Gas Law Problems

Gas Law Problems
Boyles Law
1. A gas occupies 12.3 litres 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 litres 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 litres 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 litres 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?
16. 4.00 L of a gas are under a pressure of 6.00 atm. What is the volume of the gas at 2.00 atm?
17. A gas occupies 25.3 mL at a pressure of 790.5 mm Hg. Determine the volume if the pressure is reduced to 0.804 atm.
18. A sample of gas has a volume of 12.0 L and a pressure of 1.00 atm. If the pressure of gas is increased to 2.00 atm, what is the new volume of the gas?
19. A container of oxygen has a volume of 30.0 mL and a pressure of 4.00 atm. If the pressure of the oxygen gas is reduced to 2.00 atm and the temperature is kept constant, what is the new volume of the oxygen gas?
20. A tank of nitrogen has a volume of 14.0 L and a pressure of 760.0 mm Hg. Find the volume of the nitrogen when its pressure is changed to 400.0 mm Hg while the temperature is held constant.
21. A 40.0 L tank of ammonia has a pressure of 8.00 atm. Calculate the volume of the ammonia if its pressure is changed to 12.0 atm while its temperature remains constant.
22. Two hundred litres of helium at 2.00 atm and 28.0 °C is placed into a tank with an internal pressure of 600.0 kPa. Find the volume of the helium after it is compressed into the tank when the temperature of the tank remains 28.0 °C.
23. You are now wearing scuba gear and swimming under water at a depth of 66.0 ft. You are breathing air at 3.00 atm and your lung volume is 10.0 L. Your scuba gauge indicates that your air supply is low so, to conserve air, you make a terrible and fatal mistake: you hold your breath while you surface. What happens to your lungs? Why?
24. Solve Boyle's Law equation for V2.
25. Boyle's Law deals what quantities?
a. pressure/temperature
b. pressure/volume
c. volume/temperature
d. volume temperature/pressure
26. A 1.5 litre flask is filled with nitrogen at a pressure of 12 atmospheres. What size flask would be required to hold this gas at a pressure of 2.0 atmospheres?
27. 300 mL of O2 are collected at a pressure of 645 mm of mercury. What volume will this gas have at one atmosphere pressure?
28. How many cubic feet of air at standard conditions (1.00 atm.) are required to inflate a bicycle tire of 0.50 cu. ft. to a pressure of 3.00 atmospheres?
29. How much will the volume of 75.0 mL of neon change if the pressure is lowered from 50.0 torr to 8.00 torr?
30. A tank of helium has a volume of 50.0 litres and is under a pressure of 136.1 atm.. This gas is allowed to flow into a blimp until the pressure in the tank drops to 2.722 atm. and the pressure in the blimp is 2.04 atm.. What will be the volume of the blimp?
31. What pressure is required to compress 196.0 litres of air at 1.00 atmosphere into a cylinder whose volume is 26.0 litres?
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.
Charles Law
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?
53. When 50.0 litres of oxygen at 20.0 °C is compressed to 5.00 litres, what must the new temperature be to maintain constant pressure?
54. If 15.0 litres of neon at 25.0 °C is allowed to expand to 45.0 litres, what must the new temperature be to maintain constant pressure?
55. 3.50 litres of a gas at 727.0 K will occupy how many litres at 153.0 K?
Guy-Lussac’s Law
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.
Dalton’s Law of Partial Pressure
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 litres 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?
Combined Gas Laws
Solve for x in the usual manner of cross-multiplying and dividing.
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 litres 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 litres 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 litres 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 litres of hydrogen, originally at 25.0 °C and 750.0 mm of mercury, are heated until a volume of 20.0 litres and a pressure of 3.50 atmospheres is reached. What is the new temperature?
85. A gas balloon has a volume of 106.0 litres 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?
89. A sample of gas occupies 50.0 L at 15.0 °C and 640.0 mm Hg pressure. What is the volume at STP?
90. A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 litres to 35.0 litres by moving a large piston within a cylinder. If the original pressure was 1.00 atm, what would the final pressure be?
91. The pressure of a gas is reduced from 1200.0 mm Hg to 850.0 mm Hg as the volume of its container is increased by moving a piston from 85.0 mL to 350.0 mL. What would the final temperature be if the original temperature was 90.0 °C?
92. If a gas is heated from 298.0 K to 398.0 K and the pressure is increased from 2.230 x 103 mm Hg to 4.560 x 103 mm Hg what final volume would result if the volume is allowed to change from an initial volume of 60.0 litres?
Combined Gas Law which requires Dalton's Law also
IMPORTANT NOTE: A gas collected over water is always considered to be saturated with water vapor. The vapor pressure of water varies with temperature and must be looked up in a reference source.
93. 690.0 mL of oxygen are collected over water at 26.0 °C and a total pressure of 725.0 mm of mercury. What is the volume of dry oxygen at 52.0 °C and 800.0 mm pressure?
94. 400.0 mL of hydrogen are collected over water at 18.0 °C and a total pressure of 740.0 mm of mercury.
a) What is the partial pressure of H2?
b) What is the partial pressure of H2O?
c) What is the volume of DRY hydrogen at STP?
95. 45.0 mL of wet argon gas is collected at 729.3 mm Hg and 25.0 °C. What would be the volume of this dry gas at standard conditions?
96. 19.1 L of He gas is collected over water at 681.3 mm Hg and 18.5 °C. What would be the volume of this dry gas at standard conditions?
97. 407 mL of H2 gas is collected over water at 785.3 mm Hg and 23.5 °C. What would be the volume of this dry gas at standard conditions?
98. 93.0 mL of O2 gas is collected over water at 0.930 atm and 10.0 °C. What would be the volume of this dry gas at standard conditions?
99. 6.12 L of wet xenon gas is collected at 2.00 x 105 Pa and 80.0 °C. What would be the volume of this dry gas at standard conditions?
100. A sample of oxygen collected over water when the atmospheric pressure was 1.002 atm and the room temperature, 25.5 °C occupied 105.8 mL. What would be the volume of this dry gas at standard conditions?
101. 1.000 L of hydrogen gas is collected over water at 30.0 °C at a pressure of 831.8 mm Hg. Find the volume of dry hydrogen collected at STP.
102. 50.6 mL of a gas is collected over water at 18.0 °C and 755.5 mm Hg pressure. What is the volume of dry gas at STP?
103. Write the combined gas law in equation form. Solve the combined gas law for V2.
Ideal Gas Equation
104. How many moles of gas are contained in 890.0 mL at 21.0 °C and 750.0 mm Hg pressure?
105. 1.09 g of H2 is contained in a 2.00 L container at 20.0 °C. What is the pressure in this container in mm Hg?
106. Calculate the volume 3.00 moles of a gas will occupy at 24.0 °C and 762.4 mm Hg.
107. What volume will 20.0 g of Argon occupy at STP?
108. How many moles of gas would be present in a gas trapped within a 100.0 mL vessel at 25.0 °C at a pressure of 2.50 atmospheres?
109. How many moles of a gas would be present in a gas trapped within a 37.0 litre vessel at 80.00 °C at a pressure of 2.50 atm?
110. If the number of moles of a gas are doubled at the same temperature and pressure, will the volume increase or decrease?
111. What volume will 1.27 moles of helium gas occupy at STP?
112. At what pressure would 0.150 mole of nitrogen gas at 23.0 °C occupy 8.90 L?
113. What volume would 32.0 g of NO2 gas occupy at 3.12 atm and 18.0 °C?
114. Find the volume of 2.40 mol of gas whose temperature is 50.0 °C and whose pressure is 2.00 atm.
115. Calculate the molecular weight of a gas if 35.44 g of the gas stored in a 7.50 L tank exerts a pressure of 60.0 atm at a constant temperature of 35.5 °C
116. How many moles of gas are contained in a 50.0 L cylinder at a pressure of 100.0 atm and a temperature of 35.0 °C?
117. Determine the number of moles of Krypton contained in a 3.25 litre gas tank at 5.80 atm and 25.5 °C. If the gas is Oxygen instead of Krypton, will the answer be the same? Why or why not?
118. Determine the number of grams of carbon dioxide in a 450.6 mL tank at 1.80 atm and minus 50.5 °C. Determine the number of grams of oxygen that the same container will contain under the same temperature and pressure.
119. Determine the volume of occupied by 2.34 grams of carbon dioxide gas at STP.
120. A sample of argon gas at STP occupies 56.2 litres. Determine the number of moles of argon and the mass in the sample.
121. At what temperature will 0.654 moles of neon gas occupy 12.30 litres at 1.95 atmospheres?
122. A 30.6 g sample of gas occupies 22.4 L at STP. What is the molecular weight of this gas?
123. A 40.0 g gas sample occupies 11.2 L at STP. Find the molecular weight of this gas.
124. A 12.0 g sample of gas occupies 19.2 L at STP. What is the molecular weight of this gas?
125. 96.0 g. of a gas occupies 48.0 L at 700.0 mm Hg and 20.0 °C. What is its molecular weight?
126. 20.83 g. of a gas occupies 4.167 L at 79.97 kPa at 30.0 °C. What is its molecular weight?
127. At STP 3.00 litres of an unknown gas has a mass of 9.50 grams. Calculate its molar mass.
128. At STP 0.250 litre of an unknown gas has a mass of 1.00 gram. Calculate its molar mass.
129. At STP 150.0 mL of an unknown gas has a mass of 0.250 gram. Calculate its molar mass.
130. 1.089 g of a gas occupies 4.50 L at 20.5 °C and 0.890 atm. What is its molar mass?
131. 0.190 g of a gas occupies 250.0 mL at STP. What is its molar mass? What gas is it? Hint - calculate molar mass of the gas.
132. If 9.006 grams of a gas are enclosed in a 50.00 litre vessel at 273.15 K and 2.000 atmospheres of pressure, what is the molar mass of the gas? What gas is this?
133. What is the value of and units on R? What is R called ("A letter" is not the correct answer!)?
134. A 50.00 litre tank at minus 15.00 °C contains 14.00 grams of helium gas and 10.00 grams of nitrogen gas.
a. Determine the moles of helium gas in the tank.
b. Determine the moles of nitrogen gas in the tank.
c. Determine the mole fraction of helium gas in the tank.
d. Determine the mole fraction of nitrogen gas in the tank.
e. Determine the partial pressure of helium gas in the tank.
f. Determine the partial pressure of nitrogen gas in the tank.
g. Determine the total pressure of the mixture in the tank.
h. Determine the volume that the mixture will occupy at STP.

Answers
1. (40.0 mm Hg) (12.3 litres) = (60.0 mm Hg) (x); x = 8.20 L, note three significant figures!!
2. (1.00 atm) ( 3.60 litres) = (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 litres) (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 litres) = (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)
16. (6.00 atm) (4.00 L) = (2.00 atm) (x)
17. (790.5 mm Hg) (25.3 mL) = ( 0.804 atm) (x)
This is wrong!! You MUST change one of the pressures units so both are the same. I'll change the mm Hg to atm:
(790.5 mm Hg / 760.0 mm Hg/atm) (25.3 mL) = ( 0.804 atm) (x)
18. (1.00 atm) (12.0 L) = (2.00 atm) (x)
19. (4.00 atm) (30.0 mL) = (2.00 atm) (x)
20. (760.0 mm Hg) (14.0 L) = (400.0 mm Hg) (x)
21. (8.00 atm) (40.0 L ) = (12.0 atm) (x)
22. (2.00 atm) (200.0 L) = (600.0 kPa) (x)
This is wrong. The pressure units must be the same. I'll change the atm to kPa. You could go the other way if you want, the answer would be the same.
(2.00 atm x 101.325 kPa/atm) (200.0 L) = (600.0 kPa) (x)
In fact, here's the problem with the kPa changed to atm:
(2.00 atm) (200.0 L) = (600.0 kPa / 101.325 kPa/atm) (x)
23. Your lungs will "explode." As you go up towards the surface, the pressure on your body and lungs becomes less. The air in your lungs expands. What would happen then is that the alveoli and small capillaries would rupture, causing massive bleeding in the lungs. You'd die. No, your body would not swell up and burst, like a balloon.
24. V2 = (P1V1) / P2
25. B
26. (12 atmospheres) (1.5 litre) = (2.0 atmospheres) (x)
27. (645 mm Hg) (300 mL) = (one atmosphere) (x)
This is wrong. I will change atm to mm Hg. (645 mm Hg) (300 mL) = (760 mm Hg) (x)
28. (1.00 atm.) (x) = (3.00 atmospheres) (0.50 cu. ft.)
29. (50.0 torr) ( 75.0 mL) = (8.00 torr) (x)
30. (133.38 atm.) (50.0 litres) = (2.04 atm.) (x)
Important point: 136.1 – 2.722 atm = 133.38 atm flowed out of tank. 136.1 is not used in calculation.
31. (1.00 atmosphere) (196.0) = (x) (26.0 litres) use 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.
48. Two. Or doubled.
49. V2 = (V1 times T2) / T1
50. c. volume/temperature
51. 540.0 mL / 273.0 K = x / 373.0 K
52. 2500.0 mL / 303.0 =x / 283.0 K
53. 50.0 litres / 293.0 K = 5.00 litres / x
54. 15.0 litres / 298.0 K = 45.0 litres / x
55. 3.50 litres / 1000.0 K = x / 426.0 K
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
65. 9.00 atm.
66. PHe = 0.300 x 4.00 atm = 1.20 atm. PAr = 4.00 - 1.20
67. 760.0 mmHg minus 55.3 mmHg
68. 96.00 kPa minus 12.33 kPa
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) ]
89. x = [ (640) (50) (273) ] / [ (288) (760) ]
90. A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 litres to 35.0 litres by moving a large piston within a cylinder. If the original pressure was 1.00 atm, what would the final pressure be?
91. The pressure of a gas is reduced from 1200.0 mm Hg to 850.0 mm Hg as the volume of its container is increased by moving a piston from 85.0 mL to 350.0 mL. What would the final temperature be if the original temperature was 90.0 °C?
92. If a gas is heated from 298.0 K to 398.0 K and the pressure is increased from 2.230 x 103 mm Hg to 4.560 x 103 mm Hg what final volume would result if the volume is allowed to change from an initial volume of 60.0 litres?
Combined Gas Law which requires Dalton's Law also
IMPORTANT NOTE: A gas collected over water is always considered to be saturated with water vapor. The vapor pressure of water varies with temperature and must be looked up in a reference source.
93. Pdry = 725.0 - 25.2 = 699.8 mmHg x = [ (699.8) (690.0) (325) ] / [ (299) (800.0) ]
94.
a) Pdry = 740.0 - 15.5 = 724.5 mmHg
b) 15.5 mmHg
c) x = [ (724.5 mm Hg) (400.0 mL) (273 K) ] / [ (291 K) (760.0 mm Hg)
95. Pdry = 729.3 - 23.8 = 705.5 x = [ (705.5) (45.0) (273) ] / [ (298) (760) ]
96. The 18.5 °C requires an interpolation between 18 °C and 19 °C. This gives 16.0 mmHg.
Pdry = 681.3 - 16.0 = 665.3 x = [ (665.3 mm Hg) (19.1 L) (273 K) ] / [ (291.5 K) (760 mm Hg) ]
97. The 23.5 °C requires an interpolation between 23 °C and 24 °C. This gives 21.7 mmHg.
Pdry = 785.3 - 21.7 = 763.6 x = [ (763.6 mm Hg) (407 mL) (273 K) ] / [ (296.5 K) (760 mm Hg) ]
98. 93.0 mL of O2 gas is collected over water at 0.930 atm and 10.0 °C. What would be the volume of this dry gas at standard conditions?
99. 6.12 L of wet xenon gas is collected at 2.00 x 105 Pa and 80.0 °C. What would be the volume of this dry gas at standard conditions?
100. A sample of oxygen collected over water when the atmospheric pressure was 1.002 atm and the room temperature, 25.5 °C occupied 105.8 mL. What would be the volume of this dry gas at standard conditions?
101. 1.000 L of hydrogen gas is collected over water at 30.0 °C at a pressure of 831.8 mm Hg. Find the volume of dry hydrogen collected at STP.
102. 50.6 mL of a gas is collected over water at 18.0 °C and 755.5 mm Hg pressure. What is the volume of dry gas at STP?
103. V2 = (P1 V1 T2 ) / (T1 P2 )
104. n = PV / RT n = [ (750.0 mmHg / 760.0 mmHg atm¯1) (0.890 L) ] / (0.08206 L atm mol¯1 K¯1) (294.0 K)
Please note the division of 750 by 760. That is done in order to convert the pressure from mmHg to atm., because the value for R contains atm. as the pressure unit. If we used mmHg, the pressure units would not cancel and we need to have them cancel because we require mol. only to be in the answer.
105. P = nRT / V
P = [ (1.09 g / 2.02 g mol¯1) (0.08206 L atm mol¯1 K¯1) (293.0 K) ] / 2.00 L
Multiply the answer (which is in atm) by 760.0 mmHg atm¯1 to get mmHg
106. V = nRT / P
V = [ (3.00 mol) (0.08206 L atm mol¯1 K¯1) (297.0 K) ] / (762.4 mmHg / 760.0 mmHg atm¯1
107. V = nRT / P
V = [ (20.0 g / 40.0 g mol¯1) (0.08206 L atm mol¯1 K¯1) (273.0 K) ] / (1.00 atm)
108. n = PV / RT n = [ (2.50 atm) (0.1000 L) ] / [ (0.08206 L atm mol¯1 K¯1) (298.0 K) ]
109. n = PV / RT n = [ (2.50 atm) (37.0 L) ] / [ (0.08206 L atm mol¯1 K¯1) (353.0 K) ]
110. The volume would increase. In fact, it would double.
111. V = nRT / P
V = [ (1.27 mol) (0.08206 L atm mol¯1 K¯1) (273.0 K) ] / 1.00 atm or (22.4 L / 1.00 mol) = (x / 1.27 mol)
112. P = nRT / V
P = [ (0.150 mol) (0.08206 L atm mol¯1 K¯1) (296.0 K) ] / 8.90 L
113. V = nRT / P
V = [ (32.0 g / 46.0 g mol¯1) (0.08206 L atm mol¯1 K¯1) (291.0 K) ] / 3.12 atm
114. V = nRT / P
V = [ 2.40 mol) (0.08206 L atm mol¯1 K¯1) (323.0 K) ] / 2.00 atm
115. n = PV / RT n = [ (60.0 atm) (7.50 L) ] / [ (0.08206 L atm mol¯1 K¯1) (308.5 K) ]
Divide 35.44 g by the number of moles calculated to get the molecular weight.
116. n = PV / RT n = [ (100.0 atm) (5.00 L) ] / [ (0.08206 L atm mol¯1 K¯1) (308.0 K) ]
117. n = PV / RT n = [ (5.80 atm) (3.25 L) ] / [ (0.08206 L atm mol¯1 K¯1) (298.5 K) ]
The moles of gas would be the same if the gas was switched to oxygen. Since the temperature and pressure would be the same, the same volume will contain the same number of molecules of gas, i.e. moles of gas. This is Avogadro's Hypothesis.
118. n = PV / RT n = [ (0.4506 atm) (1.80 L) ] / [ (0.08206 L atm mol¯1 K¯1) (222.5 K) ]
This calculates the number of moles of CO2. Multiply the moles by the molecular weight of CO2 to get the grams. Under the same set of conditions, the moles of oxygen would be the same, so multiply the calculated moles by the molecular weight of O2 to get the grams.
119. V = nRT / P
V = [ (2.34 g / 44.0 g mol¯1) (0.08206 L atm mol¯1 K¯1) (273.0 K) ] / 1.00 atm
120. n = PV / RT n = [ (1.00 atm) (56.2 L) ] / [ (0.08206 L atm mol¯1 K¯1) (273.0 K) ] Multiply the moles by the atomic weight of Ar to get the grams.
121. T = PV / nR T = [ (1.95 atm) (12.30 L) ] / [ (0.654 mol) (0.08206 L atm mol¯1 K¯1) ]
122. Since one mole of gas occupies 22.4 L at STP, the molecular weight of the gas is 30.6 g mol¯1
123. 11.2 L at STP is one-half molar volume, so there is 0.5 mol of gas present. Therefore, the molecular weight is 80.0 g mol¯1
124. This problem, as well as the two just above can be solved with PV = nRT. You would solve for n, the number of moles. Then you would divide the grams given by the mole calculated.
Since it is at STP, we can also use a ratio method (see prob. 111)
(19.2 L / 12.0 g) = (22.4 L / x )
125. n = PV / RT n = [ (700.0 mmHg / 760.0 mmHg atm¯1) (48.0 L) ] / [ (0.08206 L atm mol¯1 K¯1) (293.0 K) ]
Then, divide the grams given (96.0) by the moles just calculated above. This will be the molecular weight.
126. n = PV / RT n = [ (79.97 kPa / 101.325 kPa atm¯1) (4.167 L) ] / [ (0.08206 L atm mol¯1 K¯1) (303.0 K) ]
Then, divide the grams given (20.83) by the moles just calculated above. This will be the molecular weight.
127. (3.00 L / 9.50 g) = (22.4 L / x )
128. (0.250 L / 1.00 g) = (22.4 L / x )
129. (0.150 L / 0.250 g) = (22.4 L / x )
130. n = PV / RT n = [ (0.890 atm) (4.50 L) ] / [ (0.08206 L atm mol¯1 K¯1) (293.5 K) ]
Then, divide the grams given (1.089) by the moles just calculated above. This will be the molecular weight.
131. n = PV / RT n = [ (1.00 atm) (0.2500 L) ] / [ (0.08206 L atm mol¯1 K¯1) (273.0 K) ]
Then, divide the grams given (0.190) by the moles just calculated above. This will be the molecular weight.
132. If 9.006 grams of a gas are enclosed in a 50.00 litre vessel at 273.15 K and 2.000 atmospheres of pressure, what is the molar mass of the gas? What gas is this? n = PV / RT n = [ (2.000 atm) (50.00 L) ] / [ (0.08206 L atm mol¯1 K¯1) (273.15 K) ]
Then, divide the grams given (9.006) by the moles just calculated above. This will be the molecular weight.
The answer (2.019 g mol¯1) is approximately that of hydrogen gas, H2
133. 0.08206 L atm mol¯1 K¯1
The gas constant.
134.
a. 14.0 g / 4.00 g mol¯1 = 3.50 mol
b. 49.0 g / 28.0 g mol¯1 = 1.75 mol
c. 3.50 mol / 5.25 mol
d. 1.75 mol / 5.25 mol
e. Use PV = nRT to determine the total pressure in the container.
P = nRT / V
P = [ (5.25 mol) (0.08206 L atm mol¯1 K¯1) (258.0 K) ] / 50.00 L
The total pressure times the mole fraction of He will give helium's partial pressure.
f. The total pressure times the mole fraction of N2 will give nitrogen's partial pressure. Since it is the only other value, you could subtract helium's answer from the total.
g. Already done to answer part e.
h. V = nRT / P
V = [ (5.25 mol) (0.08206 L atm mol¯1 K¯1) (273.0 K) ] / 1.00 atm