Sample Problem Using Boyle's Law to Calculate Volume Problem A weather balloon with a volume of 2.00 x 103 L at a pressure of 96.3 kPa rises to an altitude of 1.00 x 103 m, where the atmospheric pressure is measured to be 60.8 kPa. Assuming there is no change in temperature or amount of gas, what is the final volume of the weather balloon? What Is Required? You need to find the volume, V2, after the pressure on the balloon has decreased. What Is Given? You know the pressure and volume for the first set of conditions and the pressure for the final set of conditions. P, = 96.3 kPa V, = 2.00 x 103 P2 = 60.8 kPa You know the temperature does not change. Plan Your Strategy Act on Your Strategy Pressure and volume are changing, at constant temperature and amount of gas. Therefore, use the equation for Boyle's law. Isolate the variable V*2 by dividing each side of the equation byP 2 . PiVi J>2 Substitute numbers and units for the known variables in the formula and solve. Make certain that the same units for pressure are used in the equation. PjVl Pi (96.3>Pa)(2.00 x 10 3 L) = 3.17 X 10 3 L According to Boyle's law, when the amount and temperature of a gas are constant, there is an inverse relationship between the pressure and volume of a gas: V a — Alternative Solution Plan Your Strategy Act on Your Strategy According to Boyle's law, a decrease in pressure will cause an increase in volume. Determine the ratio of the initial pressure and the final pressure that is greater than 1 . P, = 96.3 kPa P2 = 60.8 kPa pressure ratio ^ 1 is ^^ To find the final volume, multiply the initial volume of the balloon by the ratio of the two pressures that is greater than ] . V2 = V, x pressure ratio Q£T P 00 x 10 3 L) I1 Xx "I LD-i 96 3 kftr -60.8J<PS- = 3.17 x 10 3 L Check Your Solution The units cancel out to leave the correct unit of volume, L. You would expect the volume to increase when the pressure decreases, which is represented by the value determined. Chapter 11 Properties of Gases • MHR 513 Practice Problems Note: Assume that the temperature and amount of gas are constant in all of the following problems. 1. 1.00 L of a gas at 1.00 atm pressure is compressed to 0.437 L. What is the new pressure of the gas? 2. A container with a volume of 60.0 mL holds a sample of gas. The gas is at a pressure of 99.5 kPa. If the container is compressed to one-quarter of its volume, what is the pressure of the gas in the container? 2.29 atm 398 kPa 24 atm 0.27 L 1.3 x 102kPa 440 L 7. 14.3 mL 2 8. 1.1 X 10 L 3 9. 1.73 x 10 L 3. Atmospheric pressure on the peak of Mount Everest can be as low as 0.20 atm. If the volume of an oxygen tank is 10.0 L, at what pressure must the tank be filled so that the gas inside would occupy a volume of 1.2 x 103 L at this pressure? 2 10. a. 3.6 x 10 L b. 5.6 x 102 min 6. The volume of carbon dioxide in a fire extinguisher is 25.5 L. The pressure of the gas in this can is 260 psi. What is the volume of carbon dioxide released when sprayed if the room pressure is 15 psi? 7. A 50.0 mL sample of hydrogen gas is collected at standard atmospheric pressure. What is the volume of the gas if it is compressed to a pressure of 3.50 atm? 8. A portable air compressor has an air capacity of 15.2 L and an interior pressure of 110 psi. If all the air in the tank is released, what volume will that air occupy at an atmospheric pressure of 102 kPa? 4. If a person has 2.0 x 102 mL of trapped intestinal gas at an atmospheric pressure of 0.98 atm, what would the volume of gas be (in litres) at a higher altitude that has an atmospheric pressure of 0.72 atm? 9. A scuba tank with a volume of 10.0 L holds air at a pressure of 1.75 X 104 kPa. What volume of air at an atmospheric pressure of 101 kPa was compressed into the tank if the temperature of the air in the tank is the same as the temperature of the air before it was compressed? 5. Decaying vegetation at the bottom of a pond contains trapped methane gas. 5.5 x 102 mL of gas are released. When the gas rises to the surface, it now occupies 7.0 x 102 mL. If the surface pressure is 101 kPa, what was the pressure at the bottom of the pond? 10. An oxygen tank has a volume of 45 L and is pressurized to 1200 psi. a. What volume of gas would be released at 765 torr? b. If the flow of gas from the tank is 6.5 L per minute, how long would the tank last? Kinetic Molecular Theory and Boyle's Law Co to scienceontario to find out more Figure 11.12 The kinetic molecular theory can explain the relationship between pressure and volume. (d} and d2 represent average distances of molecules from the container wall.) 514 M H R - U n i t 5 Gases Pressure on the walls of a gas-filled container is caused by collisions of gas molecules with the walls. Each collision of a gas molecule exerts a force on the wall. The average force exerted by all the gas molecules divided by the surface area of the container is equivalent to the pressure on the walls of the container. Examine Figure 11.12 to see what happens when you change the external pressure on the gas. The containers have pistons that will move until the external pressure and the internal pressure are equal. If you increase the external pressure, the piston will move down, reducing the volume available to the gas molecules. The gas molecules are now closer together and collide with one another and the walls of the container more often. As the number of collisions over time increases, the average force exerted by all the molecules increases; thus, the gas pressure increases. If the temperature remains constant and no gas escapes or enters, the decrease in the volume of the container will be inversely proportional to the increase in the gas pressure. Pe<t increases, fand n fixed Higher Pia causes lower V, which causes more collisions, increasing the pressure ""til Pga$ = Pm The Kelvin Temperature Scale and Absolute Zero In 1848, twenty-five years after the death of Jacques Charles, Scottish physicist Lord Kelvin (William Thomson, 1824-1907) interpreted the significance of the extrapolated temperature at zero volume of a gas. Kelvin suggested that -273.15°C was the lowest possible temperature, or absolute zero. He then established a new temperature scale based on absolute zero as the starting point on the scale. The temperature scale was named the Kelvin scale in his honour. Figure 11.14 compares the Celsius and Kelvin temperature scales. The name of a unit in the Kelvin scale is the kelvin (K). The size of the kelvin is the same as the size of a Celsius degree, but the term "degree" is not used when reporting temperatures on the Kelvin scale. As well, the starting points for these two temperature scales are different. Notice that there are no negative values on the Kelvin scale. What would happen if you tried to calculate a temperature that is twice as warm as -5°C? Mathematically, the answer would be -10°C, but this is a colder temperature. When mathematical manipulations are involved in studying gas behaviour, you need to convert temperatures from the Celsius scale to the Kelvin scale. absolute zero the lowest theoretical temperature, equivalent to-273.15°C; the temperature at which the volume of a gas approaches zero For converting Celsius to kelvin: K = °C + 273.15 For converting kelvin to Celsius: °C = K — 273.15 The rounded-off value of 273 is often used as the conversion factor relating K and °C. Celsius Scale Kelvin Scale 100T 100° 80° 373 K •'''imsmmHamaaammmm^ Boiling Water 60° 333 40° 20° 313 273 K 0"C 00 293 273 Freezing Water -20° IT 373 353 253 -40° 233 -60° 213 -80° 193 -100° 173 -120° 153 -140° 133 -160° 113 -180° 93 IT -200° 73 -220° 53 -240° — -260° 33 -273 T OK 13 Absolute Zero .-. Figure 11.14 There are 273 temperature units between absolute zero and the freezing temperature of water on the Celsius and Kelvin scales. There are also TOO temperature units between the freezing and boiling temperatures of water on both scales. Apply If you double a Celsius temperature, by how much does the Kelvin temperature increase? Chapter 11 Properties of Gases • MHR 517 Learning Check 13. What is the relationship between the temperature and volume of a gas at constant pressure and amount? 14. What is absolute zero, and what is its significance? 15. Examine the graph in Figure 11.14. What do all the graph lines have in common? 16. Make the following temperature conversions, a. 27.3°C t o K c. 373.2 K to °C b.-25°CtoK d. 23.5Kto°C Activity 17. Why is it necessary to keep the pressure of a gas constant when studying the relationship between temperature and volume of a gas? 18. A teacher pours liquid nitrogen at a temperature of 77 K over a balloon. Predict the changes that would occur to the balloon. Analyzing the Temperature-Volume Relationship of a Gas In this activity, you will use data from the table below and the graph that you construct from them to analyze the relationship between the temperature of a gas and its volume and to infer the importance of the Kelvin temperature scale. Procedure 1. Copy and complete the data table. For the second column, you must calculate the Kelvin equivalent. For the last two columns, you must calculate the quotient of volume divided by temperature. Volume versus Temperature Data Tempe- TempeVolume Volume (cm3) Volume (cm3) rature rature 3 (cm ) Temperature ("C) Temperature (K) (°C) (K> 8 29.5 | 20 30.8 30 32.1 40 34 50 37 60 42 70 49 Materials • graph paper • ruler 1 2. Draw one graph using the data from columns 1 and 4. Draw a second graph using the data from columns 2 and 5. Questions 1. Use a Venn diagram to describe how the two graphs that you drew are similar and how they are different. 2. What is the x-intercept on each graph? What does each represent? 3. Analyze your calculated values of -= (°C) and -= (K) in the data table. What do you notice about the values? 4. How do the values of j (°C) compare to the values of j (K)? Explain the significance of these sets of data. 5. Based on the data in this activity, what relationship seems to exist between the volume and temperature of a gas, when pressure and amount of gas remain constant? How is that relationship affected by the temperature scale that is used? Charles's Law and the Kelvin Temperature Scale Charles's law a gas law stating that the volume of a fixed amount of gas at a constant pressure is directly proportional to the Kelvin temperature ofthegas: Voc T 518 MHR-Unit5 Gases You learned at the start of this section that the volume of a gas is proportional to its temperature, when pressure and amount of gas are constant. This relationship between temperature and volume has become known as Charles's law. This law is often stated in terms of a directly proportional relationship between temperature and volume. This statement only holds true, however, if the temperature is expressed in Kelvin units. To understand why, examine the graphs in Figure 11.5. Both graphs show that the plot of temperature versus volume is a straight line, but notice that Graph A—in which temperature is in degrees Celsius—does not show a direct proportion. The graph of the line does not pass through the origin, and doubling the temperature does not double the volume. Graph B does show a direct proportion; temperature is in kelvins, and the graph of the line passes through the origin. A temperature of 0 K corresponds to 0 ml. Doubling the temperature doubles the volume. Sample Problem Using Charles's Law to Calculate Volume of a Gas Problem A balloon inflated with air in a room in which the temperature of the air is 295 K has a volume of 650 mL. The balloon is put into a refrigerator at 277 K and left long enough for the air in the balloon to reach the same temperature as the air in the refrigerator. Predict the volume of the balloon, assuming that the amount of air has not changed and the air pressure in the room and in the refrigerator are the same. What Is Required? You need to find the volume, V2, of the balloon after it has been cooled to 277 K. What Is Given? You know the volume and temperature of the air sample for the first set of conditions and the temperature for the second set of conditions: V, = 650 ml T, = 295 K T2 = 277 K Plan Your Strategy Act on Your Strategy Temperature and volume are changing at constant pressure and amount of gas. Therefore, use the equation for Charles's law. V~i Isolate the variable V2 by multiplying each side of the equation by T2 and rearranging the equation. ^-(r2) = ^CK) V2 r, T2 n <i v Substitute numbers and units for the known variables in the formula and solve. Since the lowest number of significant digits in values in the question is two, the final volume is reported to two significant digits. ±^ V v > v ^ T, (650 mL)(277X) (295XJ = 610mL According to Charles's law, when the amount and pressure of a gas are constant, there is a directly proportional relationship between the volume of the gas and its Kelvin temperature: Vex T Alternative Solution Plan Your Strategy Act on Your Strategy According to Charles's law, a decrease in temperature will cause a decrease in volume. Determine the ratio of the initial temperature and the final temperature that is less than 1. T2 = 277 K TI = 295 K To find the final volume, multiply the initial volume of the balloon by the ratio of the two Kelvin temperatures that is less than 1. V2 = Vi x temperature ratio 277 K £73 IX = 610 ml Check Your Solution Volume units remain when the other units cancel out. Because the temperature decreases, the volume is expected to decrease. The answer represents a lower value for the volume. 520 MHR-Unit5 Gases Sample Problem Using Charles's Law to Calculate Temperature of a Gas Problem A birthday balloon is filled to a volume of 1.50 L of helium gas in an air-conditioned room at 294 K. The balloon is then taken outdoors on a warm sunny day and left to float as a decoration. The volume of the balloon expands to 1.55 L. Assuming that the pressure and amount of gas remain constant, what is the air temperature outdoors in kelvins? What Is Required? You need to find the outdoor air temperature, 72, in K. What Is Given? You know the volume and temperature of the air sample for the initial set of conditions and the volume for the final set of conditions: V, = 1.50 L Ti = 294 K V 2 = 1.55 L Act on Your Strategy Plan Your Strategy Temperature and volume are changing at constant pressure and amount of gas. Therefore, use the equation for Charles's law. V2 T2 Isolate the variable T2 by multiplying each side of the equation first by T2 and then by -rp-. V2r, Substitute numbers and units for the known variables in the formula and solve. Since the number of significant digits in values in the question is three, the final volume is reported to three significant digits. r,= V, _ (1.55/K294K) (1.50,0 = 304K Alternative Solution Act on Your Strategy Plan Your Strategy According to Charles's law, a decrease in temperature will cause a decrease in volume. Determine the ratio of the initial volume and the final volume that is greater than 1 . V, = 1.50 L V 2 = 1.55 L To find the final temperature, multiply the initial temperature T2 = T] x volume ratio of the balloon by the ratio of the two volumes that is greater pod m v L-J§~± UJ1K}X 1.50J/ than 1. = 304K Check Your Solution The unit for the answer is kelvins. When the other units cancel out, kelvins remain. Because the volume of the balloon had increased, you would expect that the temperature had increased. The answer represents an increase in temperature. Chapter 11 Properties of Gases-MHR 521 Practice Problems Note: Assume that the pressure and amount of gas are constant in all of the problems except question 20. 11. A gas has a volume of 6.0 L at a temperature of 250 K. What volume will the gas have at 450 K? 12. A syringe is filled with 30.0 mL of air at 298.15 K. If the temperature is raised to 353.25 K, what volume will the syringe indicate? 13. The temperature of a 2.25 L sample of gas decreases from 35.0°C to 20.0°C. What is the new volume? 14. A balloon is inflated with air in a room in which the air temperature is 27°C. When the balloon is placed in a freezer at -20.0°C, the volume is 80.0 L. What was the original volume of the balloon? 15. At a summer outdoor air temperature of 30.0°C, a particular size of bicycle tire has an interior volume of 685 cm3. The bicycle has been left outside in the winter and the outdoor air temperature drops to -25.0°C. Assuming the tire had been filled with air in the summer, to what volume would the tire be reduced at the winter air temperature? 16. At 275 K, a gas has a volume of 25.5 mL. What is its temperature if its volume increases to 50.0 mL? 11. 11L 12. 35.5 mL 13. 1.29 L 14. 95 L 15.561cm 3 16. 539 K 17. 308 K 18. 27°C 19. 1.25 times room temperature 20. -214°C Figure 11.16 When the temperature of a gas increases, the speed of the gas molecules increases. The gas molecules collide with the walls of the container more frequently, thus increasing the pressure. If the external pressure remains the same, the gas pushes the piston up and increases the volume of the container. 522 MHR-Unit5 Gases 17. A sealed syringe contains 37.0 mL of trapped air. The temperature of the air in the syringe is 295 K. The sun shines on the syringe, causing the temperature of the air inside it to increase. If the volume increases to 38.6 mL, what is the new temperature of the air in the syringe? 18. A beach ball is inflated to a volume of 25 L of air in the cool of the morning at 15°C. During the afternoon, the volume changes to 26 L. What was the Celsius air temperature in the afternoon? 19. The volume of a 1.50 L balloon at room temperature increases by 25.0 percent when placed in a hot-water bath. How does the temperature of the water bath compare with room temperature? 20. Compressed gases can be condensed when they are cooled. A 5.00 x 102 mL sample of carbon dioxide gas at room temperature (assume 25.0°C) is compressed by a factor of four, and then is cooled so that its volume is reduced to 25.0 mL. What must the final temperature be (in °Q? (Hint: Use both Boyle's law and Charles's law to answer the question.) Kinetic Molecular Theory and Charles's Law Applying the kinetic molecular theory to Charles's law is shown in Figure 11.16. The Kelvin temperature of a gas is directly proportional to the average kinetic energy of the gas molecules. An object's kinetic energy is related to its speed (£k = — mv1). As the temperature of a gas increases, the molecules move at higher speeds. As a result, they collide with the walls of the container and with one another more frequently and with greater force. Therefore, they exert a greater pressure on the walls of the container. If, however, the external pressure on the gas stays the same, the gas pressure causes the container to increase in size. As the volume of the container gets larger, the gas molecules must travel farther to collide with the walls of the container and with one another. As the collisions become less frequent, the pressure drops. The process continues until the pressure inside the container is once again equal to the external pressure. V increases • ..Jk. Higher T increases speed and thus collision frequency: V > Pa, V increases until m Go to scienceontario to find out more Developing a Mathematical Expression for Gay-Lussac's Law Gay-Lussac's law states that the pressure of a fixed amount of gas, at constant volume, is directly proportional to its Kelvin temperature. The relationship can be expressed as P a T, where T is given in kelvins. Using the general expression for a straight line (y = rax + b) and applying the same mathematical treatment used for Charles's law, a mathematical expression for Gay-Lussac's law is "i _ "2 Ti ~ T2 For this equation, PL and 7\ represent the initial pressure and temperature conditions and P2 and T2 represent the final pressure and temperature conditions. The relationship applies as long as the volume and amount of a gas are constant and the temperature is expressed in kelvins. The following Sample Problem and Practice Problems will reinforce your understanding of Gay-Lussac's law. Sample Problem Using Gay-Lussac's Law To Calculate Pressure of a Gas Problem The pressure of the oxygen gas inside a canister with a fixed volume is 5.0 atm at 298 K. What is the pressure of the oxygen gas inside the canister if the temperature changes to 263 K? Assume the amount of gas remains constant. What is Required? You need to find the new pressure, P2, of the oxygen gas inside the canister resulting from a decrease in temperature: What is Given? You know the initial pressure of the oxygen gas in the canister, as well as the initial and final air temperatures: PI = 5.0 atm T! = 298 K T2 = 263 K Act on Your Strategy Plan Your Strategy Temperature and pressure are changing at constant volume and amount of gas. Therefore, use the equation for Gay-Lussac's law. Isolate the variable P2 by multiplying each side of the equation by T, PL_P T, 2 T, fa) = fyfft Pft - p TT- P2 Substitute numbers and units for the known variables in the formula and solve. Since the lowest number of significant digits in values in the question is two, the final pressure is reported to two significant digits. 524 MHR-Unit5 Gases Pz PiT2 T, (5.0atm)(263K) 298 K = 4.4 atm According to Gay-Lussac's law, when the amount and volume of a gas are constant, there is a directly proportional relationship between the pressure of the gas and its Kelvin temperature: PCX T Alternative Solution Act on Your Strategy Plan Your Strategy According to Gay-Lussac's law, a decrease in temperature will cause a decrease in pressure. Determine the ratio of the initial temperature and the final temperature that is less than 1. T, = 298 K T2 = 263 K To find the final pressure, multiply the initial pressure of the gas by the ratio of the two temperatures that is less than 1. PI = P} x temperature ratio Jf.'l \f = 4.4 atm Check Your Solution The result shows the expected decrease in pressure. With kelvin units cancelling out, the remaining unit, atm, is a pressure unit. Practice Problems Note: Assume that the volume and amount of gas are constant in all of the following problems. 21. A gas is at 105 kPa and 300.0 K. What is the pressure of the gas at 120.0K? 22. The pressure of a gas in a sealed canister is 350.0 kPa at a room temperature of 298 K. The canister is placed in a refrigerator and the temperature of the gas is reduced to 278 K. What is the new pressure of the gas in the canister? 23. A propane barbeque tank is filled in the winter at -15.0°C to a pressure of 2500 kPa. What will the pressure of the propane become in the summer when the air temperature rises to 20.0°C? 24. A rubber automobile tire contains air at a pressure of 370 kPa at 15.0°C. As the tire heats up, the temperature of the air inside the tire rises to 60.0°C. What would the new pressure in the tire be? 25. A partially filled aerosol can has an internal pressure of 14.8 psi when the temperature is 20.0°C. a. What would the pressure in the can be, in kPa, if it were placed into an incinerator for disposal, which would have the effect of raising the temperature inside the can to 1800°C? b. Approximately how many times higher is that new pressure compared to standard atmospheric pressure? 26. A sealed can of gas is left near a heater, which causes the pressure of the gas to increase to 1.4 atm. What was the original pressure of the gas if its temperature change was from 20.0°C to 90.0°C? 27. Helium gas in a 2.00 L cylinder has a pressure of 1.12 atm. When the temperature is changed to 310.0 K, that same gas sample has a pressure of 2.56 atm. What was the initial temperature of the gas in the cylinder? 28. A sample of neon gas is contained in a bulb at 150°C and 350 kPa. If the pressure drops to 103 kPa, find the new temperature, in °C. 29. A storage tank is designed to hold a fixed volume of butane gas at 2.00 x 102 kPa and 39.0°C. To prevent dangerous pressure buildup, the tank has a relief valve that opens at 3.50 x 102 kPa. At what Celsius temperature does the valve open? 30. If a gas sample has a pressure of 30.7 kPa at 0.00°C, by how many degrees Celsius does the temperature have to increase to cause the pressure to double? 24. 430 kPa 21. 42.0 kPa 22. 327 kPa 23. 2800 kPa 25. a. 720 kPa 26. 27. 28. 29. 30. b. about 7 times higher 1.1 atm 136 K -150°C 273°C 273°C Chapter 11 Properties of Gases • MHR 525 Chapter 11 SELF-ASSESSMENT Select the letter of the best answer below. 1. •:wfr Which of the following statements best explains why gases can be easily compressed? a. Molecules of a gas exhibit random translational motion. b. Molecules of a gas have negligible intermolecular forces. c. Molecules of a gas have small amounts of space between them. d. Molecules of a gas are in constant motion. e. Molecules of a gas have little volume. J E fi 0 bJ X a II E M 2 3 OH V ._ ^ | !-! 5 li ~ -< \ * ~ r*j <^ ! IN 00 .i «- •- M g JD 1 IN . . (N (N , CM 2. •:fl'fr Which of the following best describes a gas? a. It assumes the volume and shape of the container, and it has weak intermolecular attractions. b. It assumes the volume and shape of the container, and it has strong intermolecular attractions. c. It has a distinct shape and volume, and it has strong intermolecular attractions. d. It has a distinct volume and assumes the shape of the container, and it has moderate intermolecular attractions. O e. It has a distinct volume and assumes the shape of the container, but it lacks intermolecular attractions. 5 ol r* o' *> in 111 a g£ -^rs' 3. •:«»» Which of the following assumptions are made by the kinetic molecular theory of gases? I. Gas molecules move randomly in all directions. II. Gas molecules exhibit negligible intermolecular forces. III. Gas molecules have negligible volume. rt oo* M IA 01 M ^ >0 tO a. I and II only b. I and III only -- 01 O 0s w a 19 "" 1 I g « * -s E o °- S "' S 2 C r^ m J u 5 IA u <T3 E1 g °3 ^ t^ IT) \O d. II only e. I, II, and III C. I only 4. BLfllJfr Which of the following represents the greatest pressure? 10 ^ 0 TJ a. 2.5 atm b. 200 kPa C. 960 mmHg d. 21 psi e. 790 torr 5. iilfi'a Which description best describes the situation shown in the diagram below? •% _1kg 1kg 80°C 536 MHR-Unit5 Gases a. a gas expanding as temperature increases and pressure remains constant b. a gas expanding as temperature and pressure rema constant c. a gas contracting as temperature increases and pressure remains constant d. a gas expanding as pressure remains constant e. a gas expanding as temperature and pressure cham 6. •»• A sample of gas is in a sealed flexible container at a fixed temperature. If the pressure on the containe is reduced by half, the volume will a. increase by a factor of 2 b. increase by a factor of 4 c. increase by a factor of 1 d. decrease by a factor of 2 e. decrease by a factor of 4 7. •!«'• A sample of nitrogen gas is placed in a sealed 2 L flexible container. Which of the following will occur if the temperature of the gas is increased? I. The pressure of the gas will increase. II. The volume of the gas will decrease. III. The speed of the gas molecules will increase. a. I and II b. I and III c. I only d. II only e. I, II, and III 8. •«• What temperature on the Kelvin scale corresponds to — 35°C? a. 238 d. 333 b. 293 e. 35 c. 308 9- •:««» Identify the choice that best describes this statement: "The volume of a fixed amount of gas is directly proportional to its temperature at a constant pressure." a. Boyle's law b. Charles's law C. Gay-Lussac's law d. kinetic molecular theory of gases e. Avogadro's law 10. •«• A sample of argon gas is stored in a container with a fixed volume at 1.00 atm of pressure. The temperature, in K, of the gas is doubled. What is the new pressure of the gas assuming the amount of gas is constant? a. 0.5 atm d. 3.00 atm b. 1.00 atm e. 4.00 atm c. 2.00 atm Use sentences and diagrams, as appropriate, to answer the questions below. Pressure and Volume Measurements of a Gas 12. flCSP Each of the following observations relate to properties of a gas. Name the property observed, and explain the observation. a. Gaseous oxygen and carbon dioxide are placed in a sealed flask. After several minutes, the gases are evenly distributed through the flask. b. Air bubbled through water in a fish tank rises to the surface and is released above the water. 15. •:«»» Torricelli and Pascal performed many important studies of atmospheric pressure. a. What is meant by the terms atmospheric pressure and standard atmospheric pressure? p( Explain what the term "millimetres of mercury" refers to and how it relates to the discovery of atmospheric pressure. 16. VUUP Determine the following conversions. a. 551 kPatopsi b. 6.0 psi to mrnHg C. 0.52 atm to kPa d. 902 mbar to mmHg and kPa 17. QdS An investigation to verify Boyle's law was conducted. The data for the investigation are shown in the table at the top of the next column. 100 2.5 X 102 75 3.3 X 102 50 25 15 5.0 X 102 l . O X 103 X a. Calculate the missing value for volume, indicated as x in the table. b. Use the data to construct a graph that shows the *S relationship between pressure and volume of a gas. I Explain how it demonstrates Boyle's law. / C 18. 13. •:«»• In your own words, describe the kinetic molecular theory of gases and its assumptions. flESfe When food is being preserved by canning, a jar is filled with very hot food, leaving a space at the top of the jar. A rubber seal is placed on top of the jar, and the lid is screwed shut. After several minutes, a "pop" is heard, and the metal lid is observed to be dented inward. Explain these observations using the kinetic molecular theory and the properties of gases. Volume (ml) Pressure (kPa) 11. flQB Construct a graphic organizer to compare the properties of gases with those of liquids. Include major similarities and differences. A sample of neon has a volume of 239 mL at 202.7 kPa of pressure. What is the pressure when the volume is 500 mL? Assume the amount and temperature of the gas are constant. 19. •«• What is absolute zeroj Dewribe a series of experiments that could be performed that would permit you to be able to determine its value. 2Q. flOB Use a graphic organizer to compare and contrast the Kelvin and Celsius scales. 21. mum A sample of gas is heated from 273 K to 290 K. If its original volume was 2.0 x 102 mL, what is the volume after being heated? Assume amount and pressure of the gas are constant. 22. mum A ball filled with air has a volume of 3.4 L at 25°C. What is its volume at 3.0°C, assuming constant amount and pressure of air? 23. C3B Xenon gas is placed in a light bulb at 300 atm and 20°C. When the bulb is in use, the bulb temperature rises to 85°C. What does the pressure in kPa become when the bulb is in use, assuming the amount and volume of gas are constant? 24. mam Air in a ball has a pressure of 11.0 psi and a temperature of 25.0°C. The temperature of the air rises to 45.0°C. Calculate the new pressure of the air, assuming a constant amount and volume of air. 25. flOB Describe one common occurrence or technology that illustrates each of the gas laws that you learned about in this chapter. (JUULotfULA jH^JL&B Self-Check If you missed question ... 1 2 Review 11.1 11.1 section (s)... 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 | 20 21 I 22 23 24 ! 1 \ n.il.ii.2 11.2 11.2 11.2, 11.3 11.3 11.3 11.1 11.1 11.1 11.3 25 | | 11.1 11.2 11.2 11.2 11.2 11.2U1.3 11.3lll.3 11.3 11.3J11.2, 1 11.2 \ 11.3 | Chapter 11 Properties of Gases • MHR 537