Boyles Law Problems
1. A gas occupies 12.3 liters at a pressure of 40.0 mm Hg. What is the volume when the
pressure is increased to 60.0 mm Hg?
2. If a gas at 25.0 °C occupies 3.60 liters at a pressure of 1.00 atm, what will be its
volume at a pressure of 2.50 atm?
3. To what pressure must a gas be compressed in order to get into a 3.00 cubic foot tank
the entire weight of a gas that occupies 400.0 cu. ft. at standard pressure?
4. A gas occupies 1.56 L at 1.00 atm. What will be the volume of this gas if the pressure
becomes 3.00 atm?
5. A gas occupies 11.2 liters at 0.860 atm. What is the pressure if the volume becomes
15.0 L?
6. 500.0 mL of a gas is collected at 745.0 mm Hg. What will the volume be at standard
pressure?
7. Convert 350.0 mL at 740.0 mm of Hg to its new volume at standard pressure.
8. Convert 338 L at 63.0 atm to its new volume at standard pressure.
9. Convert 273.15 mL at 166.0 mm of Hg to its new volume at standard pressure.
10. Convert 77.0 L at 18.0 mm of Hg to its new volume at standard pressure.
11. When the pressure on a gas increases, will the volume increase or decrease?
12. If the pressure on a gas is decreased by one-half, how large will the volume change
be?
13. A gas occupies 4.31 liters at a pressure of 0.755 atm. Determine the volume if the
pressure is increased to 1.25 atm.
14. 600.0 mL of a gas is at a pressure of 8.00 atm. What is the volume of the gas at 2.00
atm?
15. 400.0 mL of a gas are under a pressure of 800.0 torr. What would the volume of the
gas be at a pressure of 1000.0 torr?
Gas Law Problems - Boyle's Law Answers: Part One
I tried to put the answers in the form of P1V1 = P2V2. They don't have to be in that order,
except that the sub ones must be paired on one side of the equals sign and the sub twos
must be paired on the other.
Some answers at the start are provided. Work the rest out. Show units on all values, not
just the answer!! Pay attention to sig figs.
1. (40.0 mm Hg) (12.3 liters) = (60.0 mm Hg) (x); x = 8.20 L, note three significant
figures!!
2. (1.00 atm) ( 3.60 liters) = (2.50 atm) (x); x = 1.44 L
3. ( 400.0 cu. ft) (1.00 atm) = (x) (3.00 cubic foot); x = 133 atm
4. (1.56 L) (1.00 atm) = (3.00 atm) (x); 0.520 L
5. (11.2 liters) (0.860 atm) = (x) (15.0 L); x = 0.642 atm
6. ( 745.0 mm Hg) (500.0 mL) = (760.0 mm Hg) (x)
7. (740.0 mm Hg) (350.0 mL) = (760.0 mm Hg) (x)
8. (63.0 atm) (338 L) = (1.00 atm) (x)
9. (166.0 mm Hg) (273.15 mL) = (760.0 mm Hg) (x)
10. (18.0 mm Hg) (77.0 L) = (760.0 mm Hg) (x)
11. Volume will decrease.
12. It will double in size.
13. (0.755 atm) (4.31 liters) = (1.25 atm) (x)
14. (8.00 atm) (600.0 mL) = (2.00 atm) (x)
15. (800.0 torr) (400.0 mL) = (1000.0 torr) (x)
Charles Law Problems
Abbreviations
atm - atmosphere
mm Hg - millimeters of mercury
(milliliter)
torr - another name for mm Hg
(liter) = 1000 mL
Pa - Pascal (kPa = kilo Pascal)
K - Kelvin
°C - degrees Celsius
101.325 kPa = 101,325 Pa
Conversions
K = °C + 273
1 cm3 (cubic centimeter)
1 dm3 (cubic decimeter)
Standard Conditions
0.00 °C = 273 K
1.00 atm = 760.0 mm Hg
=
=
1 mL
1 L
=
32. Calculate the decrease in temperature when 2.00 L at 20.0 °C is compressed to 1.00
L.
33. 600.0 mL of air is at 20.0 °C. What is the volume at 60.0 °C?
34. A gas occupies 900.0 mL at a temperature of 27.0 °C. What is the volume at 132.0
°C?
35. What change in volume results if 60.0 mL of gas is cooled from 33.0 °C to 5.00 °C?
36. Given 300.0 mL of a gas at 17.0 °C. What is its volume at 10.0 °C?
37. A gas occupies 1.00 L at standard temperature. What is the volume at 333.0 °C?
38. At 27.00 °C a gas has a volume of 6.00 L. What will the volume be at 150.0 °C?
39. At 225.0 °C a gas has a volume of 400.0 mL. What is the volume of this gas at 127.0
°C?
40. At 210.0 °C a gas has a volume of 8.00 L. What is the volume of this gas at -23.0 °C?
41. The temperature of a 4.00 L sample of gas is changed from 10.0 °C to 20.0 °C. What
will the volume of this gas be at the new temperature if the pressure is held constant?
42. Carbon dioxide is usually formed when gasoline is burned. If 30.0 L of CO2 is
produced at a temperature of 1.00 x 103 °C and allowed to reach room temperature (25.0
°C) without any pressure changes, what is the new volume of the carbon dioxide?
43. A 600.0 mL sample of nitrogen is warmed from 77.0 °C to 86.0 °C. Find its new
volume if the pressure remains constant.
44. What volume change occurs to a 400.0 mL gas sample as the temperature increases
from 22.0 °C to 30.0 °C?
45. A gas syringe contains 56.05 milliliters of a gas at 315.1 K. Determine the volume
that the gas will occupy if the temperature is increased to 380.5 K
46. A gas syringe contains 42.3 milliliters of a gas at 98.15 °C. Determine the volume
that the gas will occupy if the temperature is decreased to -18.50 °C.
47. When the temperature of a gas decreases, does the volume increase or decrease?
48. If the Kelvin temperature of a gas is doubled, the volume of the gas will increase by
____.
49. Solve the Charles' Law equation for V2.
50. Charles' Law deals with what quantities?
a. pressure/temperature
b. pressure/volume
c. volume/temperature
d. volume/temperature/pressure
51. If 540.0 mL of nitrogen at 0.00 °C is heated to a temperature of 100.0 °C what will be
the new volume of the gas?
52. A balloon has a volume of 2500.0 mL on a day when the temperature is 30.0 °C. If
the temperature at night falls to 10.0 °C, what will be the volume of the balloon if the
pressure remains constant?
Gas Law Problems - Charles' Law Answers:
I used V1 / T1 = V2 / T2 to set up the answers.
32. (2.00 L) / 293.0 K) = (1.00 L) / (x); x = 146.5 K
33. (600.0 mL) / (293.0) = (x) / (333.0 K); x = 682 mL
34. (900.0 mL) / (300.0 K) = (x) / (405.0 K); x = 1215 mL
35. (60.0 mL) / (306.0 K) = (x) / (278.00 K)
36. (300.0) / (290.0 K) = (x) / (283.0 K)
37. (1.00 L ) / (273.0 K) = (x) / (606.0 K)
38. (6.00 L) / (300.00 K) = (x) / (423.0 K)
39. (400.0 mL) / (498.0 K) = (x) / (400.0 K)
40. (8.00 L) / (483.0 K) = (x) / (250.0 K)
41. (4.00 L) / (283.0 K) = (x) / (293.0 K)
42. (30.0 L ) / (1273 K) = (x) / (298.0 K)
43. (600.0 mL) / (350.0 K) = (x) / (359.0 K)
44. 400.0 mL / 295.0 K = x / 303.0K
45. 56.05 milliliters / 315.1 K = x / 380.5 K
46. 42.3 milliliters / 371.15 K = x / 254.50 K
47. Decrease.
50. c. volume/temperature
48. Two. Or doubled.
51. 540.0 mL / 273.0 K = x / 373.0 K
49. V2 = (V1 times T2) / T1
52. 2500.0 mL / 303.0 =x / 283.0 K
Gas Law Problems- Gay-Lussac's Law
Abbreviations
atm - atmosphere
mm Hg - millimeters of mercury
(milliliter)
torr - another name for mm Hg
(liter) = 1000 mL
Pa - Pascal (kPa = kilo Pascal)
K - Kelvin
°C - degrees Celsius
101.325 kPa = 101,325 Pa
Conversions
K = °C + 273
1 cm3 (cubic centimeter)
1 dm3 (cubic decimeter)
Standard Conditions
0.00 °C = 273 K
1.00 atm = 760.0 mm Hg
=
=
1 mL
1 L
=
56. Determine the pressure change when a constant volume of gas at 1.00 atm is heated
from 20.0 °C to 30.0 °C.
57. A gas has a pressure of 0.370 atm at 50.0 °C. What is the pressure at standard
temperature?
58. A gas has a pressure of 699.0 mm Hg at 40.0 °C. What is the temperature at standard
pressure?
59. If a gas is cooled from 323.0 K to 273.15 K and the volume is kept constant what
final pressure would result if the original pressure was 750.0 mm Hg?
60. If a gas in a closed container is pressurized from 15.0 atmospheres to 16.0
atmospheres and its original temperature was 25.0 °C, what would the final temperature
of the gas be?
61. A 30.0 L sample of nitrogen inside a rigid, metal container at 20.0 °C is placed inside
an oven whose temperature is 50.0 °C. The pressure inside the container at 20.0 °C was at
3.00 atm. What is the pressure of the nitrogen after its temperature is increased?
62. A sample of gas at 3.00 x 103 mm Hg inside a steel tank is cooled from 500.0 °C to
0.00 °C. What is the final pressure of the gas in the steel tank?
63. The temperature of a sample of gas in a steel container at 30.0 kPa is increased from 100.0 °C to 1.00 x 103 °C. What is the final pressure inside the tank?
64. Calculate the final pressure inside a scuba tank after it cools from 1.00 x 103 °C to
25.0 °C. The initial pressure in the tank is 130.0 atm.
Gas Law Answers - Gay-Lussac's Law
Gay-Lussac's Law is P1 / T1 = P2 / T2
56. 1.00 atm / 293 K = x / 303 K; x = 1.03 atm.
57. 0.370 atm / 323 K = x / 273 K; x = 0.313 atm.
58. 699.0 mm Hg / 313 K = 760 mm Hg / x
59. 750.0 mm Hg / 323.0 K = x / 273.15 K
60. 15.0 atm / 298 K = 16.0 atm / x
61. 3.00 atm. / 293 K = x / 323
62. 3.00 x 103 mm Hg / 773 K = x / 273
63. 30.0 kPa / 173 K = x / 1273
64. 130.0 atm. /1273 K = x / 298 K
COMBINED GAS LAW PROBLEMS
73. A gas has a volume of 800.0 mL at minus 23.00 °C and 300.0 torr. What would the
volume of the gas be at 227.0 °C and 600.0 torr of pressure?
74. 500.0 liters of a gas are prepared at 700.0 mm Hg and 200.0 °C. The gas is placed
into a tank under high pressure. When the tank cools to 20.0 °C, the pressure of the gas is
30.0 atm. What is the volume of the gas?
75. What is the final volume of a 400.0 mL gas sample that is subjected to a temperature
change from 22.0 °C to 30.0 °C and a pressure change from 760.0 mm Hg to 360.0 mm
Hg?
76. What is the volume of gas at 2.00 atm and 200.0 K if its original volume was 300.0 L
at 0.250 atm and 400.0 K.
77. At conditions of 785.0 torr of pressure and 15.0 °C temperature, a gas occupies a
volume of 45.5 mL. What will be the volume of the same gas at 745.0 torr and 30.0 °C?
78. A gas occupies a volume of 34.2 mL at a temperature of 15.0 °C and a pressure of
800.0 torr. What will be the volume of this gas at standard conditions?
79. The volume of a gas originally at standard temperature and pressure was recorded as
488.8 mL. What volume would the same gas occupy when subjected to a pressure of
100.0 atm and temperature of minus 245.0 °C?
80. At a pressure of 780.0 mm Hg and 24.2 °C, a certain gas has a volume of 350.0 mL.
What will be the volume of this gas under STP
81. A gas sample occupies 3.25 liters at 24.5 °C and 1825 mm Hg. Determine the
temperature at which the gas will occupy 4250 mL at 1.50 atm.
82. If 10.0 liters of oxygen at STP are heated to 512 °C, what will be the new volume of
gas if the pressure is also increased to 1520.0 mm of mercury?
83. What is the volume at STP of 720.0 mL of a gas collected at 20.0 °C and 3.00 atm
pressure?
84. 2.00 liters of hydrogen, originally at 25.0 °C and 750.0 mm of mercury, are heated
until a volume of 20.0 liters and a pressure of 3.50 atmospheres is reached. What is the
new temperature?
85. A gas balloon has a volume of 106.0 liters when the temperature is 45.0 °C and the
pressure is 740.0 mm of mercury. What will its volume be at 20.0 °C and 780 .0 mm of
mercury pressure?
86. If the absolute temperature of a given quantity of gas is doubled and the pressure
tripled, what happens to the volume of the gas?
87. 73.0 mL of nitrogen at STP is heated to 80.0 °C and the volume increase to 4.53 L.
What is the new pressure?
88. 500.0 mL of a gas was collected at 20.0 °C and 720.0 mm Hg. What is its volume at
STP?
COMBINED GAS LAW ANSWERS
73. x = [ (300 torr) (800 mL) (500 K) ] / [ (250 K) (600 torr) ]; x = 800.0 mL
Keep in mind that torr = mmHg.
74. x = [ (700/760) (500) (293) ] / [ (473) (30) ]; x = 9.51 L
Note that this problem mixes pressure units. The 700/760 fraction converts mmHg to
atm.
75. x = [ (760 mm Hg) (400 mL) (303 K) ] / [ (295 K) (360 mm Hg) ]; x = 867.3 mL
76. x = [ (0.25) (300) (200) ] / [ (400) (2) ]
77. x = [ (785) (45.5) (303) ] / [ (288) (745) ]
78. x = [ (800) (34.2) (273) ] / [ (288) (760) ]
79. x = [ (1) (488.8) (28) ] / [ (273) (100) ]
80. x = [ (780) (350) (273) ] / [ (297.2) (760) ]
81. x = [ (1.50 atm) (4.25 L) (297.5 K) ] / [(1825 mm Hg/760 atm mmHg¯1) (3.25 L) ]
Note that we had to change units around somewhat. Also note that a temperature was
solved for rather than the usual volume.
82. x = [ (760) (10) (785) ] / [ (273) (1520) ]
83. x = [ (3.00 atm) (720.0 mL) (273 K) ] / [ (293 K) (1.00 atm)
84. x = [ (3.50) (20) (298) ] / [ (750/760) (2.00) ]
85. x = [ (740 mmHg) (106 L) (293 K) ] / [ (318 K) (780 mmHg) ]
86. x = [ (1) (10) (2) ] / [ (1) (3) ]
87. x = [ (1.00 atm) (73.0 mL) (353 K) ] / [ (273 K) (4530 mL) ]
88. x = [ (720) (500) (273) ] / [ (293) (760) ]
Gas Law Problems- Dalton's Law
65. A container holds three gases: oxygen, carbon dioxide, and helium. The partial
pressures of the three gases are 2.00 atm, 3.00 atm, and 4.00 atm, respectively. What is
the total pressure inside the container?
66. A container with two gases, helium and argon, is 30.0% by volume helium. Calculate
the partial pressure of helium and argon if the total pressure inside the container is 4.00
atm.
67. If 60.0 L of nitrogen is collected over water at 40.0 °C when the atmospheric pressure
is 760.0 mm Hg, what is the partial pressure of the nitrogen?
68. 80.0 liters of oxygen is collected over water at 50.0 °C. The atmospheric pressure in
the room is 96.00 kPa. What is the partial pressure of the oxygen?
69. A tank contains 480.0 grams of oxygen and 80.00 grams of helium at a total pressure
of 7.00 atmospheres. Calculate the following.
a) How many moles of O2 are in the
tank?
b) How many moles of He are in the
tank?
c) Total moles of gas in tank.
d) Mole fraction of O2.
e) Mole fraction of He.
f) Partial pressure of O2.
g) Partial pressure of He.
70. A tank contains 5.00 moles of O2, 3.00 moles of neon, 6.00 moles of H2S, and 4.00
moles of argon at a total pressure of 1620.0 mm Hg. Complete the following table
Moles
O2
Ne
H2S
Ar
Total
18.00
Mole fraction
1
Partial pressure
1620
Vapor Pressure Data for H2O
If you need to convert to atmospheres, divide by 101.325. If you need to convert to
millimeters of mercury, divide by 101.325, then multiply by 760.0
Temperature
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
kPa
0.61129
0.65716
0.70605
0.75813
0.81359
0.8726
0.93537
1.0021
1.073
1.1482
1.2281
1.3129
1.4027
1.4979
1.5988
1.7056
1.8185
1.938
2.0644
2.1978
2.3388
2.4877
2.6447
2.8104
2.985
3.169
3.3629
3.567
3.7818
4.0078
4.2455
4.4953
4.7578
5.0335
5.3229
5.6267
5.9453
6.2795
6.6398
6.9969
7.3814
7.784
8.2054
8.6463
9.1075
9.5895
10.094
10.62
11.171
11.745
Temperature
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
kPa
12.344
12.97
13.623
14.303
15.012
15.752
16.522
17.324
18.159
19.028
19.932
20.873
21.851
22.868
23.925
25.022
26.163
27.347
28.576
29.852
31.176
32.549
33.972
35.448
36.978
38.563
40.205
41.905
43.665
45.487
47.373
49.324
51.342
53.428
55.585
57.815
60.119
62.499
64.958
67.496
70.117
72.823
75.614
78.494
81.465
84.529
87.688
90.945
94.301
97.759