1. Chapter 19. Thermal Properties of Matter Physics, 6th Edition
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?
PV1 (200 kPa)(4.00 m3 )
PV1 = PV2 ;
1 2 P2 = 1
= P2 = 400 kPa
V2 (2.00 m3 )
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?
PV1 (300 kPa)(2.6 m3 )
PV1 = PV2 ;
1 2 V2 = 1
= V2 = 7.72 m3
P2 (101 kPa)
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. [ T1 = 200 + 2730 = 293 K ]
V1 V2 TV2 (293 K)(212 cm 3 )
= ; T2 = 1 = ; T2 = 310.6 K
T1 T2 V1 200 cm3
tC = 310.60 – 2730 = 37.60C; 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; T2 = 250 + 2730 = 298 K; V1 = 400 mL
V1 V2 T2V1 (298 K)(400 mL)
= ; V2 = = V2 = 363 mL
T1 T2 T1 328 K
2

2. Chapter 19. Thermal Properties of Matter Physics, 6th Edition
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;
P P2 PT2 (241.3 kPa)(352 K)
1
= ; P2 = 1
= = 283.1 kPa
T1 T2 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? [ P2 = 2P1 ]
P P2 P2T1 (2 P )(300 K) ; T2 = 600 K
1
= ; T2 = = 1
T1 T2 P1 P1
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?
PV1 PV2 m1 P2 (2.00 kg)(450 kPa)
1
= 2 ; m2 = = ; m2 = 1.80 kg
m1T1 m2T2 P1 500 kPa
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
PV1 PV2 PV1T2 (200 kPa)(5 L)(333 K)
1
= 2 ; V2 = 1
=
T1 T2 T1 P2 (298 K)(120 kPa)
V2 = 9.31 L
3

3. Chapter 19. Thermal Properties of Matter Physics, 6th Edition
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? [ T1 = 200 + 2730 = 293 K ]
PV1 PV2 PV2T1 (1500 kPa)(0.3 m3 )(293 K)
1
= 2 ; T2 = 2
= ; T2 = 651 K
T1 T2 PV1
1 (101.3 kPa)(2.0 m 3 )
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 ]
PV1 PV2 PV1T2 (600 kPa)(6 L)(280 K)
1
= 2 ; P2 = 1
= ; P2 = 1020 kPa
T1 T2 TV21 (330 K)(3 L)
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 T2V1 (363 K)(0.8 L)
= ; V2 = = ; V2 = 1.03 L
T1 T2 T1 283 K
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; T2 = 2730 + 500 = 323 K
P P2 PT2 (44.7 lb/in.2 )(323 K)
1
= ; P2 = 1
= = 52.1 lb/in.2
T1 T2 T1 (277 K)
P2 = 52.1 lb/in.2 – 14.7 lb/in.2 = 37.4 lb/in.2; P2 = 37.4 lb/in.2
4

4. Chapter 19. Thermal Properties of Matter Physics, 6th Edition
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?
PV1 PV2 PV2T1 (600 kPa)(4 L)(300 K)
1
= 2 ; T2 = 2
= T2 = 1200 K
T1 T2 PV1
1 (300 kPa)(2 L)
Molecular Mass and the Mole
19-14. How many moles are contained in 600 g of air (M = 29 g/mole)?
m 600 g
n= = ; n = 20.7 mol
M 29 g/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?
m 400 g N
n= = ; n = 14.3 mol n=
M 28 g/mol 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)?
m
n= ; m = nM = (4 mol)(29 g/mol); m = 116 g
M
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?
m
n= ; m = nM = (3 mol)(2 g/mol); m = 6.00 g
M
m
n= ; m = nM = (3 mol)(29 g/mol); m = 87.0 g
M
5

5. Chapter 19. Thermal Properties of Matter Physics, 6th Edition
*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 mH N A (4 g)(6.023 x 1023 molecules/mol)
n= = ; NH = =
M NA MH 2 g/mol
NH = 1.20 x 1024 molecules of H2
mH 4g mO 4g
nH = = ; nH = 2 mol nO = = ; nH = 0.125 mol
M H 2 g/mol M O 32 g/mol
*19-19. What is the mass of one molecule of oxygen (M = 32 g/mol)?
m N NM (1 molecule)(32 g/mol)
n= = ; m= = ; m = 5.31 x 10-23 g
M NA N A 6.023 x 1023 molecules/mol
 1 kg 
m = 5.31 x 10-23g  ; m = 5.31 x 10-26 kg
 1000 g 
*19-20. The molecular mass of CO2 is 44 g/mol. What is the mass of a single molecule of CO2?
m N NM (1 molecule)(44 g/mol)
n= = ; m= = ; m = 7.31 x 10-23 g
M NA N A 6.023 x 1023 molecules/mol
 1 kg 
m = 7.31 x 10-23g  ; m = 7.31 x 10-26 kg
 1000 g 
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 (300, 000 Pa)(0.026 m3 )
PV = nRT ; T= =
nR (3 mol)(8.314 J/mol ⋅ K)
T = 313 K; tC = 3130 – 2730; tC = 39.70C
6

6. Chapter 19. Thermal Properties of Matter Physics, 6th Edition
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? [ T = 270 + 2730 = 300 K; V = 16 L = 16 x 10-3 m3 ]
m mRT (200 g)(8.314 J/mol ⋅ K)(300 K)
PV = RT ; P= = ; P = 1.08 x 106 Pa
M MV (29 g/mol)(16 x 10-3m 3 )
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 ]
m MPV (28 g/mol)(202,000 Pa)(2 m3 )
PV = RT ; m= = ;
M RT (353 K)(8.314 J/mol ⋅ K)
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 ]
m mRT (8 g)(8.314 J/mol ⋅ K)(273 K)
PV = RT ; V= = ; V = 6.40 x 10-3 m3
M 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?
N NRT (2 x 1023 molecules)(8.314 J/mol ⋅ K)(300 K)
PV = RT ; P= =
NA 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 ]
m MPV (28 g/mol)(500,000 Pa)(2 m3 )
PV = RT ; m= = ; m = 11.2 kg
M RT (300 K)(8.314 J/mol ⋅ K)
7

7. Chapter 19. Thermal Properties of Matter Physics, 6th Edition
19-27. How many moles of gas are contained in a volume of 2000 cm3 at conditions of standard
temperature and pressure (STP)?
PV (101,300 Pa)(2 x 10-3m 3 )
PV = nRT ; n= = ; n = 0.0893 mol
RT (273 K)(8.314 J/mol ⋅ K)
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 ]
m mRT (0.27 g)(8.314 J/mol ⋅ K)(613 K)
PV = RT ; P= =
M 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:
10.64 mmHg
Relative Humidity = = 0.608; 60.8%
17.4 mmHg
19-30. The dew point is 200C. What is the relative humidity when the air temperature is 240C?
17.5 mmHg
Re lative Humidity = ; Rel. hum. = 78.1%
22.4 mmHg
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 From Table
8

8. Chapter 19. Thermal Properties of Matter Physics, 6th Edition
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 ; x = 25.4 mm Hg
31.8 mmHg
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 Rel Hum. = 61.8%
28.3 mmHg
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
PV1 PV2 PV2T1 (230 kPa)(10 L)(280 K) ; T2 = 526 K
1
= 2 ; T2 = 2
=
T1 T2 PV1
1 (102 kPa)(12 L m3 )
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 ]
PV2 (84.7 lb/in.2 )(2.8 ft 3 )
PV1 = PV2 ;
1 2 V2 = 2
= ; V2 = 16.1 ft3.
P 1 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?
nRT (0.230 mol)(8.314 J/mol ⋅ K)(300 K) N
P= = ; P = 191 kPa n=
V 3 x 10-3 m3 NA
N = nNA = (0.230 mol)(6.023 x 1023 molecules/mol) = 1.39 x 1023 molecules
9

9. Chapter 19. Thermal Properties of Matter Physics, 6th Edition
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 10-3 m3 )
n= = ; n = 0.481 mol
RT (8.314 J/mol ⋅ K)(300 K)
m
n= ; m = nM = (0.481 mol)(4 g/mol) ; m = 1.92 g
M
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
m PVM (45.7 lb/in.2 )(5 x 10−3m 3 )
PV = 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?
22.4 mmHg
x= = 0.889 Rel. hum. = 88.9%
25.2 mmHg
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

10. Chapter 19. Thermal Properties of Matter Physics, 6th Edition
*19-41. What is the density of oxygen gas (M = 32 g/mol) at a temperature of 230C and
atmospheric pressure?
m m PM (101,300 Pa)(32 g/mol)
PV = RT ; ρ= = = = 1320 g/m3
M 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 ]
PVM (1 atm)(5 L)(44 g/mol)
At 1 atm: m = = = 8.932 g
RT (0.0821 L ⋅ atm/mol ⋅ K)(300 K)
PVM (60 atm)(5 L)(44 g/mol)
At 60 atm: m = = = 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
m m PM ρ1 MP / RT1 PT2
PV = RT ; ρ= = = 1
= 1
M V RT ρ 2 MP2 / RT2 P2T1
ρ1 PT2 ρ1 P2T1
= 1 ; ρ2 =
ρ 2 P2T1 PT2
1
ρ1 P2T1 (1.25 kg/m3 )(18 atm)(273 K)
ρ2 = = ; ρ = 18.4 kg/m3
PT2
1 (1 atm)(333 K)
11

11. Chapter 19. Thermal Properties of Matter Physics, 6th Edition
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
PV1 PV2 PV1T2 (2801 kPa)(14 L)(238 K)
1
= 2 ; V2 = 1
= ; V2 = 310 L
T1 T2 P2T1 (101.3 kPa)(297 K)
(b) P2 = P3 = 1 atm; T2 = 238 K, V2 = 310 L, T3 = 297 K
PV2 PV3 V2T3 (310 L)(297 K)
2
= 3 ; V3 = = ; V3 = 387 L
T2 T3 T2 238 K
*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
PV1 PV2 m2 P2T1 (401.3 kPa)(300 K)
1
= 2 ; = = ;
m1T1 m2T2 m1 PT2 (501.3 kPa)(288 K)
1
m2
= 0.834 ; Mass remaining = 83.4%
m1
12

12. Chapter 19. Thermal Properties of Matter Physics, 6th Edition
*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; R = 0.0821 atm L/mol K
m PVM (1 atm)(2 L)(28 g/mol)
PV = RT ; m= = ; m = 1.705 g
M RT (0.0821 L ⋅ atm/mol ⋅ K)(400 K)
(b) This same mass remains when it returns to T = 270+ 2730 = 300 K, therefore,
mRT (1.705 g)(0.0821 L ⋅ atm/mol ⋅ K)(300 K)
P= = ; 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?
m mRT (8 g )(0.0821 L ⋅ atm/mol ⋅ K)(300 K)
PV = RT ; V = = ; V = 0.308 L
M MP (64 g/mol)(10 atm)
m N1 N A m (6.023 x 1023 molecules/mol)(8 g)
n= = ; N1 = =
M NA 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:
P N1 N1 ( 12 P ) N1 7.529 x 1022 molecules
1
= and N2 = 1
= =
P2 N 2 P 1 2 2
N2 = 3.765 x 1022 molecules; Molecules lost = N1 – N2 = 3.765 x 1022 molecules
3.764 x 1022 molecules
time = ; time = 371 s
1022 molecules/s
13

13. Chapter 19. Thermal Properties of Matter Physics, 6th Edition
*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 ]
mRT1 (2 g)(0.0821 L ⋅ atm/mol ⋅ K)(330 K)
V1 = = ; V1 = 1.13 L
MP1 (4 g/mol)(12 atm)
MPV2 (4 g/mol)(7 atm)(1.129 L)
m2 = 2
= ; m2 = 1.328 g
RT2 (0.0821 L ⋅ atm/mol ⋅ K)(290 K)
m1 = 2.00 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.
m1 RT1 m1 RT1 (0.97 m1 ) RT2
V1 = = V2 ; = ; T1 = 0.97T2
MP 1 MP 1 MP2
T1 300 K
T2 = = ; T2 = 309 K
0.97 0.97
14