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Gas Laws Practice Problems
1) Work out each problem on scratch paper.
2) Click ANSWER to check your answer.
3) Click NEXT to go on to the next problem.
P1V1T2 = P2V2T1
CLICK TO START
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Helium occupies 3.8 L at -45°C. What volume will it occupy at 45°C?
ANSWER
QUESTION #1
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ANSWER #1
V1 = 3.8 L
T1 = -45°C = 228 K
V2 = ?
T2 = 45°C = 318 K
NEXT
CHARLES’ LAW
P1V1T2 = P2V2T1
V2 = 5.3 L
BACK TO PROBLEM
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Ammonia gas occupies a volume of 450. mL at 720. mm Hg. What volume will it occupy at standard pressure?
ANSWER
QUESTION #2
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ANSWER #2
V1 = 450. mL
P1 = 720. mm Hg
V2 = ?
P2 = 760. mm Hg
NEXT
BOYLE’S LAW
P1V1T2 = P2V2T1
V2 = 426 mL
BACK TO PROBLEM
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A gas at STP is cooled to -185°C. What pressure in atmospheres will it have at this temperature (volume remains constant)?
ANSWER
QUESTION #3
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ANSWER #3
P1 = 1 atm
T1 = 273 K
P2 = ?
T2 = -185°C = 88 K
NEXT
GAY-LUSSAC’S LAW
P1V1T2 = P2V2T1
P2 = 0.32 atm
BACK TO PROBLEM
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A gas occupies 1.5 L at 850 mm Hg and 15°C. At what pressure will this gas occupy 2.5 L at 30.0°C?
ANSWER
QUESTION #4
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ANSWER #4
V1 = 1.5 L
P1 = 850 mm Hg
T1 = 15°C = 288 K
P2 = ?
V2 = 2.5 L
T2 = 30.0°C = 303 K
NEXT
COMBINED
GAS LAW
P1V1T2 = P2V2T1
P2 = 540 mm Hg
BACK TO PROBLEM
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Chlorine gas has a pressure of 1.05 atm at 25°C. What pressure will it exert at 75°C?
ANSWER
QUESTION #5
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ANSWER #5
P1 = 1.05 atm
T1 = 25°C = 298 K
P2 = ?
T2 = 75°C = 348 K
NEXT
GAY-LUSSAC’S
LAW
P1V1T2 = P2V2T1
P2 = 1.23 atmBACK TO PROBLEM
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A gas occupies 256 mL at 720 torr and 25°C. What will its volume be at STP?
ANSWER
QUESTION #6
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ANSWER #6
V1 = 256 mL
P1 = 720 torr
T1 = 25°C = 298 K
V2 = ?
P2 = 760. torr
T2 = 273 K
NEXT
COMBINED
GAS LAW
P1V1T2 = P2V2T1
V2 = 220 mL
BACK TO PROBLEM
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At 27°C, fluorine occupies a volume of 0.500 dm3. To what temperature in degrees Celsius should it be lowered to bring the volume to 200. mL?
ANSWER
QUESTION #7
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ANSWER #7
T1 = 27ºC = 300. K
V1 = 0.500 dm3
T2 = ?°C
V2 = 200. mL = 0.200 dm3
NEXT
CHARLES’ LAW
P1V1T2 = P2V2T1
T2 = -153°C
(120 K)BACK TO PROBLEM
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A gas occupies 125 mL at 125 kPa. After being heated to 75°C and depressurized to 100.0 kPa, it occupies 0.100 L. What was the original temperature of the gas?
ANSWER
QUESTION #8
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ANSWER #8
V1 = 125 mL
P1 = 125 kPa
T2 = 75°C = 348 K
P2 = 100.0 kPa
V2 = 0.100 L = 100. mL
T1 = ?
NEXT
COMBINED
GAS LAW
P1V1T2 = P2V2T1
T1 = 544 K
(271°C)BACK TO PROBLEM
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ANSWER
QUESTION #9
A 3.2-L sample of gas has a pressure of 102 kPa. If the volume is reduced to 0.65 L, what pressure will the gas exert?
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ANSWER #9
V1 = 3.2 L
P1 = 102 kPa
V2 = 0.65 L
P2 = ?
NEXT
BOYLE’S LAW
P1V1T2 = P2V2T1
P2 = 502 kPa
BACK TO PROBLEM
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A gas at 2.5 atm and 25°C expands to 750 mL after being cooled to 0.0°C and depressurized to 122 kPa. What was the original volume of the gas?
ANSWER
QUESTION #10
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ANSWER #10
P1 = 2.5 atm
T1 = 25°C = 298 K
V2 = 750 mL
T2 = 0.0°C = 273 K
P2 = 122 kPa = 1.20 atm
V1 = ?
EXIT
COMBINED
GAS LAW
P1V1T2 = P2V2T1
V1 = 390 mLBACK TO PROBLEM
Ideal Gas Law & Gas Stoichiometry
1) Work out each problem on scratch paper.
2) Click ANSWER to check your answer.
3) Click NEXT to go on to the next problem.
PV = nRT
R = 0.0821 Latm/molK = 8.315 dm3kPa/molK
CLICK TO START
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Tire of car contains nitrogen, oxygen, carbon dioxide and argon. The total pressure of the tire is 93.6 kPa. The partial pressures of
nitrogen, oxygen & carbon dioxide are 15.4 kPa, 90 mmHg and .356 atm. resp. What is the partial pressure exerted by argon?
Given
P-total= 93.6KPa
PN = 15.4 kPa
PO = 90 mmHg
PCO2 = .356 atm
Solve :
Ptotal = P1 + P2 + P3 + P4
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How many grams of CO2 are produced from 75 L of CO at 35°C and 96.2 kPa?
2CO + O2 2CO2
ANSWER
QUESTION #1
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Find the new molar volume:
n = 1 mol
V = ?
P = 96.2 kPa
T = 35°C = 308 K
R = 8.315 dm3kPa/molK
CONTINUE...
PV = nRT
V = 26.6 L/mol
BACK TO PROBLEM
ANSWER #1
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ANSWER #1 (con’t)
75 LCO
1 molCO
26.6 L CO
= 120 g CO2
2 molCO2
2 molCO
44.01g CO2
1 molCO2
2CO + O2 2CO2 75 L ? g
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How many moles of oxygen will occupy a volume of 2.5 L at 1.2 atm and 25°C?
ANSWER
QUESTION #2
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n = ?
V = 2.5 L
P = 1.2 atm
T = 25°C = 298 K
R = 0.0821 Latm/molK
NEXT
PV = nRT
n = 0.12 mol
BACK TO PROBLEM
ANSWER #2
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What volume will 56.0 grams of nitrogen (N2) occupy at 96.0 kPa and 21°C?
ANSWER
QUESTION #3
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V = ?
n = 56.0 g = 2.00 mol
P = 96.0 kPa
T = 21°C = 294 K
R = 8.315 dm3kPa/molK
NEXT
PV = nRT
V = 50.9 dm3
BACK TO PROBLEM
ANSWER #3
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What volume of NH3 at STP is produced if 25.0 g of N2 is reacted with excess H2?
N2 + 3H2 2NH3
ANSWER
QUESTION #4
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25.0 gN2
NEXTBACK TO PROBLEM
ANSWER #4
1 molN2
28.02 g N2
= 40.0 L NH3
N2 + 3H2 NH3
2 molNH3
1 molN2
22.4 LNH3
1 molNH3
25.0 g ? L
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What volume of hydrogen is produced from 25.0 g of water at 27°C and 1.16 atm?
2H2O 2H2 + O2
ANSWER
QUESTION #5
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Find the new molar volume:
n = 1 mol
V = ?
P = 1.16 atm
T = 27°C = 300. K
R = 0.0821 Latm/molK
CONTINUE...
PV = nRT
V = 21.2 L/mol
BACK TO PROBLEM
ANSWER #5
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ANSWER #5 (con’t)
25.0 gH2O
1 molH2O
18.02 g H2O
= 29.4 L H2
2 molH2
2 molH2O
21.2 LH2
1 molH2
2H2O 2H2 + O2 25.0 g ? L
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How many atmospheres of pressure will be exerted by 25 g of CO2 at 25°C and 0.500 L?
ANSWER
QUESTION #6
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P = ?
n = 25 g = 0.57 mol
T = 25°C = 298 K
V = 0.500 L
R = 0.0821 Latm/molK
NEXT
PV = nRT
P = 28 atm
BACK TO PROBLEM
ANSWER #6
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How many grams of CaCO3 are required to produce 45.0 L of CO2 at 25°C and 2.3 atm?
CaCO3 + 2HCl CO2 + H2O + CaCl2
ANSWER
QUESTION #7
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Find the new molar volume:
n = 1 mol
V = ?
P = 2.3 atm
T = 25°C = 298 K
R = 0.0821 Latm/molK
CONTINUE...
PV = nRT
V = 11 L/mol
BACK TO PROBLEM
ANSWER #7
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ANSWER #7
45.0dm3 CO2
1 molCO2
11 dm3 CO2
= 410 g CaCO3
CaCO3 + 2HCl CO2 + H2O + CaCl2
1 molCaCO3
1 molCO2
100.09 gCaCO3
1 molCaCO3
? g 45.0 dm3
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Find the number of grams of CO2 that exert a pressure of 785 torr at 32.5 L and 32°C.
ANSWER
QUESTION #8
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n = ?
P = 785 torr = 1.03 atm
V = 32.5 L
T = 32°C = 305 K
R = 0.0821 Latm/molK
NEXT
PV = nRT
n = 1.34 mol
59.0 g CO2
BACK TO PROBLEM
ANSWER #8
