity distribution will spread.

3. The average molecular velocity will increase, and thedistribution of the molecular velocities will spread.

4. The average molecular velocity will stay the same, and thedistribution of the molecular velocities will stay the same.


ANSWERChoice 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


QUESTIONReal gases exhibit their most “ideal” behavior at which relative conditions?

1. Low temperatures and low pressures2. High temperatures and high pressures3. High temperatures and low pressures4. Low temperatures and high pressures


ANSWERChoice 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

chapter 5 gases. copyright © houghton mifflin company. all rights reserved.crs question, 5–2...

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