Physics, 6th EditionChapter 19. Thermal Properties of Matter Chapter 19. Thermal Properties of Matter General Gas Laws: Initial and Final States 19-1. An ideal gas occupies a volume of 4.00 m3 at 200 kPa absolute pressure. What will be the new pressure if the gas is slowly compressed to 2.00 m3 at constant temperature? PV 1 1 = PV 2 2; PV (200 kPa)(4.00 m3 ) 1 1 = V2 (2.00 m3 ) P2 = P2 = 400 kPa 19-2. The absolute pressure of a sample of ideal gas is 300 kPa at a volume of 2.6 m3. If the pressure decreases to 101 kPa at constant temperature, what is the new volume? PV 1 1 = PV 2 2; V2 = PV (300 kPa)(2.6 m3 ) 1 1 = P2 (101 kPa) V2 = 7.72 m3 19-3. Two hundred cubic centimeters of an ideal gas at 200C expands to a volume of 212 cm3 at constant pressure. What is the final temperature. V1 V2 = ; T1 T2 [ T1 = 200 + 2730 = 293 K ] TV (293 K)(212 cm 3 ) 1 2 T2 = = ; V1 200 cm3 tC = 310.60 – 2730 = 37.60C; T2 = 310.6 K tC = 37.60C 19-4. The temperature of a gas sample decreases from 550C to 250C at constant pressure. If the initial volume was 400 mL, what is the final volume? T1 = 550 + 2730 = 328 K; V1 V2 = ; T1 T2 V2 = T2 = 250 + 2730 = 298 K; T2V1 (298 K)(400 mL) = T1 328 K 2 V1 = 400 mL V2 = 363 mL A steel cylinder contains an ideal gas at 270C.3 kPa)(352 K) 1 2 = = 283.00 kg – 1. The absolute pressure of a sample of gas initially at 300 K doubles as the volume remains constant. What is the new temperature? P1 P2 = . The gauge pressure is 140 kPa. P1 P2 = .3 kPa. Leaked = 200 g 19-8. T1 T2 T2 = [ P2 = 2P1 ] P2T1 (2 P1 )(300 K) . Five liters of a gas at 250C has an absolute pressure of 200 kPa. Thermal Properties of Matter 19-5.Physics. T2 = 790 + 2730 = 352 K. but the absolute pressure decreases from 500 kPa down to 450 kPa.80 kg = 0.31 L 3 . what is the new gauge pressure? T1 = 270 + 2730 = 300 K.3 kPa.00 kg of an ideal gas. How many grams of the gas leaked out over night? PV PV 1 1 = 2 2.1 kPa T1 300 K Gauge Pressure = 283. = P1 P1 T2 = 600 K 19-7.200 kg. what is the final volume? T1 = 250 + 2730 = 298 K. If the temperature of the container increases to 790C. T1 T2 V2 = PV (200 kPa)(5 L)(333 K) 1 1T2 = T1 P2 (298 K)(120 kPa) V2 = 9. 6th Edition Chapter 19.00 kg)(450 kPa) = .80 kg Lost = 2.1 kPa – 101. Amt. P1 = 140 kPa + 101.3 kPa = 241. Overnight the temperature and volume remain constant. P2 = 182 kPa 19-6. T2 = 600 + 2730 = 333 K PV PV 1 1 = 2 2. T1 T2 P2 = PT (241. m1T1 m2T2 m2 = m1 P2 (2. P1 500 kPa m2 = 1. If the absolute pressure reduces to 120 kPa and the temperature increases to 600C. A steel cylinder contains 2. 7 lb/in.2 + 14. what is the temperature of the discharged air? PV PV 1 1 = 2 2.2 – 14. T2 = 900 + 2730 = 363 K V1 V2 = . What will be the new pressure if the same sample of gas is placed into a 3-L container at 70C.7 lb/in. What is the new volume? T1 = 100 + 2730 = 283 K. P1 P2 = .2 = 37.2 = 44. An air compressor takes in 2 m3 of air at 200C and one atmosphere (101. what is the new gauge pressure? P1 = 30 lb/in.4 lb/in.2 T1 (277 K) P2 = 52. T1 283 K V2 = 1. T2 = 70 + 2730 = 280 K ] PV PV 1 1 = 2 2.2. The inside of an automobile tire is under a gauge pressure of 30 lb/in. T1 T2 P2 = PV (600 kPa)(6 L)(280 K) 1 1T2 = . TV (330 K)(3 L) 1 2 P2 = 1020 kPa 19-11.1 lb/in. T1 T2 P2 = T2 = 2730 + 500 = 323 K PT (44. 6th Edition Chapter 19. If 0. T1 T2 T2 = [ T1 = 200 + 2730 = 293 K ] PV (1500 kPa)(0.1 lb/in.8 L) = .0 m 3 ) 1 1 T2 = 651 K 19-10. If the compressor discharges into a 0.2 T1 = 2730 + 40 = 277 K.3 m3 )(293 K) 2 2T1 = .4 lb/in.2 )(323 K) 1 2 = = 52. Thermal Properties of Matter 19-9. the inside air temperature rises to 500C.7 lb/in. PV (101.3-m3 tank at an absolute pressure of 1500 kPA. After several hours.3 kPa)(2. A 6-L tank holds a sample of gas under an absolute pressure of 600 kPa and a temperature of 570C. [ T1 = 570 + 2730 = 330 K. Assuming constant volume.2 at 40C.8 –L of a gas at 100C is heated to 900C at constant pressure.2 .7 lb/in.3 kPa) pressure. T1 T2 V2 = T2V1 (363 K)(0. 4 P2 = 37.03 L 19-12.Physics. How many moles of gas are there in 400 g of nitrogen gas (M = 28 g/mole)? How many molecules are in this sample? n= m 400 g = . M m = nM = (4 mol)(29 g/mol).3 mol)(6. A 2-L sample of gas has an absolute pressure of 300 kPa at 300 K.7 mol 19-15.Physics.0 moles of hydrogen? How many grams of air (M = 29 g/mol) are there in 3. If both pressure and volume experience a two-fold increase.023 x 1023molecules/mol) . M 28 g/mol n = 14. Thermal Properties of Matter 19-13. What is the mass of a 4-mol sample of air (M = 29 g/mol)? n= m . m = 87. 6th Edition Chapter 19.0 g 5 . T1 T2 T2 = PV (600 kPa)(4 L)(300 K) 2 2T1 = PV (300 kPa)(2 L) 1 1 T2 = 1200 K Molecular Mass and the Mole 19-14. M 29 g/mol n = 20. what is the final temperature? PV PV 1 1 = 2 2. m = 6. N = 8.00 g n= m .60 x 1024 molecules 19-16. m = 116 g 19-17. M m = nM = (3 mol)(2 g/mol). How many moles are contained in 600 g of air (M = 29 g/mole)? n= m 600 g = .0 moles of air? n= m . How many grams of hydrogen gas (M = 2 g/mol) are there in 3. M m = nM = (3 mol)(29 g/mol).3 mol n= N NA N = nN A = (14. How many molecules of hydrogen gas (M = 2 g/mol) are needed to have the same mass as 4 g of oxygen (M = 32 g/mol)? How many moles are in each sample? m N n= = . M H 2 g/mol nO = nH = 2 mol mO 4g = .314 J/mol ⋅ K) T = 313 K. 6 tC = 39. 000 Pa)(0.20 x 1024 molecules of H2 nH = mH 4g = . 1000 g m = 5. What is the mass of one molecule of oxygen (M = 32 g/mol)? n= m N = .125 mol *19-19. N A 6. M NA m= NM (1 molecule)(44 g/mol) = .31 x 10-23g .31 x 10-23 g m = 7.31 x 10-23 g m = 5.026 m3 ) = nR (3 mol)(8. M NA m= NM (1 molecule)(32 g/mol) = . Three moles of an ideal gas have a volume of 0.31 x 10-23g .023 x 1023molecules/mol 1 kg m = 7. What is the temperature of the gas in degrees Celsius? PV = nRT . 6th Edition Chapter 19.Physics.31 x 10-26 kg *19-20. M NA mH N A (4 g)(6. T= PV (300. The molecular mass of CO2 is 44 g/mol. M O 32 g/mol nH = 0.31 x 10-26 kg The Ideal Gas Law 19-21.026 m3 and a pressure of 300 kPa. What is the mass of a single molecule of CO2? n= m N = . tC = 3130 – 2730.70C . Thermal Properties of Matter *19-18. N A 6. 1000 g m = 7.023 x 1023molecules/mol 1 kg m = 5.023 x 1023 molecules/mol) NH = = MH 2 g/mol NH = 1. RT (353 K)(8. what is the mass of gas in the tank? [ T = 270 + 2730 = 300 K ] PV = m RT .40 x 10-3 m3.3 kPa ] PV = m RT .314 J/mol ⋅ K)(300 K) = N AV (6. M m = 3854 g. V = 6.08 x 106 Pa 19-23.314 J/mol ⋅ K)(300 K) = . 6th Edition Chapter 19. P = 101. MV (29 g/mol)(16 x 10-3m 3 ) P = 1.314 J/mol ⋅ K) 7 . M m= MPV (28 g/mol)(500.000 Pa)(2 m3 ) = . m = 3. A 16-L tank contains 200 g of air (M = 29 g/mol) at 270C. M V= mRT (8 g)(8. V = 16 L = 16 x 10-3 m3 ] P= mRT (200 g)(8. How many kilograms of nitrogen gas (M = 28 g/mol) will occupy a volume of 2000 L at an absolute pressure of 202 kPa and a temperature of 800C? [ T = (80 + 273) = 353 K ] MPV (28 g/mol)(202.40 L 19-25. NA P= NRT (2 x 1023 molecules)(8. What is the absolute pressure of this sample? PV = m RT . What volume is occupied by 8 g of nitrogen gas (M = 28 g/mol) at standard temperature and pressure (STP)? [ T = 273 K. What is the absolute gas pressure? PV = N RT . If the temperature is 270C. A 2 m3 tank holds nitrogen gas (M = 28 g/mole) under a gauge pressure of 500 kPa.40 x 10-3 m3 MP (28 g/mol)(101.85 kg 19-24. Thermal Properties of Matter 19-22.000 Pa)(2 m3 ) m= = .300 Pa) V = 6.023 x 1023molecules/mol)(2 x 10-3 m3 ) P = 414 kPa 19-26. V = 6.2 kg RT (300 K)(8. M [ T = 270 + 2730 = 300 K.314 J/mol ⋅ K)(273 K) = .314 J/mol ⋅ K) m PV = RT . m = 11. A 2-L flask contains 2 x 1023 molecules of air (M = 29 g/mol) at 300 K.Physics. If the air temperature is 200C and the dew point is 120C. and the relative humidity is found as follows: Relative Humidity = 10. 17.8% 19-30. 22. RT (273 K)(8. What is the dew point? From the table.8 mmHg Dew point = 23. How many moles of gas are contained in a volume of 2000 cm3 at conditions of standard temperature and pressure (STP)? PV = nRT . what is the relative humidity? Remember that the actual vapor pressure at a given temperature is the same as the saturated vapor pressure for the dew-point temperature.55 x 108 Pa Humidity 19-29.3 x 10-6 m3 ) P = 2.0893 mol 19-28. What is its absolute pressure assuming the water vapor is an ideal gas? [ T = 3400 + 2730 = 613 K ] PV = m RT . A 0.64 mmHg = 0. M P= mRT (0. What is the relative humidity when the air temperature is 240C? Re lative Humidity = 17. saturation vapor pressure is 28. at 280C.30 cm3 cylinder contains 0.608.3 mmHg) = 21.314 J/mol ⋅ K) n = 0.3 mmHg. Thermal Properties of Matter 19-27.27 g)(8. hum.314 J/mol ⋅ K)(613 K) = MV (18 g/mol)(0.300 Pa)(2 x 10-3m 3 ) n= = . PV (101. vapor pressure = 0.4 mmHg 60. 6th Edition Chapter 19.5 mmHg .4 mmHg Rel. The table values for 200C and 120C are used here. = 78.50C 8 From Table .27 g of water vapor (M = 18 g/mol) at 3400C.77(28.1% 19-31. The dew point is 200C.Physics. The relative humidity is 77 percent when the air temperature is 280C. V2 = 10 L = 10 x 10-3 m3.8 ft 3 ) 2 2 = . T1 T2 T2 = PV (230 kPa)(10 L)(280 K) . T1 = 70 + 2730 = 280 K PV PV 1 1 = 2 2.230 mol)(8. P1 14. A 3-L container is filled with 0.7 lb/in. P = 191 kPa V 3 x 10-3 m3 n= N NA N = nNA = (0. What volume of air at one atmosphere of pressure is required to fill this tire if there is no change in temperature or volume? [ P2 = 70 lb/in.314 J/mol ⋅ K)(300 K) = . A tractor tire contains 2. V2 = 16.8 mm Hg ] x = 0. Find its temperature when the volume reduces to 10 L and the pressure increases to 230 kPa.8% Challenge Problems 19-34. Thermal Properties of Matter 19-32.230 mol of an ideal gas at 300 K.023 x 1023 molecules/mol) = 1.4 mm Hg 19-33.3 mmHg Rel Hum.618 28. A sample of gas occupies 12 L at 70C and at an absolute pressure of 102 kPa. vapor pressure is 31. What is the pressure of water vapor in the air on a day when the temperature is 860F and the relative humidity is 80 percent? [ For 860F the sat.7 lb/in. 2 2T1 = PV (102 kPa)(12 L m3 ) 1 1 T2 = 526 K 19-35. = 61.1 ft3.5 mmHg = 0.39 x 1023 molecules 9 . V1 = 12 L = 12 x 10-3 m3. The air temperature in a room during the winter is 280C. What is the pressure of the gas? How many molecules are in this sample of gas? P= nRT (0. V2 = PV (84.2 + 14. 31.80 .2 )(2.7 lb/in. 6th Edition Chapter 19.2 ] PV 1 1 = PV 2 2.7 lb/in.2 = 84.Physics.8 mmHg x = 25.8 ft3 of air at a gauge pressure of 70 lb/in.2 19-36. What is the relative humidity if moisture first starts forming on a window when the temperature of its surface is at 200C? 17.2.230 mol)(6. Thermal Properties of Matter 19-37. M n = 0.2 + 14.2 )(5 x 10−3m 3 ) RT . How many moles of helium gas (M = 4 g/mol) are there in a 6-L tank when the pressure is 2 x 105 Pa and the temperature is 270C? What is the mass of the helium? PV (2 x 106 Pa)(6 x 10-3m 3 ) n= = .7 lb/in.314 J/mol ⋅ K)(300 K) n= m . the dewpoint is: 67.Physics. The air temperature inside a car is 260C. m = 1.2 = 45.2 Assume the volume of the tire is 5000 cm3 and its temperature is 270C? P = 31 lb/in. m = = .8 mmHg) = 17.16 x 10-8 kg 19-39. The lens in a sensitive camera is clear when the room temperature is 71.7 lb/in. What is the lowest temperature of the lens if it is not to become foggy from moisture? At 71. RT (8. The dew point is 240C.889 25.481 mol m = nM = (0.7 lb/in.2.9% 19-40.88(19.92 g 19-38.4 mmHg = 0.4 mmHg Thus. V = 5000 cm3 = 5 L PV = m PVM (45. saturated pressure is 19. from tables. What is the relative humidity inside the car? x= 22.16 x 10-5 g or m = 9.60F and the relative humidity is 88 percent.60C. M RT (8.2 mmHg Rel. 6th Edition Chapter 19. hum. How many grams of air (M = 29 g/mol) must be pumped into an automobile tire if it is to have a gauge pressure of 31 lb/in.80F 10 .481 mol)(4 g/mol) .314 J/mol ⋅ K)(300 K) m = 9. = 88.8 mmHg Actual pressure = 0. 32 kg/m3 *19-42. How many grams of CO2 can be added to the tank if the maximum absolute pressure is 60 atm and there is no change in temperature? [ 5000 cm3 = 5 L ] At 1 atm: m = PVM (1 atm)(5 L)(44 g/mol) = = 8. T1 = 273 K.0821 L ⋅ atm/mol ⋅ K)(300 K) At 60 atm: m = PVM (60 atm)(5 L)(44 g/mol) = = 535.Physics. The density of an unknown gas at STP is 1.0821 L ⋅ atm/mol ⋅ K)(300 K) Mass added = 535. Thermal Properties of Matter *19-41. Added = 527 g *19-43.932 g. A 5000-cm3 tank is filled with carbon dioxide (M = 44 g/mol) at 300 K and 1 atm of pressure. T2 = 600 + 2730 = 333 K PV = m RT .25 kg/m3 )(18 atm)(273 K) = .314 J/mol ⋅ K)(296 K) ρ = 1.9 g – 8.932 g RT (0. ρ = 18. ρ 2 P2T1 ρ2 = ρ1 MP1 / RT1 PT = = 1 2 ρ 2 MP2 / RT2 P2T1 ρ2 = ρ1 P2T1 PT 1 2 ρ1 P2T1 (1. M ρ= m PM (101.25 kg/m3. M ρ= m PM = V RT ρ1 PT = 1 2. What is the density of this gas at 18 atm and 600C? P1 = 1 atm. What is the density of oxygen gas (M = 32 g/mol) at a temperature of 230C and atmospheric pressure? PV = m RT .300 Pa)(32 g/mol) = = = 1320 g/m3 V RT (8. 6th Edition Chapter 19.4 kg/m3 PT (1 atm)(333 K) 1 2 11 .9 g RT (0. P2 = 18 atm. m1T1 m2T2 m2 = 0. V1 = 14 L T1 = 240 + 2730 = 297 K. The next morning it is noticed that the gauge pressure is only 300 kPa and that the temperature is 150C. During the night a leak develops in the tank.3 kPa = 401.3 kPa = 501.3 kPa = 2801. P2 = 101. T3 = 297 K PV PV 2 2 = 3 3. T1 T2 V2 = PV (2801 kPa)(14 L)(238 K) 1 1T2 = .3 kPa. m1 m2 P2T1 (401. A tank with a capacity of 14 L contains helium gas at 240C under a gauge pressure of 2700 kPa.3 kPa)(297 K) V2 = 310 L (b) P2 = P3 = 1 atm.3 kPa)(288 K) 1 2 Mass remaining = 83. m1 PT (501. T2 = 238 K.3 kPa.Physics.834 . One evening when the temperature is 270C.3 kPa)(300 K) = = . the gauge at the top of the tank indicates a pressure of 400 kPa. P2T1 (101. Thermal Properties of Matter Critical Thinking Questions *19-44. P2 = 300 kPa + 101. T2 238 K V3 = 387 L *19-45 A steel tank is filled with oxygen. What is the final volume of the balloon? (a) P1 = 2700 kPa + 101. T2 T3 V3 = V2T3 (310 L)(297 K) = . 6th Edition Chapter 19. T2 = -350 + 2730 = 238 K PV PV 1 1 = 2 2. V2 = 310 L. T2 = 2730 + 150 = 288 K.4% 12 . (A) What will be the volume of a balloon filled with this gas if the helium expands to an internal absolute pressure of 1 atm and the temperature drops to -350C? (b) Now suppose the system returns to its original temperature (240C).3 kPa T1 = 2730 + 270 = 300 K.3 kPa. What percentage of the original gas remains in the tank? P1 = 400 kPa + 101. V1 = V2 PV PV 1 1 = 2 2. 0821 L ⋅ atm/mol ⋅ K)(300 K) RT . A 2-L flask is filled with nitrogen (M = 28 g/mol) at 270C and 1 atm of absolute pressure.308 L M MP (64 g/mol)(10 atm) n= m N1 = . M m= R = 0. PV = m RT . 1022 molecules/s 13 time = 371 s .529 x 1022 molecules in original sample Now.Physics.529 x 1022 molecules N2 = = = P1 2 2 N2 = 3. P= mRT (1.0821 L ⋅ atm/mol ⋅ K)(300 K) = . P = 0. how long will it take to reduce the pressure by one-half? PV = m mRT (8 g )(0. A stopcock at the top of the flask is opened to the air and the system is heated to a temperature of 1270C. What mass of nitrogen is in the flask? What is the final pressure? (a) The mass remaining in the flask after heating to 1270C is found as follows: T = 1270 + 2730 = 400 K. m = 1.765 x 1022 molecules.750 atm MV (28 g/mol)(2 L) *19-47. Molecules lost = N1 – N2 = 3.0821 L ⋅ atm/mol ⋅ K)(400 K) (b) This same mass remains when it returns to T = 270+ 2730 = 300 K.764 x 1022 molecules . M NA N1 = N A m (6. 6th Edition Chapter 19. therefore. What is the volume of 8 g of sulfur dioxide (M = 64 g/mol) if it has an absolute pressure of 10 atm and a temperature of 300 K? If 1020 molecules leak from this volume every second.765 x 1022 molecules time = 3. V = = .705 g RT (0. V = 0. the pressure P is proportional to the number of molecules so that: P1 N1 = P2 N 2 and N1 ( 12 P1 ) N1 7.705 g)(0.0821 atm L/mol K PVM (1 atm)(2 L)(28 g/mol) = . Thermal Properties of Matter *19-46.023 x 1023 molecules/mol)(8 g) = M 64 g/mol N1 = 7. The stopcock is then closed and the system is allowed to return to 270C. MP1 (4 g/mol)(12 atm) m2 = MPV (4 g/mol)(7 atm)(1.672 g *19-49. RT2 (0. How many grams of helium have leaked out of the container? [ T1 = 570 + 2730 = 330 K. V1 = m1 RT1 = V2 . 6th Edition Chapter 19.0821 L ⋅ atm/mol ⋅ K)(290 K) m1 = 2.13 L m2 = 1. Assume V1 = V2 T1 = ? Note: The pressure inside a balloon will always be equal to the pressure of the atmosphere. A flask contains 2 g of helium (M = 4 g/mol) at 570C and 12 atm absolute pressure. What must be the temperature of the air in a hot-air balloon in order that the mass of the air be 0. MP1 T2 = m1 RT1 (0.Physics.00 g. The temperature then decreases to 170C and the pressure falls to 7 atm. T2 = 270 + 2730 = 300 K.0821 L ⋅ atm/mol ⋅ K)(330 K) = . Thermal Properties of Matter *19-48.97 14 T2 = 309 K .00 – 1. 0.129 L) 2 2 = .97m1.97 times that of an equal volume of air at a temperature of 270C? m2 = 0.328 = 0. so P1 = P2.97 0. V1 = 1. T2 = 290 K ] V1 = mRT1 (2 g)(0. since that is the pressure applied to the surface of the balloon.328 g mass lost = 2.97T2 MP1 MP2 T1 300 K = . T1 = 0.97 m1 ) RT2 = .