Solucionario Capitulo 19 - Paul E. Tippens

Physics, 6th Edition

Chapter 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



19-5. A steel cylinder contains an ideal gas at 270C. The gauge pressure is 140 kPa. If the temperature of the container increases to 790C, what is the new gauge pressure? T1 = 270 + 2730 = 300 K; T2 = 790 + 2730 = 352 K; P1 = 140 kPa + 101.3 kPa = 241.3 kPa; P1 P2 = ; T1 T2

P2 =

PT (241.3 kPa)(352 K) 1 2 = = 283.1 kPa T1 300 K

Gauge Pressure = 283.1 kPa – 101.3 kPa;

P2 = 182 kPa

19-6. 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 = ; T1 T2

T2 =

[ P2 = 2P1 ]

P2T1 (2 P1 )(300 K) ; = P1 P1

T2 = 600 K

19-7. A steel cylinder contains 2.00 kg of an ideal gas. Overnight the temperature and volume remain constant, but the absolute pressure decreases from 500 kPa down to 450 kPa. How many grams of the gas leaked out over night? PV PV 1 1 = 2 2; m1T1 m2T2

m2 =

m1 P2 (2.00 kg)(450 kPa) = ; P1 500 kPa

m2 = 1.80 kg

Lost = 2.00 kg – 1.80 kg = 0.200 kg; Amt. Leaked = 200 g 19-8. Five liters of a gas at 250C has an absolute pressure of 200 kPa. If the absolute pressure reduces to 120 kPa and the temperature increases to 600C, what is the final volume? T1 = 250 + 2730 = 298 K; T2 = 600 + 2730 = 333 K PV PV 1 1 = 2 2; T1 T2

V2 =

PV (200 kPa)(5 L)(333 K) 1 1T2 = T1 P2 (298 K)(120 kPa) V2 = 9.31 L

3



19-9. An air compressor takes in 2 m3 of air at 200C and one atmosphere (101.3 kPa) pressure. If the compressor discharges into a 0.3-m3 tank at an absolute pressure of 1500 kPA, what is the temperature of the discharged air? PV PV 1 1 = 2 2; T1 T2

T2 =

[ T1 = 200 + 2730 = 293 K ]

PV (1500 kPa)(0.3 m3 )(293 K) 2 2T1 = ; PV (101.3 kPa)(2.0 m 3 ) 1 1

T2 = 651 K

19-10. A 6-L tank holds a sample of gas under an absolute pressure of 600 kPa and a temperature of 570C. What will be the new pressure if the same sample of gas is placed into a 3-L container at 70C.

[ T1 = 570 + 2730 = 330 K; T2 = 70 + 2730 = 280 K ]

PV PV 1 1 = 2 2; T1 T2

P2 =

PV (600 kPa)(6 L)(280 K) 1 1T2 = ; TV (330 K)(3 L) 1 2

P2 = 1020 kPa

19-11. If 0.8 –L of a gas at 100C is heated to 900C at constant pressure. What is the new volume? T1 = 100 + 2730 = 283 K; T2 = 900 + 2730 = 363 K V1 V2 = ; T1 T2

V2 =

T2V1 (363 K)(0.8 L) = ; T1 283 K

V2 = 1.03 L

19-12. The inside of an automobile tire is under a gauge pressure of 30 lb/in.2 at 40C. After several hours, the inside air temperature rises to 500C. Assuming constant volume, what is the new gauge pressure? P1 = 30 lb/in.2 + 14.7 lb/in.2 = 44.7 lb/in.2 T1 = 2730 + 40 = 277 K; P1 P2 = ; T1 T2

P2 =

T2 = 2730 + 500 = 323 K

PT (44.7 lb/in.2 )(323 K) 1 2 = = 52.1 lb/in.2 T1 (277 K)

P2 = 52.1 lb/in.2 – 14.7 lb/in.2 = 37.4 lb/in.2;

4

P2 = 37.4 lb/in.2



19-13. A 2-L sample of gas has an absolute pressure of 300 kPa at 300 K. If both pressure and volume experience a two-fold increase, what is the final temperature? PV PV 1 1 = 2 2; 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. How many moles are contained in 600 g of air (M = 29 g/mole)? n=

m 600 g = ; M 29 g/mol

n = 20.7 mol

19-15. 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 28 g/mol

n = 14.3 mol

n=

N NA

N = nN A = (14.3 mol)(6.023 x 1023molecules/mol) ; N = 8.60 x 1024 molecules 19-16. What is the mass of a 4-mol sample of air (M = 29 g/mol)? n=

m ; M

m = nM = (4 mol)(29 g/mol);

m = 116 g

19-17. How many grams of hydrogen gas (M = 2 g/mol) are there in 3.0 moles of hydrogen? How many grams of air (M = 29 g/mol) are there in 3.0 moles of air? n=

m ; M

m = nM = (3 mol)(2 g/mol);

m = 6.00 g

n=

m ; M

m = nM = (3 mol)(29 g/mol);

m = 87.0 g

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*19-18. 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 NA

mH N A (4 g)(6.023 x 1023 molecules/mol) NH = = MH 2 g/mol NH = 1.20 x 1024 molecules of H2

nH =

mH 4g = ; M H 2 g/mol

nO =

nH = 2 mol

mO 4g = ; M O 32 g/mol

nH = 0.125 mol

*19-19. What is the mass of one molecule of oxygen (M = 32 g/mol)? n=

m N = ; M NA

m=

NM (1 molecule)(32 g/mol) = ; N A 6.023 x 1023molecules/mol

 1 kg  m = 5.31 x 10-23g  ;  1000 g 

m = 5.31 x 10-23 g

m = 5.31 x 10-26 kg

*19-20. The molecular mass of CO2 is 44 g/mol. What is the mass of a single molecule of CO2? n=

m N = ; M NA

m=

NM (1 molecule)(44 g/mol) = ; N A 6.023 x 1023molecules/mol

 1 kg  m = 7.31 x 10-23g  ;  1000 g 

m = 7.31 x 10-23 g

m = 7.31 x 10-26 kg

The Ideal Gas Law 19-21. Three moles of an ideal gas have a volume of 0.026 m3 and a pressure of 300 kPa. What is the temperature of the gas in degrees Celsius? PV = nRT ;

T=

PV (300, 000 Pa)(0.026 m3 ) = nR (3 mol)(8.314 J/mol ⋅ K)

T = 313 K; tC = 3130 – 2730;

6

tC = 39.70C



19-22. A 16-L tank contains 200 g of air (M = 29 g/mol) at 270C. What is the absolute pressure of this sample? PV =

m RT ; M

[ T = 270 + 2730 = 300 K; V = 16 L = 16 x 10-3 m3 ] P=

mRT (200 g)(8.314 J/mol ⋅ K)(300 K) = ; MV (29 g/mol)(16 x 10-3m 3 )

P = 1.08 x 106 Pa

19-23. 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)(202,000 Pa)(2 m3 ) m= = ; RT (353 K)(8.314 J/mol ⋅ K)

m PV = RT ; M

m = 3854 g;

m = 3.85 kg

19-24. What volume is occupied by 8 g of nitrogen gas (M = 28 g/mol) at standard temperature and pressure (STP)? [ T = 273 K, P = 101.3 kPa ] PV =

m RT ; M

V=

mRT (8 g)(8.314 J/mol ⋅ K)(273 K) = ; V = 6.40 x 10-3 m3 MP (28 g/mol)(101,300 Pa) V = 6.40 x 10-3 m3;

V = 6.40 L

19-25. A 2-L flask contains 2 x 1023 molecules of air (M = 29 g/mol) at 300 K. What is the absolute gas pressure? PV =

N RT ; NA

P=

NRT (2 x 1023 molecules)(8.314 J/mol ⋅ K)(300 K) = N AV (6.023 x 1023molecules/mol)(2 x 10-3 m3 ) P = 414 kPa

19-26. A 2 m3 tank holds nitrogen gas (M = 28 g/mole) under a gauge pressure of 500 kPa. If the temperature is 270C, what is the mass of gas in the tank? [ T = 270 + 2730 = 300 K ] PV =

m RT ; M

m=

MPV (28 g/mol)(500,000 Pa)(2 m3 ) = ; m = 11.2 kg RT (300 K)(8.314 J/mol ⋅ K)

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19-27. How many moles of gas are contained in a volume of 2000 cm3 at conditions of standard temperature and pressure (STP)? PV = nRT ;

PV (101,300 Pa)(2 x 10-3m 3 ) n= = ; RT (273 K)(8.314 J/mol ⋅ K)

n = 0.0893 mol

19-28. A 0.30 cm3 cylinder contains 0.27 g of water vapor (M = 18 g/mol) at 3400C. What is its absolute pressure assuming the water vapor is an ideal gas? [ T = 3400 + 2730 = 613 K ] PV =

m RT ; M

P=

mRT (0.27 g)(8.314 J/mol ⋅ K)(613 K) = MV (18 g/mol)(0.3 x 10-6 m3 ) P = 2.55 x 108 Pa

Humidity 19-29. If the air temperature is 200C and the dew point is 120C, 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. The table values for 200C and 120C are used here, and the relative humidity is found as follows: Relative Humidity =

10.64 mmHg = 0.608; 17.4 mmHg

60.8%

19-30. The dew point is 200C. What is the relative humidity when the air temperature is 240C? Re lative Humidity =

17.5 mmHg ; 22.4 mmHg

Rel. hum. = 78.1%

19-31. The relative humidity is 77 percent when the air temperature is 280C. What is the dew point? From the table, at 280C, saturation vapor pressure is 28.3 mmHg. vapor pressure = 0.77(28.3 mmHg) = 21.8 mmHg Dew point = 23.50C

8

From Table



19-32. 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. vapor pressure is 31.8 mm Hg ] x = 0.80 ; 31.8 mmHg

x = 25.4 mm Hg

19-33. The air temperature in a room during the winter is 280C. What is the relative humidity if moisture first starts forming on a window when the temperature of its surface is at 200C? 17.5 mmHg = 0.618 28.3 mmHg

Rel Hum. = 61.8%

Challenge Problems 19-34. A sample of gas occupies 12 L at 70C and at an absolute pressure of 102 kPa. Find its temperature when the volume reduces to 10 L and the pressure increases to 230 kPa. V1 = 12 L = 12 x 10-3 m3; V2 = 10 L = 10 x 10-3 m3; T1 = 70 + 2730 = 280 K PV PV 1 1 = 2 2; T1 T2

T2 =

PV (230 kPa)(10 L)(280 K) ; 2 2T1 = PV (102 kPa)(12 L m3 ) 1 1

T2 = 526 K

19-35. A tractor tire contains 2.8 ft3 of air at a gauge pressure of 70 lb/in.2. 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.2 + 14.7 lb/in.2 = 84.7 lb/in.2 ] PV 1 1 = PV 2 2;

V2 =

PV (84.7 lb/in.2 )(2.8 ft 3 ) 2 2 = ; V2 = 16.1 ft3. P1 14.7 lb/in.2

19-36. A 3-L container is filled with 0.230 mol of an ideal gas at 300 K. What is the pressure of the gas? How many molecules are in this sample of gas? P=

nRT (0.230 mol)(8.314 J/mol ⋅ K)(300 K) = ; P = 191 kPa V 3 x 10-3 m3

n=

N NA

N = nNA = (0.230 mol)(6.023 x 1023 molecules/mol) = 1.39 x 1023 molecules 9



19-37. 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= = ; RT (8.314 J/mol ⋅ K)(300 K) n=

m ; M

n = 0.481 mol

m = nM = (0.481 mol)(4 g/mol) ;

m = 1.92 g

19-38. 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.2 Assume the volume of the tire is 5000 cm3 and its temperature is 270C? P = 31 lb/in.2 + 14.7 lb/in.2 = 45.7 lb/in.2; V = 5000 cm3 = 5 L PV =

m PVM (45.7 lb/in.2 )(5 x 10−3m 3 ) RT ; m = = ; M RT (8.314 J/mol ⋅ K)(300 K) m = 9.16 x 10-5 g or m = 9.16 x 10-8 kg

19-39. The air temperature inside a car is 260C. The dew point is 240C. What is the relative humidity inside the car? x=

22.4 mmHg = 0.889 25.2 mmHg

Rel. hum. = 88.9%

19-40. The lens in a sensitive camera is clear when the room temperature is 71.60F and the relative humidity is 88 percent. What is the lowest temperature of the lens if it is not to become foggy from moisture? At 71.60C, saturated pressure is 19.8 mmHg Actual pressure = 0.88(19.8 mmHg) = 17.4 mmHg Thus, from tables, the dewpoint is: 67.80F

10



*19-41. What is the density of oxygen gas (M = 32 g/mol) at a temperature of 230C and atmospheric pressure? PV =

m RT ; M

ρ=

m PM (101,300 Pa)(32 g/mol) = = = 1320 g/m3 V RT (8.314 J/mol ⋅ K)(296 K)

ρ = 1.32 kg/m3

*19-42. A 5000-cm3 tank is filled with carbon dioxide (M = 44 g/mol) at 300 K and 1 atm of pressure. 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) = = 8.932 g RT (0.0821 L ⋅ atm/mol ⋅ K)(300 K)

At 60 atm: m =

PVM (60 atm)(5 L)(44 g/mol) = = 535.9 g RT (0.0821 L ⋅ atm/mol ⋅ K)(300 K)

Mass added = 535.9 g – 8.932 g; Added = 527 g

*19-43. The density of an unknown gas at STP is 1.25 kg/m3. What is the density of this gas at 18 atm and 600C? P1 = 1 atm; T1 = 273 K; P2 = 18 atm; T2 = 600 + 2730 = 333 K PV =

m RT ; M

ρ=

m PM = V RT

ρ1 PT = 1 2; ρ 2 P2T1 ρ2 =

ρ1 MP1 / RT1 PT = = 1 2 ρ 2 MP2 / RT2 P2T1 ρ2 =

ρ1 P2T1 PT 1 2

ρ1 P2T1 (1.25 kg/m3 )(18 atm)(273 K) = ; ρ = 18.4 kg/m3 PT (1 atm)(333 K) 1 2

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Critical Thinking Questions *19-44. A tank with a capacity of 14 L contains helium gas at 240C under a gauge pressure of 2700 kPa. (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). What is the final volume of the balloon? (a) P1 = 2700 kPa + 101.3 kPa = 2801.3 kPa; P2 = 101.3 kPa; V1 = 14 L T1 = 240 + 2730 = 297 K; T2 = -350 + 2730 = 238 K PV PV 1 1 = 2 2; T1 T2

V2 =

PV (2801 kPa)(14 L)(238 K) 1 1T2 = ; P2T1 (101.3 kPa)(297 K)

V2 = 310 L

(b) P2 = P3 = 1 atm; T2 = 238 K, V2 = 310 L, T3 = 297 K PV PV 2 2 = 3 3; T2 T3

V3 =

V2T3 (310 L)(297 K) = ; T2 238 K

V3 = 387 L

*19-45 A steel tank is filled with oxygen. One evening when the temperature is 270C, the gauge at the top of the tank indicates a pressure of 400 kPa. During the night a leak develops in the tank. The next morning it is noticed that the gauge pressure is only 300 kPa and that the temperature is 150C. What percentage of the original gas remains in the tank? P1 = 400 kPa + 101.3 kPa = 501.3 kPa;

P2 = 300 kPa + 101.3 kPa = 401.3 kPa

T1 = 2730 + 270 = 300 K; T2 = 2730 + 150 = 288 K; V1 = V2 PV PV 1 1 = 2 2; m1T1 m2T2 m2 = 0.834 ; m1

m2 P2T1 (401.3 kPa)(300 K) = = ; m1 PT (501.3 kPa)(288 K) 1 2 Mass remaining = 83.4%

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*19-46. A 2-L flask is filled with nitrogen (M = 28 g/mol) at 270C and 1 atm of absolute pressure. A stopcock at the top of the flask is opened to the air and the system is heated to a temperature of 1270C. The stopcock is then closed and the system is allowed to return to 270C. 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; PV =

m RT ; M

m=

R = 0.0821 atm L/mol K

PVM (1 atm)(2 L)(28 g/mol) = ; m = 1.705 g RT (0.0821 L ⋅ atm/mol ⋅ K)(400 K)

(b) This same mass remains when it returns to T = 270+ 2730 = 300 K, therefore, P=

mRT (1.705 g)(0.0821 L ⋅ atm/mol ⋅ K)(300 K) = ; P = 0.750 atm MV (28 g/mol)(2 L)

*19-47. 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, how long will it take to reduce the pressure by one-half? PV =

m mRT (8 g )(0.0821 L ⋅ atm/mol ⋅ K)(300 K) RT ; V = = ; V = 0.308 L M MP (64 g/mol)(10 atm) n=

m N1 = ; M NA

N1 =

N A m (6.023 x 1023 molecules/mol)(8 g) = M 64 g/mol

N1 = 7.529 x 1022 molecules in original sample Now, the pressure P is proportional to the number of molecules so that: P1 N1 = P2 N 2

and

N1 ( 12 P1 ) N1 7.529 x 1022 molecules N2 = = = P1 2 2

N2 = 3.765 x 1022 molecules; Molecules lost = N1 – N2 = 3.765 x 1022 molecules time =

3.764 x 1022 molecules ; 1022 molecules/s 13

time = 371 s



*19-48. A flask contains 2 g of helium (M = 4 g/mol) at 570C and 12 atm absolute pressure. The temperature then decreases to 170C and the pressure falls to 7 atm. How many grams of helium have leaked out of the container? [ T1 = 570 + 2730 = 330 K; T2 = 290 K ] V1 =

mRT1 (2 g)(0.0821 L ⋅ atm/mol ⋅ K)(330 K) = ; MP1 (4 g/mol)(12 atm)

m2 =

MPV (4 g/mol)(7 atm)(1.129 L) 2 2 = ; RT2 (0.0821 L ⋅ atm/mol ⋅ K)(290 K) m1 = 2.00 g;

V1 = 1.13 L

m2 = 1.328 g

mass lost = 2.00 – 1.328 = 0.672 g

*19-49. What must be the temperature of the air in a hot-air balloon in order that the mass of the air be 0.97 times that of an equal volume of air at a temperature of 270C? m2 = 0.97m1; T2 = 270 + 2730 = 300 K; Assume V1 = V2

T1 = ?

Note: The pressure inside a balloon will always be equal to the pressure of the atmosphere, since that is the pressure applied to the surface of the balloon, so P1 = P2. V1 =

m1 RT1 = V2 ; MP1 T2 =

m1 RT1 (0.97 m1 ) RT2 = ; T1 = 0.97T2 MP1 MP2

T1 300 K = ; 0.97 0.97

14

T2 = 309 K

Solucionario Capitulo 19 - Paul E. Tippens

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