Tough Gas Laws Worksheet (1).doc 662l5c
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Name
Period
Date
The Gas Laws – Ch. 10 (p.313-322) 1. The gas left in a used aerosol can is at a pressure of 1 atm at 27C. If this can is thrown into a fire, what is the internal pressure of the gas when its temperature reaches 927C? GIVEN
T1= 300K P1= 101 KPA T2= 1200K P2= X ANSWER:
GAS LAW
WORK
BOYLE FORMULA
101 / 300 = X / 1200 X = 404
P1/T1=P2/T2
404 KPA OR 4 ATM
2. A sample of carbon dioxide occupies a volume of 3.50 L at 125 kPa. What pressure would the gas exert if the volume were decreased to 2.00 L? GIVEN
V1 = 3.50 L P1 = 125 KPA V2 = 2.00 L P2 = X ANSWER:
GAS LAW
WORK
CHARLES FORMULA
(125) (3.50) = (2.00) (X) X = 218.75 KPA
P1V1 = P2V2
218.75 KPA
3. A sample of propane occupies 250.0 L at 125 kPa and 38C. Find its volume at 100.0 kPa and 95C. GIVEN
V1 P1
= 250.0 L = 125 KPA T1 = 311 K V2 = x P2 = 100.0 kPa T2 = 368 K ANSWER:
GAS LAW
WORK
COMBINED FORMULA
(125) (250.0) / 311 = (100.0) (X) / 368 X = 369.77 P1V1/T1=P2V2/T2
369.77 L
The Gas Laws – Ch. 6
CHEM
4. Oxygen gas is at a temperature of 40C when it occupies a volume of 2.3 L. To what temperature in Celsius should it be raised to occupy a volume of 6.5 dm 3? GIVEN
T1 = 313 K V1 = 2.3 L T2 = X V2 = 6.5 L ANSWER:
GAS LAW
CHARLES FORMULA
V1/T1 = V2/T2
WORK
2.3 / 313 = 6.5 / X X = 884 884 – 273 = 611 C
611 C
5. Fluorine exerts a pressure of 900. torr. When the pressure is changed to 1.5 atm, its volume is 250. mL. What was the original volume? GIVEN
1 TORR = 133 PA 900 TORR = 120 KPA P1 = 120 KPA V1 = X P2 = 151.5 V2 = 250 ML ANSWER:
GAS LAW
WORK
BOYLE FORMULA
P1 V1
(120) (X) = (151.5) (250) X = 315.625 ML
= P2 V2
315.625 ML
6. The volume of a gas is 200.0 mL at 275 K and 92.1 kPa. Find its volume at STP. GIVEN
V1 = 200 ML T1 = 275 K P1 = 92.1 KPA V2 = X T2 = 293 K P2 = 101 KPA ANSWER:
GAS LAW
WORK
COMBINED FORMULA
(92.1) (200) / 275 = (101) (X) / 293 X = 194 ML P1V1/T1=P2V2/T2
194 ML
7. A sample of N2 occupies a volume of 250 mL at 25C. What volume will it occupy at 95C? GIVEN
V1 = 250 ML T1 = 298 K V2 = X T2 = 368 K ANSWER:
GAS LAW
WORK
CHARLES FORMULA
250 / 298 = X / 368 X = 308 ML
V1/T1 = V2/T2
308 ML
The Gas Laws – Ch. 6
CHEM
Period
Date
The Gas Laws – Ch. 10 (p.313-322) 1. The gas left in a used aerosol can is at a pressure of 1 atm at 27C. If this can is thrown into a fire, what is the internal pressure of the gas when its temperature reaches 927C? GIVEN
T1= 300K P1= 101 KPA T2= 1200K P2= X ANSWER:
GAS LAW
WORK
BOYLE FORMULA
101 / 300 = X / 1200 X = 404
P1/T1=P2/T2
404 KPA OR 4 ATM
2. A sample of carbon dioxide occupies a volume of 3.50 L at 125 kPa. What pressure would the gas exert if the volume were decreased to 2.00 L? GIVEN
V1 = 3.50 L P1 = 125 KPA V2 = 2.00 L P2 = X ANSWER:
GAS LAW
WORK
CHARLES FORMULA
(125) (3.50) = (2.00) (X) X = 218.75 KPA
P1V1 = P2V2
218.75 KPA
3. A sample of propane occupies 250.0 L at 125 kPa and 38C. Find its volume at 100.0 kPa and 95C. GIVEN
V1 P1
= 250.0 L = 125 KPA T1 = 311 K V2 = x P2 = 100.0 kPa T2 = 368 K ANSWER:
GAS LAW
WORK
COMBINED FORMULA
(125) (250.0) / 311 = (100.0) (X) / 368 X = 369.77 P1V1/T1=P2V2/T2
369.77 L
The Gas Laws – Ch. 6
CHEM
4. Oxygen gas is at a temperature of 40C when it occupies a volume of 2.3 L. To what temperature in Celsius should it be raised to occupy a volume of 6.5 dm 3? GIVEN
T1 = 313 K V1 = 2.3 L T2 = X V2 = 6.5 L ANSWER:
GAS LAW
CHARLES FORMULA
V1/T1 = V2/T2
WORK
2.3 / 313 = 6.5 / X X = 884 884 – 273 = 611 C
611 C
5. Fluorine exerts a pressure of 900. torr. When the pressure is changed to 1.5 atm, its volume is 250. mL. What was the original volume? GIVEN
1 TORR = 133 PA 900 TORR = 120 KPA P1 = 120 KPA V1 = X P2 = 151.5 V2 = 250 ML ANSWER:
GAS LAW
WORK
BOYLE FORMULA
P1 V1
(120) (X) = (151.5) (250) X = 315.625 ML
= P2 V2
315.625 ML
6. The volume of a gas is 200.0 mL at 275 K and 92.1 kPa. Find its volume at STP. GIVEN
V1 = 200 ML T1 = 275 K P1 = 92.1 KPA V2 = X T2 = 293 K P2 = 101 KPA ANSWER:
GAS LAW
WORK
COMBINED FORMULA
(92.1) (200) / 275 = (101) (X) / 293 X = 194 ML P1V1/T1=P2V2/T2
194 ML
7. A sample of N2 occupies a volume of 250 mL at 25C. What volume will it occupy at 95C? GIVEN
V1 = 250 ML T1 = 298 K V2 = X T2 = 368 K ANSWER:
GAS LAW
WORK
CHARLES FORMULA
250 / 298 = X / 368 X = 308 ML
V1/T1 = V2/T2
308 ML
The Gas Laws – Ch. 6
CHEM
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