The Combined As Law

Name: ______Hour: ____ Date: ______Chemistry: The Combined Gas Law

Solve the following problems. As always, include enough work and show the units to ensure full credit.

1. The pressure of a gas changes from 120 kPa to 50 kPa. The volume changes from 45 L to 40 L. If the initial temperature is 81oC, what is the final temperature in oC?

2. A sample of nitrogen goes from 21 m3 to 14 m3 and its pressure increases from 100 kPa to 150 kPa. The final temperature is 300 K. What was the initial temperature in Kelvins?

3. A sample of argon goes from 500 K to 350 K and its pressure changes from 280 kPa to 380 kPa. If the initial volume is 18 dm3, what is the final volume?

4. A sample of neon experiences a pressure drop from 75 kPa to 53 kPa. The temperature increases from 27oC to 93oC. If the initial volume is 12 L, what is the final volume?

5. The volume of a sample of helium increases from 5 L to 25 L and its temperature drops from 2000 K to 1750 K. If the initial pressure is 1500 mm Hg, what is the final pressure?

6. The temperature of a gas increases from 212oC to 380oC. The volume goes from 30 dm3 to 18 dm3. If the final pressure is 1.85 atm, what was the initial pressure?

Answers: 1. –142oC 2. 300 K 3. 9.3 dm3 4. 20.7 L 5. 262.5 mm Hg 6. 0.82 atm Chemistry: The Combined Gas Law KEY

Solve the following problems. As always, include enough work and show the units to ensure full credit.

1. The pressure of a gas changes from 120 kPa to 50 kPa. The volume changes from 45 L to 40 L. If the initial temperature is 81oC, what is the final temperature in oC?

P1  120 kPa P2  50 kPa P1V1 P2V2 120 kPa45 L 50 kPa40 L V1  45 L V2  40 L = = o T1 T2 354 K T2 T1  81 C  273  354 K T2  x K o T2  131 K  K - 273  C 131 K - 273  oC oC  - 142o C

2. A sample of nitrogen goes from 21 m3 to 14 m3 and its pressure increases from 100 kPa to 150 kPa. The final temperature is 300 K. What was the initial temperature in Kelvins?

P1  100 kPa P2  150 kPa P V P V 100 kPa 21 dm3 3 3 3 1 1 2 2    150 kPa14 dm  V1  21 dm V2  14 dm = = T1 T2 T1 300 K T1  x K T2  300 K

T1  300 K

3. A sample of argon goes from 500 K to 350 K and its pressure changes from 280 kPa to 380 kPa. If the initial volume is 18 dm3, what is the final volume?

P1  280 kPa P2  380 kPa P V P V 3 380 kPa V 3 3 1 1 2 2 280 kPa18 dm    2  V1  18 dm V2  x dm = = T1 T2 500 K 350 K T1  500 K T2  350 K

3 V2  9.3 dm

4. A sample of neon experiences a pressure drop from 75 kPa to 53 kPa. The temperature increases from 27oC to 93oC. If the initial volume is 12 L, what is the final volume?

P1  75 kPa P2  53 kPa P V P V 3 1 1 2 2 75 kPa12 L 53 kPaV2  V1  12 L V2  x dm = = T T o o 1 2 300 K 366 K T1  27 C  273  300 K T2  93 C  273  366 K

V2  20.7 L

5. The volume of a sample of helium increases from 5 L to 25 L and its temperature drops from 2000 K to 1750 K. If the initial pressure is 1500 mm Hg, what is the final pressure?

P1  1500 mm Hg P2  x mm Hg P1V1 P2V2 1500 mm Hg5 L P2 25 L V1  5 L V2  25 L = = T1 T2 2000 K 1750 K T1  2000 K T2  1750 K

P2  262.5 mm Hg

6. The temperature of a gas increases from 212oC to 380oC. The volume goes from 30 dm3 to 18 dm3. If the final pressure is 1.85 atm, what was the initial pressure?

P1  x atm P2  1.85 atm P V P V 3 3 3 3 1 1 2 2 P130 dm  1.85 atm18 dm  V  30 dm V  18 dm = = 1 2 T T o o 1 2 485 K 653 K T1  212 C  273  485 K T2  380 C  273  653 K

P1  0.82 atm The Combined Gas Law

P1  120 kPa P2  50 kPa P1V1 P2V2 120 kPa45 L 50 kPa40 L V1  45 L V2  40 L = = o T1 T2 354 K T2 T1  81 C  273  354 K T2  x K o T2  131 K  K - 273  C 131 K - 273  oC oC  - 142o C 2.

P1  100 kPa P2  150 kPa P V P V 100 kPa 21 dm3 3 3 3 1 1 2 2    150 kPa14 dm  V1  21 dm V2  14 dm = = T1 T2 T1 300 K T1  x K T2  300 K

T1  300 K 3. P  280 kPa P  380 kPa 1 2 P V P V 3 3 3 3 1 1 2 2 280 kPa18 dm  380 kPax dm  V1  18 dm V2  x dm = = T1 T2 500 K 350 K T1  500 K T2  350 K

3 V2  9.3 dm 4.

P1  75 kPa P2  53 kPa P V P V 3 1 1 2 2 75 kPa12 L 53 kPax L V1  12 L V2  x dm = = T T o o 1 2 300 K 366 K T1  27 C  273  300 K T2  93 C  273  366 K

V2  20.7 L 5.

P1  1500 mm Hg P2  x mm Hg P1V1 P2V2 1500 mm Hg5 L x kPa25 L V1  5 L V2  25 L = = T1 T2 2000 K 1750 K T1  2000 K T2  1750 K

P2  262.5 mm Hg 6.

P1  x atm P2  1.85 atm P V P V 3 3 3 3 1 1 2 2 x atm30 dm  1.85 atm18 dm  V  30 dm V  18 dm = = 1 2 T T o o 1 2 485 K 653 K T1  212 C  273  485 K T2  380 C  273  653 K

P1  0.82 atm

Answers: 1. –142oC 2. 300 K 3. 9.3 dm3 4. 20.7 L 5. 262.5 mm Hg 6. 0.82 atm