Transcript Gas Laws Practice Problems

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