Chapter 9 The Gaseous State

Chapter 9 The Gaseous State 9.1 (a) effusion; (b) Boyle s law; (c) combined gas law; (d) ideal gas; (e) pressure; (f) Dalton s law of partial pressures; (g) molar volume; (h) ideal gas constant, R 9. (a) diffusion; (b) Charles s law; (c) Gay-Lussac s law of combining volumes; (d) ideal gas law; (e) barometer; (f) kinetic-molecular theory; (g) standard temperature and pressure (STP); (h) Avogadro s hypothesis 9.3 The y-axis units and label are missing (in both the graph and the data tables), x-axis units are missing, the points should not be connected by lines (a straight best fit line that comes closest to all points should be used), gridlines should be uniformly labeled (for example, 10, 0, 30, 40, 50, etc.), the x-axis maximum should be 60 instead of 10, a graph title is missing. 9.4 Each graph axis should have labels and units. In addition rather than connecting the dots, scientific graphs use trend lines which show the trend of the data. Notice the trend line is a best-fit line and does not necessarily touch any of the data points. A title should also be included. 9 1

9.5 There is a trend, but it is not linear because the data clearly are curved in an upward direction. 9.6 After graphing the data draw a best-fit line through the data points. Determining the equation that fits the line requires determining the variables m and b in the equation for a straight line: y mx + b You can do this by using two points on the best fit line. By taking the points from the best-fit line you are getting the equation that best fits all the data. 100ºC 67 g/100 ml 40ºC 34 g/100 ml Next solve the set of equations: 67 m(100) + b 34 m(40) + b m 0.55 g/(ºc 100 ml) b 1 g/100 ml Equation: y 0.55x + 1 9.7 The values for h represent the pressure applied to the gas in excess of. The length of the gas column is proportional to the volume of the gas. When the pressure is high, the volume is small. As the pressure decreases, the volume increases. Gas pressure and volume are inversely related. 9

9.8 Pressure and volume are inversely related. Because the slope is constant we can state that 1 Pressure ( h) k or P k where k is a constant. 9.9 You can estimate the vapor pressure by drawing a line vertically from the temperature and seeing where the intersection point lies on the vapor pressure axis (horizontal line). ºC 1 torr 38ºC 50 torr 9 3

9.10 There appears to be a linear trend in the data. At 0.9 g/ml the velocity is approximately 1360 m/s. 9.11 (a) 5x + 1 3; subtract 1 from both sides 5x ; divide both sides by five x 5 0.4 (b) 0.41 / x.00 ; multiply both sides by x 0.41.00 x ; divide both sides by x 0.41.00 0.06 (c) x 3 x ; add x to both sides x 3 ; divide both sides by x 16 ; take the square root of both sides x x 16 ±4 (d) x 6 x; add x to both sides 3x 6; divide both sides by 3 x 6 3 9.1 (a) 14x + 16 44 ; subtract 16 from both sides 14x 8 ; divide both sides by 14 x 8 14.0 (b) 1 x + 3 5; subtract 3 from both sides 3 1 x ; multiply both sides by 3 3 x 3 6 (c) 1x 4( 1 x) ; divide both sides by four 9 4

3x 1 x; add x to both sides 4x 1 ; divide both sides by 4 x 1 4 3 (d) 3x+ 15 4x+ 1 ; subtract 3x from both sides 15 x + 1 ; subtract 1 from both sides x 15 1 3 9.13 For temperature conversions we use the equation: T F 1.8T C + 3 To convert from Fahrenheit to Celsius we rearrange the equation: ( T F 3) T C 1.8 It is worth noting that 1.8 and 3 are exact in this equation and that the significant figures are determined by the precision of the temperature you are converting: ( 1 3) 180. (a) T C 100. C 1.8 1.8 ( 80.0 3) 48.0 (b) T C 6.7 C 1.8 1.8 ( 3.0 3) 0.0 (c) T C 0.0 C 1.8 1.8 ( 40.0 3) 7.0 (d) T C 40.0 C 1.8 1.8 9.14 For Celsius to Fahrenheit temperature conversions we use the equation: T F 1.8T C + 3 It is worth noting that 1.8 and 3 are exact in this equation and that the significant figures are determined by the precision of the temperature you are converting: T 1.8 37.0 + 3 66.6 + 3 98.6 C (a) F ( ) (b) F ( ) (c) F ( ) (d) ( ) T 1.8 1 + 3 381.6 + 3 414 C T 1.8 100.0 + 3 180.0 + 3 1.0 C T 1.8 40.0 + 3 7.0 + 3 40.0 C F 9.15 Solve the equation algebraically and then calculate for the unknown value. (a) (b) (c) (d) ( 1.00)( 1.00) 4.4 (.500)( 0.0806) ( 0.750)( 3.00) 0.116 ( 0.0806)( 37) ( 1.50)( 0.0806)( 455) ( 3.5) (.67)( 0.0806)( 3) ( 15.0) P T nr P n RT nrt 17. P nrt P 4.70 9 5

9.16 Solve the equation algebraically and then calculate for the unknown value. (a) (b) (c) (d) ( 3.55)( 1.75) 369 (.05)( 0.0806) ( 1.00)(.5) ( 0.0806)( 98) ( 3.00)( 0.0806)( 535) ( 0.15) ( 1.57)( 0.0806)( 343) ( 6.5) P T nr P n 0.90 RT nrt 1050 P nrt P 7.07 9.17 When compared to other states of matter, gases have low densities and are very compressible. They also take the shapes of their containers. We describe a gas in terms of its pressure, temperature, volume, and the number of moles of the gas in the sample. 9.18 Unlike liquids and solids, gases take the shapes of and fill their containers. Atoms and molecules in the gas phase are very much more mobile than they are in the solid or liquid phases. Compared to solids and liquids, gases have low densities and are very compressible. 9.19 The density of warm air is lower than the density of cool air. 9.0 Warm air rises because it is less dense than cool air. 9.1 As temperature decreases, gas molecules move more slowly. As a result, they collide with less force, and a larger number of molecules occupy a unit volume. As shown in the figure, at lower temperatures the gas density is higher, but the molecular velocity is lower. 9. (a) When a gas is heated, either the pressure or volume (or both) can change. If the volume changes, there should be fewer molecules in the image, but the piston should have moved upward. The gas molecules should also be moving faster as a consequence of the temperature increase. (b) The piston should have moved up to show the expansion of the gas. The image incorrectly shows that the molecules increase in size. When the volume of gas increases, the molecules do not increase in size. (c) The sample density should decrease (because the volume increased), but the size of molecules does not change. (d) Although the piston moved up, the density of molecules did not go down as it should have. 9.3 Gas pressure is the amount of force exerted by the gas particles divided by the area over which the force is force exerted. P area 9.4 When a gas molecule strikes a wall, the wall experiences a force. When we add together the forces of all the molecules striking the wall and divide by the area of the wall, we determine the pressure of the gas. 9 6

9.5 We measure absolute pressure with a barometer (Figure 9.11). Essentially, a barometer allows us to compare the pressure exerted by a column of a liquid (usually mercury) to the pressure exerted by the atmosphere. A tire gauge compares the pressure of the air inside a tire to atmospheric pressure. 9.6 (a) We commonly use many different units for measuring pressure. These would include atmospheres, torr, pascals, kilopascals, pounds per square inch, and inches, centimeters and millimeters of mercury. (b) Here are several different pressure units and their relationships to osphere (atm). 9.9 in Hg 76 cm Hg 760 mm Hg 101,35 Pascal 101.35 kpa 14.7 lb/in (also known as pounds per square inch, or psi) Because each of these values is equal to, they are also equal to each other. When we convert from inches of mercury to torr, for example, we can use the relationship 9.9 in Hg. 9.7 If the temperature doesn t change, the velocity of the gas molecules remains constant. Because the volume is larger, the density of molecules is lower. 9.8 (a) If the volume of a gas sample increases, the density of gas molecules should decrease because we are not adding any gas molecules so those that are present can spread out over a larger volume. In this image, the density of particles is the same (which is incorrect) and the molecules increase in size (which is also incorrect). When a gas expands, the space between the molecules increases but the molecules do not change size. (b) The density of the particles did not change; it should have decreased. (c) The density appears to have changed, but the piston did not move. In order for the volume to increase, the piston must move up (as is shown in figures (a) and (b)). (d) The piston should have moved up and the gas molecules should have remained the same size. 9.9 We use the following relationships to convert between the various pressure units: 9.9 in Hg 76 cm Hg 760 mm Hg 101,35 Pa 101.35 kpa 14.7 lb/in For example, to convert between pascals and mm Hg, use the relationship: 760 mm Hg 101,35 Pa. (a) Pressure in atm 745 torr 0.980 atm (b) Pressure in torr 1.3 atm 935 torr (c) Pressure in atm 90.1 mm Hg 0.119 atm 760 mm Hg 1000 Pa (d) Pressure in Pa 0.643 kpa 643 Pa 1 kpa 9 7

(e) Pressure in mm Hg (f) Pressure in torr 5 1.35 10 Pa 4 7.51 10 Pa 760 mm Hg 1.01 10 3 mm Hg 101,35 Pa 563 torr; 101,35 Pa 101,35 Pa (g) Pressure in Pa 798 torr 1.06 10 5 Pa; 10 mm (h) Pressure in mm Hg 9.3 cm Hg 93 mm Hg 1 cm 9.30 We use the following relationships to convert between the various pressure units: 9.9 in Hg 76 cm Hg 760 mm Hg 101,35 Pa 101.35 kpa 14.7 lb/in For example, to convert between pascals and millimeters of mercury, use the relationship: 760 mm Hg 101,35 Pa. (a) Pressure in torr 1.15 atm 874 torr (b) Pressure in atm 968 torr 1.7 atm 5 (c) Pressure in atm.50 10 Pa.47 atm 101,35 Pa (d) Pressure in torr 695 mm Hg 695 torr 760 mm Hg Note that the units torr and mm Hg are interchangeable: 1 torr 1 mm Hg. 101,35 Pa (e) Pressure in Pa 0.953 atm 9.66 10 4 Pa 760 mm Hg (f) Pressure in mm Hg 653 torr 653 mm Hg 101,35 Pa (g) Pressure in Pa 1545 mm Hg.060 10 5 Pa 760 mm Hg (h) Pressure in kpa 3.73 kpa 3.68 10 atm 101.35 kpa 9.31 The conversion from inches of mercury to pascals is quite lengthy. The first half of the conversion involves converting from inches to millimeters (English to metric conversion). The second half involves using the relationships among pressure units to convert from millimeters of mercury to pascals. The problem solving map looks like:.54 cm 1 in 10 mm 1 cm Pressure in Hg cm Hg mm Hg 760 mm Hg 101,35 Pa mm Hg atm Pa Pressure in Pa 30.4 in.54 cm 1 in 10 mm 1 cm 101,35 Pa 1.04 10 5 Pa 760 mm An alternate, but important, conversion method involves using the definitions of pressure given in the text. We know that 9.9 in Hg and also that 101,35 Pa. This means that 9.9 in Hg 101,35 Pa. Using this factor, we can convert from inches to pascals in one step (although with less precision): 9 8

Pressure in Pa 30.4 in Hg 101,35 Pa 1.0 10 5 Pa 9.9 in Hg 9.3 The conversion from kilopascals to inches of mercury is quite lengthy. The first half of the conversion is converting from kilopascals (kpa) to millimeters of mercury (mm Hg) (pressure conversions). The second half involves converting from millimeters of mercury to inches (English to metric conversion). The problem solving map looks like: 1000 Pa 1kPa 101, 35 Pa 760 mm Hg 10 mm 1 cm kpa Pa atm mm Hg.54 cm 1 in cm Hg in Hg P in Hg 101.19 kpa 1000 Pa 760 mm Hg 1 kpa 101,35 Pa 1 cm Hg 1 in Hg 9.881 in Hg 10 mm Hg.54 cm Hg The conversion factor.54 cm/in is exact. An alternate, but important, conversion method involves using the definitions of pressure given in the text. We know that 9.9 in Hg and also that 101,35 Pa. This means that 9.9 in Hg 101,35 Pa. To convert to kilopascals, we use the definition: 1 kpa 1000 Pa: Pressure in Hg 101.19 kpa 1000 Pa 9.9 in Hg 9.9 in Hg 1 kpa 101,35 Pa 9.33 Boyle s law tells us that, at constant temperature, as the pressure of a gas increases the volume decreases. This is an inverse relationship, so that if the pressure increases by a factor of three, the volume decreases to one third (1/3) of its original volume, assuming the temperature remains constant. 9.34 Boyle s law is an inverse relationship: As the volume of a gas sample increases, the pressure of the gas decreases, assuming that the temperature is kept constant. When the volume of a gas increases by four times its original volume, its pressure decreases to one fourth its original pressure (at constant temperature). Mathematically, we can write Boyle s law as P 1 1 P. 9.35 If the container volume decreases, the particles will collide with the container walls more frequently, causing an increase in pressure. If the container size is decreased to half its original volume, and the sample size remains constant, we expect to find twice as many molecules in the same space. The velocity of the molecules will not change as long as the temperature remains constant, so the pressure on the container walls will double. 9.36 If the volume of the container is increased, the particles will collide with the container wall less frequently causing a decrease in pressure. If the container volume is tripled, the pressure of the gas molecules against the container walls decreases to one third of its original pressure. This is because, at constant temperature, the velocity of the molecules does not change, but the frequency of their collisions with the container walls decreases. Because there are 8 gas molecules in the original picture there should be about 8 1/3 /3 molecules in the new picture. 9 9

9.37 (a) 3.60 L; (b) 6.7 ml; (c) 0.39 ml. Boyle s law describes the relationship of pressure and volume changes (P & ) on a gas sample, assuming that the sample size (number of moles) and temperature are kept constant. Boyle s law is an inverse relationship: As the volume of a gas sample increases, the pressure of the gas decreases, assuming that the temperature is kept constant. For example, in part (a) because the pressure increases by a factor of 5.00 atm/.00 atm (units must be the same), we conclude that the volume will decrease by a factor of.00 atm/5.00 atm. We can compute the final volume as follows: Final volume 3.00atm 6.00L 5.00atm 3.60 L Boyle s law also states that P 1 1 P. For part (a) we have: 1 P 1 P 6.00 L 3.00 atm 5.00 atm? We can rearrange Boyle s law and solve for as follows: 1 1 P P (3.00 atm )(6.00L) 5.00 atm 3.60 L By either applying the proportionality, or using Boyle s law, we obtain the same result. After you complete your calculation, you should make sure to evaluate your answer to ensure that it is reasonable. The effect of the pressure increase should be a volume decrease. (b) The pressure increases, so we should see a volume decrease, and we do. (60.0 torr )(40 ml) 90.0 torr 6.7 ml (c) The pressure increases, so we should see a volume decrease, and we do. (40.0 torr )(.50mL) 55 torr 0.39 ml 9.38 Boyle s law describes the relationship of pressure and volume changes (P and ) on a gas sample, assuming that the sample size (number of moles) and temperature are kept constant. Boyle s law is an inverse relationship: As the volume of a gas sample increases, the pressure of the gas decreases, assuming that the temperature is kept constant. Before we apply Boyle s law, we must make certain that the pressure units are the same. Mathematically, we can write Boyle s law as P 1 1 P. For these problems, we are solving for : 1 1 P P (a) Using Boyle's law, we get the new volume: 9 10

(.50 atm )(.50L) 75 atm 0.0086 L Because the pressure increases, the volume should decrease, and it does. (b) With a decrease in pressure we expect an increase in the volume,. (85 torr )(6.5L) 456 torr 11.3 L (c) (50.0 torr )(450mL) 30.0 torr 7.5 10 ml 9.39 According to Boyle's law, if the volume of a gas sample increases by a factor of 151 ml/405 ml, its pressure decreases by a factor of 05 ml/151 ml. For part (a) we can write: P 405 ml 151 ml 60 torr 161 torr We can obtain the same results using the mathematical expression of Boyle's law, P 1 1 P, solving for P : 1 1 P P Note: When we solve these problems, the volume units must agree so that they cancel properly. (a) P (60 torr)(405 ml ) 151 ml 161 torr Because the sample volume increases, we expect that the pressure will decrease, and it does. (b) Before we can use Boyle's law, we must convert 1.50 L into ml. 1000 ml olume in ml 1.50 L 1.50 10 3 ml 1L P 3 (0.00100 torr)(1.50 10 ml ) 15.0 ml 0.100 torr Because the volume decreases, we expect the pressure to increase, and it does. (c) P (0.83 atm)(805 L ) 37.5 L 17.9 atm Because the volume decreases, we expect pressure to increase, and it does. 9.40 According to Boyle's law, a decrease in the volume of a gas results in an increase in pressure. Mathematically, we express Boyle's law as P 1 1 P. We are looking for the final pressure, so we solve the equation for P : P 1 1 P If the units of 1 and are the same, we need not worry about unit conversions. (a) Prior to using Boyle's law, we must convert 155 ml to L so that the volume units agree. 9 11

olume in L 155 ml 1L 1000 ml 0.155 L P (845 torr)(0.155 L ) 1.55 L 84.5 torr We see that the calculated pressure is lower than the initial pressure as a result of an increase in volume. This is in agreement with the inverse relationship between pressure and volume shown by Boyle s law. (b) P (5.30 atm)(.85 L ) 4.50 L 3.36 atm Because the volume increases, the pressure decreases as predicted by Boyle s law. (c) Prior to using Boyle's law, we must convert 5500 ml to L so that the volume units agree. olume in L 5500 ml 1L 1000 ml 5.5 L P (755 torr)(.00 L ) 5.5 L 75 torr Because the volume increases, the pressure decreases as predicted by Boyle s law. 9.41 It is helpful to organize the data in table form. P 1 1 P 1.5 atm 95 L? 6.35 L Based on Boyle's law, we predict that if the volume of the gas sample decreases by a factor of 6.35 L/95 L, the pressure should increase by a factor of 95 L/6.35 L, assuming the sample size and temperature are kept constant. Solving Boyle's law for P we have: P 1 1 P P 1.5atm 95 L 6.35 L 18 atm As we predicted, the pressure is higher by a factor of 95 L/6.35 L. 9.4 It is helpful to organize the data in table form. P 1 1 P 0.945 atm 186 ml 1.76 atm? Based on Boyle's law, we predict that if the pressure of a gas sample increases by a factor of 1.76 atm/0.945 atm, its volume will decrease by a factor of 0.945 atm/1.76 atm, assuming that the sample size and temperature are kept constant. Solving Boyle's law for we have: 1 1 P P (0.945 atm )(186 ml) 1.76 atm 99.9 ml 9 1

As we predicted, the volume decreased by a factor of 0.945 atm/1.76 atm. 9.43 Organize the data in table form. P 1 1 P 1.00 atm 0.550 L 75 torr? Notice that the units of pressure for P 1 and P differ. Because, we can substitute 1.00 atm with in the Boyle's law expression: 1 1 P P ( )(0.550 L) 75 torr 0.577 L The volume of H required is larger than 0.550 L because it is being collected at a lower pressure. 9.44 Organize the data in table form. P 1 1 P.00 atm.50 L 735 torr? Notice that the units of pressure for P 1 and P differ. We must convert 735 torr to atm before using Boyle's law. Pressure in atm 735 torr 0.967 atm 1 1 P P (.00 atm )(.50 L) 0.967 atm 5.17 L The larger volume the gas occupies at the lower pressure is consistent with Boyle's law. 9.45 Charles s law states that if the pressure and sample size of a gas are kept constant, the volume and temperature (in kelvins) of the gas are directly proportional to each other (volume increases when temperature increases; volume decreases when temperature decreases). If the temperature increases, the gas will occupy a larger volume. 9.46 Charles s law indicates the direct proportionality between the absolute temperature of a gas and its volume, provided the pressure and sample size are kept constant. If the temperature increases, the volume of the gas will increase. If the temperature decreases, the volume of the gas will decrease. Mathematically, we can write Charles s law as: T T 1 1 9.47 The velocity of the gas particles increases as temperature increases. This means they strike the walls of the container with greater force. If the container volume does not increase, the gas pressure increases. To maintain a constant pressure, as stated in the problem, the volume of the container will increase. 9.48 The velocity of the gas particles decreases as the temperature decreases. This means that they strike the container walls with less force. If the container does not decrease in volume, the pressure decreases. To maintain a constant pressure, the volume of the container will decrease. 9.49 In Charles's law problems, we must always express temperature in kelvins. We can state Charles s law mathematically as: T T 1 1 9 13

1 Solving for : T T1 We see that if the temperature, T, increases, the volume, must also increase. (a) Both temperatures are given in Celsius. The conversion is: T K T C + 73.15 K. T 1 30.0 C + 73.15 303. K T 0.0 C + 73.15 73. K 6.00 L 73. K 5.41 L 303. K Because the temperature decreases, the volume also decreases. (b) Both temperatures are given in Celsius. The conversion is: T K T C + 73.15 K. T 1 60.0 C + 73.15 13.15 K (4 sig figs) T 401.0 C + 73.15 674.15 K (4 sig figs) 1 ml 674.15 K 671 ml 13.15 K Because temperature increases, the volume also increases. 47.5 L (c) 337 K 75.5 L 1 K Because the temperature increases, the volume also increases. 9.50 When we solve Charles's law problems, we must always express temperature in kelvins. We can state Charles s law mathematically as: T T 1 1 1 Solving for : T T1 We see that if the temperature, T, increases, the volume, must also increase. (a) Both temperatures are given in Celsius. The conversion is: T K T C + 73.15 K. T 1 0.0 C + 73.15 73. K T 100.0 C + 73.15 373. K 4 L 373. K 306 L 73. K Because the temperature increases, the volume also increases. (b) 15 ml 45 K 450 K 1.5 10 3 ml Because the temperature increases, the volume also increases. (c) Both temperatures are given in Celsius. The conversion is: T K T C + 73.15 K. 9 14

T 1 45 C + 73.15 318 K T 450 C + 73.15 7. 10 K 156 L 318 K 7. 10 K 3.5 10 ml Because the temperature increases, the volume also increases. 9.51 When we solve Charles's law problems, we must always express temperature in kelvins. We can state Charles s as: T T 1 1 1 Solving for T : T T 1 (a) Although we are asked to give the final temperature in Celsius, we must solve for T in kelvins and then express that temperature in degrees Celsius. T 1 0.0 C + 73.15 73. K T 140.0 ml 73. K 546 K 70.0 ml Because the volume doubles, the temperature also doubles. Convert T to Celsius: T C T K 73.15 K 546 K 73.15 73 C (b) The units for volume must be the same so they will cancel. First, we convert 85 ml to liters and 37 C to kelvins before we apply Charles s law: ll olume in liters 85 ml 0.085 L 1000 ml T 1 37 C + 73.15 36 K T 0.085 L 36 K 7.9 K.55 L Because the volume decreases, the temperature also decreases. T C T K 73.15 K 7.9 K 73.15 65. C (c) T 135 L 165 K 55 K 87.5 L T C T K 73.15 K 55 K 73.15 19 C 9.5 When we solve Charles's law problems, we must express the temperature in kelvins. We can state Charles s law mathematically as: T T 1 1 1 Solving for T : T T 1 (a) The units for volume must be the same so they will cancel. First, we convert 100.0 C to kelvins: 9 15

T 1 100.0 C + 73.15 373. K T 100.0 ml 373. K 149. K 50.0 ml Because the volume decreases, the temperature also decreases. T C T K 73.15 K 149. K 73.15 14.0 C (b) First, we convert 7.5 C to kelvins before we use Charles s law: T 1 7.5 C + 73.15 300.7 K T 148 ml 300.7 K 356 K 15 ml Because the volume increases, the temperature also increases. T C T K 73.15 K 356 K 73.15 83 C (c) T 57. L 300 K 1 10 3 K (to 1 significant figure). 13.7 L T C T K 73.15 K 1 10 3 73.15 1 10 3 C (to one significant figure) 9.53 It is often helpful to create a table like the one shown below and input the data from the problem. 1 T 1 T 0.150 ml 4. C? 6.5 C Because only temperature and volume are changing, we use Charles s law to solve for the new volume: 1 T T1 Before doing the calculation, we must convert the temperatures to kelvins: T 1 4. C + 73.15 97.4 K T 6.5 C + 73.15 335.7 K 0.150 ml 335.7 K 0.169 ml 97.4 K Because the temperature increases, the volume of the bubble also increases. 9.54 It is often helpful to create a table like the one shown below and input the data from the problem. 1 T 1 T 79.0 L 7.0 C 3.0 L? Because only temperature and volume are changing, we use Charles s law and solve for temperature: T 1 T 1 Convert 7.0 C to kelvins: T 1 7.0 C + 73.15 300. K 9 16

T 3.0 L 300. K 11 K 79.0 L Because the volume decreases, the temperature also decreases. T C T K 73.15 K 11 K 73.15 15 C 9.55 Nothing happens to the particles if the temperature, pressure, and volume are constant. If the tank is sealed so that no gas molecules can escape, then the pressure does not change. If the tank is opened, then some of the gas molecules will leave the tank to maintain an equilibrium pressure with the air outside the tank. 9.56 The pressure at sea level is higher than at 9000 ft. As the external pressure on the container increases, the volume of the container will decrease as long as the temperature stays constant. 9.57 We know that the temperature and pressure of a fixed volume of gas are directly related, so we can write an equation relating the change in pressure that occurs when the temperature of a gas sample changes. The equation is similar to that for Charles s law, except that we substitute pressure for volume (recall that volume and temperature are also directly proportional). P1 P T T 1 1 Solving for P : P T P T1 (a) Convert temperatures to kelvins: T 1 0.0 C + 73.15 73. K T 105.0 C + 73.15 378. K P 378. K 30 torr 418 torr 73. K The temperature increases, so the pressure also increases. (b) Convert temperatures to kelvins: T 1 5.0 C + 73.15 98. K T 0.0 C + 73.15 73. K P 73. K 735 torr 673 torr 98. K The temperature decreases, so the pressure also decreases. (c) P 373 K 3.5 atm 4.44 atm 73 K Because the temperature increases, the pressure also increases. 9.58 We know that temperature and pressure of a fixed volume of a gas are directly related, so we can write an equation relating the change in pressure of a gas sample when the temperature of the gas sample changes. The equation is similar to that for Charles s law, except that we substitute pressure for volume (recall that volume and temperature are also directly proportional). P1 P T T 1 9 17

1 Solving for P : P T P T 1 (a) P 315 K 55 torr 357 torr 5 K Because the temperature decreases, the pressure also decreases. (b) Convert temperatures to kelvins: T 1 5.0 C + 73.15 98. K P 06 K 895 torr 618 torr 98. K Because the temperature decreases, the pressure also decreases. (c) Convert temperatures to kelvins: T 1 150 C + 73.15 40 K T 3 C + 73.15 96 K P 96 K.74 atm 1.9 atm 40 K Because the temperature decreases, the pressure also decreases. 9.59 We know that temperature and pressure of a fixed volume of a gas are directly related, so we can write an equation that relates the change in pressure of a gas sample with a change in temperature. The equation is similar to Charles s law, except that we substitute pressure for volume (recall that volume and temperature are also directly proportional). P1 P T1 T 1 Solving for T : T P T P1 (a) Convert temperature to kelvins: T 1 30.0 C + 73.15 303. K T 915 torr 303. K 18 K 155 torr Convert T to degrees Celsius: T C T K 73.15 K 18 K 73.15 91 C (b) Convert temperature to kelvins: T 1 50.0 C + 73.15 53. K The pressures need to be expressed in the same units. Convert 104 torr to atm: Pressure in atm 104 torr 1.37 T 1.37 53. K 1.0 10 3 K 0.70 atm 9 18

Convert T to degrees Celsius: T C T K 73.15 K 1.0 10 3 K 73.15 750 C (c) T 1000.0 torr 355 K 7.10 10 K 500.0 torr Convert T to degrees Celsius: T C T K 73.15 K 7.10 10 K 73.15 437 C 9.60 We know that the temperature and pressure of a fixed volume of a gas are directly related, so we can write an equation relating the change in pressure associated with a change in the temperature of a gas sample. The equation is similar to Charles s law, except that we substitute pressure for volume (recall that volume and temperature are also directly proportional). P1 P T T 1 Solving for T : T 1 P T P (a) Convert temperature to kelvins: T 1 5.0 C + 73.15 98. K T 735 torr 1 98. K 90 K 43 torr Convert T to degrees Celsius: T C T K 73.15 K 90 K 73.15 69 C (b) T 1.0 atm 05 K 105 K.35 atm Convert T to degrees Celsius: T C T K 73.15 K 105 K 73.15 168 C (c) The pressures must be in the same units. Convert 875 torr to atm: Pressure in atm 875 torr 1.15 atm T 0.85 atm 375 K.8 10 K 1.15 atm Convert T to degrees Celsius: T C T K 73.15 K.8 10 K 73.15 0 C (There are no significant figures in the answer.) 9.61 We know that the temperature and pressure of a fixed volume of a gas are directly related, so we can write an equation relating the pressure change associated with a change in the temperature of a gas sample. The equation is similar to Charles s law, except that pressure replaces volume (recall that volume and temperature are also directly proportional). P1 P T T 1 9 19

1 Solving for P : P T P T 1 P 1 T 1 P T 7.5 atm 18.5 C? 37. C Convert temperatures to Celsius and calculate P : T 1 18.5 C + 73.15 91.7 K T 37. C + 73.15 310.4 K P 310.4 K 7.5 atm 7.7 91.7 K 9.6 We know that the temperature and pressure of a fixed volume of a gas are directly related, so we can write an equation relating the pressure change associated with a change in the temperature of a gas sample. The equation is similar to Charles s law, except that pressure replaces volume (recall that volume and temperature are also directly proportional). P1 P T T 1 P 1 T 1 P T 6.75 atm 5.0 C 1.5 atm? Convert temperature to kelvins: T 1 8.0 C + 73.15 98. K T 1.5 atm 98. K 55. K 6.75 atm Convert T to degrees Celsius: T C T K 73.15 K 55. K 73.15 17.9 C 9.63 When we use the combined gas law, it helps if we solve the equation for the desired variable before we substitute numbers into the expression. When we rearrange the equation, we try to keep the state 1 and state variables separate (it helps us to keep from mixing up the variables). We must also make sure that the units are consistent (i.e. that the pressures are in the same units and the temperatures are in kelvins). Combined gas law: (a) The missing value on the data table is : P 1 1 T T P 1 P T 1 1 P T 1 P 1 1 T 1 P T 0.50 atm.50 L 0.0 C 760.0 torr? 0.0 C We need to express the pressures in common units and convert the temperatures to kelvins. Since 760.0 torr is 1.000 atm, we can simply substitute 1.000 atm for P. First, we convert temperatures to kelvins and then solve for : T 1 0.0 C + 73.15 93. K 9 0

T 0.0 C + 73.15 73. K (0.50 atm )(.50 L) 73. K 1. L 93. K 1.000 atm (b) The value missing from the data table is T : T T 1 P P 1 1 P 1 1 T 1 P T 0.50 atm 15 L 5.0 C 100.0 torr 6.0 L? Temperature and pressure both need to be converted to appropriate units: T 1 5.0 C + 73.15 98. K P (atm) 100.0 torr 0.1316 atm T 98. K (0.1316 atm )(6.0 L ) 77.8 K (or 195.3 C) (0.50 atm )(15 L ) (c) The value missing from the data table is P : P P T T 1 1 1 P 1 1 T 1 P T 00.0 torr 455 ml 300.0 K? 00.0 ml 37 C First, we convert T to kelvins: T 37 C + 73.15 6.00 10 K (00.0 torr)(455 ml ) 6.00 10 K P 9.10 10 torr (300.0 K ) 00.0 ml 9.64 When we use the combined gas law, it helps if we solve the equation for the desired variable before we substitute numbers into the expression. When we rearrange the equation, we try to keep the state 1 and state variables separate (it helps to keep us from mixing up the variables). We must also make certain that the units are consistent (i.e. that pressures are in the same units and temperatures are in kelvins. Combined gas law: (a) The value missing from the data table is P : P P T T 1 1 1 P T 1 1 P T 1 P 1 1 T 1 P T 900.0 torr 601 ml 10.0 C? 100.0 ml 0.0 C Convert temperatures to kelvins: T 1 10.0 C + 73.15 63. K T 0.0 C + 73.15 73. K 9 1

P (900.0 torr)(601 ml ) 73. K 468 torr (63. K ) 100.0 ml (b) The value missing from the data table is T : T T 1 P P 1 1 P 1 1 T 1 P T 75.0 atm 37 ml 147 K 150.0 atm 474 ml? T 147 K (150.0 atm )(474 ml ) 588 K (75.0 atm )(37 ml ) (c) The value missing from the data table is : P 1 1 T T P 1 P 1 1 T 1 P T 760.0 torr 1.1 L 0.0 C 700.0 torr? 5.0 C The pressures and temperature units need to be changed. Because 760.0 torr 1.000 atm, we can make this substitution. Convert temperatures to kelvins: T 1 0.0 C + 73.15 73. K T 5.0 C + 73.15 98. K (760.0 torr )(1.1 L) 98. K 1.33 L 73. K 700.0 torr 9.65 We want to calculate the volume of oxygen at STP ( 1 ). Standard temperature and pressure are 73.15 K (0 C) and (by definition). The tank will hold 0.500 L at 3.50 atm and 4.5 C. This can be considered the final state of the gas. P 1 1 T 1 P T? 73.15 K 3.50 atm 0.500 L 4.5 C Convert 4.5 C to kelvins: T 4.5 C + 73.15 97.7 K Solve the combined gas law for 1. 1 P T1 T P 1 (3.50 atm )(0.500 L) 97.7 K 73.15 K 1.61 L 9.66 The initial state of the gas is.0 L at 5.0 C and 75 torr, and we are trying to determine the pressure (P ) when the gas is heated to 134 C and compressed to 4.5 L. P 1 1 T 1 P T 75 torr.0 L 5.0 C? 4.50 L 134 C Convert temperatures to kelvins: 9

T 1 5.0 C + 73.15 98. K T 134 C + 73.15 407 K Solve the combined gas law for P. P P T T 1 1 1 (75 torr)(.0 L ) 98. K 407 K 4.50 L 4.84 10 3 torr 9.67 Gay-Lussac s Law states that volumes of gases react in simple, whole number ratios when the volumes of the reactants and products are measured at the same temperature and pressure. These ratios correspond to the coefficients in the balanced chemical equation for the reaction. 9.68 The ratio of the volumes of the gaseous reactant and product corresponds to the ratio of the stoichiometric coefficients for these substances in the balanced chemical reaction. 9.69 The molar volume of all gases is approximately.414 L/mol at STP. 9.70 The volume of 1.00 mol of H gas at STP is.4 L. This is approximately true for all gases. 9.71 At STP, 1.00 mol of any gas occupies.414 L. We determine the mass of each gas sample from the number of moles and the molar mass of the gas. (a) Moles CH 4 8.6 L 0.385 mol CH 4.414 L Given that the molar mass of CH 4 (16.04 g/mol) we can calculate the mass of CH 4 : 16.04 g Mass CH 4 0.385 mol 6.17 g CH 4 (b) Convert ml to L: 1L olume in L 350.0 ml 0.3500 L 1000 ml Moles Xe 0.3500 L 1.56 10 mol Xe.414 L We use the molar mass of Xe (131.3 g/mol) to calculate the mass of Xe gas in the sample: 131.3 g Mass Xe 1.56 10 mol.050 g Xe (c) Moles CO 48.1 L.15 mol CO.414 L We use the molar mass of CO (8.01 g/mol) to calculate the mass of CO gas in the sample: 8.01 g Mass CO.15 mol 60.1 g CO 9.7 At STP, 1.00 mol of any gas occupies.414 L. We determine the mass of each gas sample from the number of moles and the molar mass of the gas. (a) Convert ml to L: 9 3

1L olume in L 135 ml 0.135 L 1000 ml Moles H 0.135 L 6.0 10 3 mol H.414 L We use the molar mass of H (.016 g/mol) to calculate the mass of H gas in the sample: Mass H 3.016 g 6.0 10 mol 0.011 g H (b) Moles N 8.96 L 0.400 mol N.414 L We use the molar mass of N (8.0 g/mol) to calculate the mass of N gas in the sample: 8.0 g Mass N 0.400 mol 11. g N (c) Moles He 0.75 L 0.033 mol He.414 L We use the molar mass of He (4.003 g/mol) to calculate the mass of He gas in the sample: 4.003 g Mass He 0.033 mol 0.13 g He 9.73 Avogadro s law tells us that because both balloons have the same volume they contain the same number of gas particles (and, therefore, the same number of moles). Because argon atoms are more massive than helium atoms, the balloon containing argon has the greater mass, and, therefore, the greater density (the volumes of the balloons are the same). 9.74 Avogadro s law tells us that both balloons contain the same number of gas molecules. The CO -filled balloon has the greater mass, because CO molecules are more massive than O molecules. Because the volumes of the balloons are the same and the mass of the CO balloon is greater, the density of the CO balloon is also greater. 9.75 We use the molar mass of each gas to calculate the number of moles of gas in each sample. Because the gases are at STP, one mole of each gas occupies.414 L. (a) 5.8 g NH 3 (17.03 g/mol) Moles NH 3 5.8 g 0.34 mol NH 3 17.03 g.414 L olume NH 3 0.34 mol 7.6 L NH 3 (b) 48 g O (3.00 g/mol) Moles O 48 g 1.5 mol O 3.00 g.414 L olume O 1.5 mol 34 L O 9 4

(c) 10.8 g He (4.003 g/mol) Moles He 10.8 g.70 mol He 4.003 g.414 L olume He.7 mol 60.5 L He 9.76 We use the molar mass of each gas to calculate the number of moles of gas in each sample. Because the gases are at STP, one mole of each gas occupies.414 L. (a). g CO (44.01 g/mol) Moles CO. g 0.050 mol CO 44.01 g.414 L olume CO 0.050 mol 1.1 L CO (b) 5.6 g N (8.0 g/mol) Moles N 5.6 g 0.0 mol N 8.0 g.414 L olume N 0.0 mol 4.5 L N (b) 145 g Ar (39.95 g/mol) Moles Ar 145 g 3.6 mol Ar 39.95 g.414 L olume Ar 3.6 mol 81 L Ar 9.77 We convert the mass of helium (45 g, given in the figure) to the equivalent number of moles, and then to volume in liters, using the molar mass of helium (4.003 g/mol) and the molar volume of a gas at STP (.414 L/mol). Once we have determined the volume, we can determine approximately how many balloons we can fill from the tank. Moles He 45 g 106 mol He 4.003 g.414 L olume He 106 mol.38 10 3 L He Because each balloon holds 1.0 L of gas, we can estimate that we can fill approximately 380 balloons. 9.78 We convert the mass of CO (370 g, given in the figure) to the equivalent number of moles, and then to volume in liters, using the molar mass of CO (44.01 g/mol) and the molar volume of a gas at STP (.414 L/mol). Once we determine the volume of CO, we can estimate the number of balloons we can fill using the volume of each balloon. Moles CO 370 g 8.4 mol CO 44.01 g.414 L olume CO 8.4 mol 1.9 10 L CO 9 5

Because each balloon holds 0.75 L, we can estimate the number of balloons we can fill from the CO in the tank: Balloons 1.9 10 L 1 balloon.5 10 balloons 0.75 L 9.79 Convert 5.0 C to kelvins: T 1 5.0 C + 73.15 98. K. For STP, T 73.15 K and P. Solve the combined gas law for. P 1 1 T T P 1 (1.00 atm )(1.0 L) 98. K 73.15 K 11.0 L 9.80 To solve this problem using the combined gas law, we can first calculate the volume of the H at STP (, 73.15 K). Then we can use the combined gas law to calculate the volume the He occupies at 30.0 C and 0.975 atm: P 1 1 T T P 1.414 L olume of gas at STP 1.50 mol 33.6 L P 1 1 T 1 P T 33.6 L 73.15 K 0.975 atm? 30.0 C Convert T to kelvins: T 30.0 C + 73.15 303. K ( )(33.6 L) 303. K 38.3 L 73.15 K 0.975 atm 9.81 An ideal gas is any gas whose behavior is described by the five postulates of kinetic-molecular theory. Ideal gases have elastic collisions (they bounce and/or collide without losing energy), travel in straight lines (because they are not attracted to other molecules), and occupy zero volume. In addition, the average kinetic energy of an ideal gas is directly proportional to the temperature of the gas. 9.8 (a) The ideal gas law is: P nrt. (b) The ideal gas law is used to relate the variables that describe the state of a gas. These variables are pressure, volume, number of moles, and temperature. Given any three of these variables, we can calculate the fourth. 9.83 To calculate the volume occupied by each gas, we first determine the number of moles of each gas and convert the temperature of the gases to kelvins. Whenever we use the ideal gas law, we must make certain that the units of P,, n, and T match the units of R that we are using (R 0.0806 L atm/(mol K)). For each gas, the temperature is T 100.0 C + 73.15 K 373.15 K (four significant figures) Rearranging the ideal gas law for volume: (a) 5.8 g NH 3 (17.03 g/mol) nrt P Moles NH 3 5.8 g 0.34 mol NH 3 17.03 g 9 6

L atm ( 0.34 mol ) 0.0806 mol 373.15 K K ( ) 15.0 atm 0.69 L NH 3 (b) 48 g O (3.00 g/mol) Moles O 48 g 1.5 mol O 3.00 g L atm ( 1.5 mol ) 0.0806 mol 373. K K ( ) 15.0 atm 3.1 L O (c) 10.8 g He (4.003 g/mol) Moles He 10.8 g.70 mol He 4.003 g L atm (.70 mol ) 0.0806 mol 373. K K ( ) 15.0 atm 5.51 L He 9.84 To calculate the volume occupied by each gas, we first determine the number of moles of each gas and convert the temperature of the gases to kelvins. Anytime we use the ideal gas law, we must make certain that the units P,, n, and T match the units of R that we are using (R 0.0806 L atm/(mol K)). For each gas, the temperature is T 75.0 C + 73.15 K 348. K nrt Rearranging the ideal gas law for volume: P (a). g CO (44.01 g/mol) Moles CO. g 0.050 mol CO 44.01 g L atm ( 0.050 mol ) 0.0806 mol 348. K K ( ) 3.5 atm 0.41 L CO (b) 5.6 g N (3.00 g/mol) Moles N 5.6 g 0.0 mol N 8.0 g L atm ( 0.0 mol ) 0.0806 mol 348. K K ( ) 3.5 atm 1.6 L N (c) 7.5 g Ar (39.95 g/mol) Moles Ar 7.5 g 0.19 mol Ar 39.95 g 9 7

L atm ( 0.19 mol ) 0.0806 mol 348. K K ( ) 3.5 atm 1.5 L Ar P 9.85 Rearranging the ideal gas law for the number of moles in the sample: n RT Because R 0.0806 L atm/(mol K), we must express pressure in atmospheres and, of course, temperature in kelvins. Pressure in atm 7 torr 0.950 atm T 87.5 C + 73.15 360.7 K (a) 7.6 L CH 4 (16.04 g/mol) P (0.950 atm )(7.6 L ) n RT L atm 0.0806 360.7 K mol K ( ) 0.45 mol 16.04 g Mass CH 4 0.45 mol 3.9 g CH 4 (b) 135 ml H (.016 g/mol) First convert ml to L: 1L olume in L 135 ml 0.135 L 1000 ml P (0.950 atm )(0.135 L ) n RT L atm 0.0806 360.7 K mol K Mass H (c) 8.96 L N (8.0 g/mol) ( ) 4.33 10 3.016 g 4.33 10 mol 8.74 10 3 g H P (0.950 atm )(8.96 L ) n RT L atm 0.0806 360.7 K mol K ( ) 0.88 mol 8.0 g Mass N 0.88 mol 8.06 g N 3 mol P 9.86 Rearranging the ideal gas law for the number of moles of gas in the sample: n RT Because R 0.0806 L atm/(mol K), we must express pressure in atmospheres, and, of course, temperature in kelvins. Pressure in atm 790 torr 1.04 atm ( significant figures) 9 8

T 54.0 C + 73.15 37. K (a) 150.0 ml Xe (131.3 g/mol) Convert 150.0 ml Xe to liters: 1L olume in L 150.0 ml 0.1500 L 1000 ml P (1.04 atm )(0.1500 L ) n RT L atm 0.0806 37. K mol K ( ) 5.8 10 3 131.3 g Mass Xe 5.8 10 mol 0.76 g Xe (b) 38.1 L CO (8.01 g/mol) P (1.04 atm )(38.1 L ) n RT L atm 0.0806 37. K mol K ( ) 1.5 mol 8.01 g Mass CO 1.5 mol 41 g CO (c).5 L O (3.00 g/mol) P (1.04 atm )(.5 L ) n RT L atm 0.0806 37. K mol K ( ) 0.097 mol 3.00 g Mass O 0.097 mol 3.1 g O 3 mol 9.87 The balloons sink or float depending on how their densities compare to the density of air. Recall that if the balloons (or any containers) have equal volume, pressure, and temperature, they contain the same number of particles (Avogadro s Law or ideal gas law). Because each CO molecule is heavier than each He atoms, and there are equal numbers of each, the CO -filled balloon will have more mass and, therefore, higher density. Why doesn t CO float in air? Air is less dense than CO. Air is primarily a mixture of N and O, both of which have lower masses than CO. The CO -filled balloon sinks because CO gas is more dense than air. 9.88 The larger balloon is at the higher temperature. The velocity of the N molecules is greater in the warmer balloon, so they collide with the walls of the balloon more often and with greater force, causing the volume to increase. The density of particles is lower in the larger balloon. 9 9

9.89 At STP, one mole of any gas occupies.414 L. If we know the molar mass of a gas, we can calculate its density (d m/, the mass of one mole divided by the volume of one mole). (a) NH 3 (17.03 g/mol). d 17.03 g.414 L (b) N (8.0 g/mol) d 8.0 g.414 L (c) N O (44.0 g/mol) d 44.0 g.414 L 0.7598 g/l 1.50 g/l 1.964 g/l 9.90 At STP, one mole of gas occupies.414 L. If we know the molar mass of the gas, we can calculate its density (d m/, the mass of one mole divided by the volume of one mole). (a) NO (30.01 g/mol) d 30.01 g.414 L (b) NO (46.01 g/mol) d 46.01 g.414 L (c) O (3.00 g/mol) d 3.00 g.414 L 1.339 g/l.053 g/l 1.48 g/l 9.91 Density is the mass of a substance divided by its volume. For gas samples, it is convenient for us to determine density by dividing the molar mass of the gas by its volume, which we calculate using the ideal gas law. T 5 C + 73.15 98. K Pressure in atm 735 torr 0.967 atm n 1.00 mol (we choose this for convenience) L atm ( 1.00 mol ) 0.0806 nrt mol K P 0.967 atm ( 98. K) 5.3 L To calculate the density, we divide the molar mass by the volume it occupies: (a) NH 3 (17.03 g/mol) One mole of NH 3 has a mass of 17.03 g and occupies a volume of 5.3 L. The density is: m 17.03 g d 0.673 g/l 5.3 L 9 30

(b) N (8.0 g/mol) One mole of N has a mass of 8.0 g and occupies a volume of 5.3 L. The density is: m 8.0 g d 1.11 g/l 5.3 L (c) N O (44.0 g/mol) One mole of N O has a mass of 44.0 g and occupies a volume of 5.3 L. The density is: m 44.0 g d 1.74 g/l 5.3 L 9.9 Density is the mass of a substance divided by volume. For gas samples, it is convenient for us to determine density by dividing the molar mass of the gas by its volume, which we calculate using the ideal gas law. T 5 C + 73.15 98. K Pressure in atm 735 torr 0.967 atm n 1.00 mol L atm ( 1.00 mol ) 0.0806 nrt mol K P 0.967 atm ( 98. K) 5.3 L To calculate the density, divide the molar mass by the volume it occupies: (a) NO (30.01 g/mol) One mole of NO has a mass of 30.01 g and occupies a volume of 5.3 L. The density is: m 30.01 g d 1.19 g/l 5.3 L (b) NO (46.01 g/mol) One mole of NO has a mass of 46.01 g and occupies a volume of 5.3 L. The density is: m 46.01 g d 1.8 g/l 5.3 L (c) O (3.00 g/mol) One mole of O has a mass of 3.00 g and occupies a volume of 5.3 L. The density is: m 3.00 g d 1.6 g/l 5.3 L 9.93 According to the ideal gas law, equal-molar amounts of every gas occupy the same volume at the same pressure and temperature. Using this law, we can calculate the number of moles of gas in the container. First, we convert pressure and temperature to units consistent with the units in the gas constant: Pressure in atm 840 torr 1. Temperature in K 50.0 C + 73.15 33. K 9 31

n P RT ( 1. ) 5.00 L ( ) L atm 0.0806 33. K mol K ( ) 0. Now we can determine the mass of each gas using its molar mass: (a) H (.016 g/mol).016 g Mass H 0. 0.4 g H mol (b) CH 4 (16.04 g/mol) 16.04 g Mass CH 4 0. 3.3 g CH 4 mol (c) SO (64.06 g/mol) 64.06 g Mass SO 0. 13 g SO mol 9.94 According to the ideal gas law, equal-molar amounts of every gas occupy the same volume at the same pressure and temperature. Using this law, we can calculate the number of moles of gas in the container. First, we convert pressure and temperature to units that are consistent with the units in the gas constant: Pressure in atm 650 torr 0.86 atm Temperature in K 5.0 C + 73.15 98. K P ( 0.86 atm )(.50 L ) n RT L atm 0.0806 98. K mol K ( ) 0.087 mol Now, we can determine the mass of each gas using its molar mass: (a) O (3.00 g/mol) 3.00 g Mass O 0.087 mol.8 g O mol (b) CO (44.01 g/mol) 44.01 g Mass CO 0.087 mol 3.8 g CO mol (c) He (4.003 g/mol) 4.003 g Mass He 0.087 mol 0.35 g He mol 9.95 Dalton s law of partial pressures states that the total pressure in a container is the sum of the pressures exerted by each of the individual gases in the container. Each gas exerts a pressure on the walls of the container as if it were the only gas in the container. Gas molecules behave in this way because they are not strongly attracted to each other. 9 3

9.96 Suppose we have a closed container that contains CO (g) and some liquid water. The total pressure inside the container is actually the sum of the pressures of the CO (g) and H O(g) (water vapor). To determine the pressure of just the CO (g) we can subtract the pressure of the water from the total pressure in the container. The vapor pressure of water depends only on the temperature so we can determine that pressure from a table in the textbook or from another reference book. Mathematically, we express Dalton s law of partial pressures for this system as: P P + P. The pressure of CO is: P P P total CO water CO total water 9.97 According to Dalton s law of partial pressures, the total pressure (78 torr) is the sum of the pressure exerted by the oxygen gas and water vapor (0.0 torr). To calculate the partial pressure of oxygen gas we subtract the water vapor pressure from the total pressure. Mathematically, Dalton s law of partial pressures looks like: P P + P total O water P P P 78 torr 0.0 torr 708 torr O total water 9.98 According to Dalton s law of partial pressures, the total pressure (74 torr) is a sum of the pressure exerted by the hydrogen gas and water vapor (15.5 torr). To calculate the partial pressure of hydrogen gas we subtract the water vapor pressure from the total pressure. Mathematically, Dalton s law of partial pressures looks like: P P + P total H water P P P 74 torr 15.5 torr 77 torr H total water 9.99 (a) We use molar mass to calculate the number of moles of N (8.0 g/mol). Moles N 78.0 g.78 mol N 8.0 g (b) We use molar mass to calculate the number of moles of Ne (0.18 g/mol). Moles Ne 4.0 g.08 mol Ne 0.18 g (c) To calculate the partial pressure of N we use the ideal gas law (P nrt). We know the number of moles of N (part (a)) and temperature (50.0 C), but the volume is missing. To calculate the volume, we take advantage of a concept from Dalton s law of partial pressures; that each gas occupies the total volume of the container. Because we know the total pressure, number of moles of N, and temperature, we can calculate the total volume. P total 3.75 atm n total.78 mol +.08 mol 4.86 mol T 50.0 C + 73.15 33. K total? L atm ( 4.86 mol ) 0.0806 nrt total mol 33. K K P 3.75 atm total ( ) 34.4 L Using the total volume, calculate the partial pressure of N using the moles of N. 9 33

L atm (.78 mol ) 0.0806 33. K nn RT mol K P N 34.4 L total ( ).15 atm (d) To calculate the partial pressure of Ne, we can do exactly as we did in (c) or we can use Dalton s law of partial pressures. Either method gives the same result. According to Dalton's law of partial pressures, the total pressure is the sum of the pressures of Ne and N. Mathematically, we can express Dalton's law of partial pressure as: P total P N +P Ne P Ne Ptotal PN 3.75 atm.15 atm 1.60 atm 9.100 (a) We use molar mass to calculate the number of moles of CO (44.01 g/mol). Moles CO 150.0 g 3.4 CO 44.01 g (b) We use molar mass to calculate the number of moles of O (3.00 g/mol). Moles O 4.0 g 0.750 mol O 3.00 g (c) To calculate the partial pressure of CO we use the ideal gas law (P nrt). We know the number of moles of CO (part (a)) and temperature (50.0 C), but the volume is missing. To calculate the volume, we take advantage of a concept from Dalton s law of partial pressures: each gas occupies the total volume of the container. Because we know the total pressure, the number of moles, and the temperature, we can calculate the total volume. P total 4.5 atm n total 3.4 + 0.750 mol 4.16 mol T 5.0 C + 73.15 98. K total? L atm ( 4.16 mol ) 0.0806 nrt total mol 98. K K P 4.5 atm total ( ) 3.9 L Using the total volume, we can calculate the partial pressure of CO using the moles of CO. L atm ( 3.4 ) 0.0806 33. K nco RT mol K P CO 3.9 L total ( ) 3.48 atm (d) To calculate the partial pressure of O, we can do exactly as we did in (c) or we can use Dalton s law of partial pressures. Either method gives the same result. According to Dalton's law of partial pressures the total pressure is the sum of the pressure of O and CO. Mathematically, we can express Dalton's law of partial pressures as: P total P CO +P O P P P O total CO 4.5 atm 3.48 atm 0.77 atm 9 34