GAS MATH

1. GAS MATH Init 2/27/2008 by Daniel R. Barnes

2. SWBAT . . . . . . solve math problems using the combined gas law.

3. V1 = 400 mL P = 200 kPa 1 V2 = ? T1 = T2 P2 = 18,000 kPa “kPa” = “kilopascal” = a unit of pressure The air inside a tennis ball in a sinking cruise ship is initially at 200 kPa and occupies 400 mL. What will its volume be when the ship has sunk enough that the pressure has gone up to 18,000 kPa? Assume that the temperature remains constant. “mL” = “milliliter” = a unit of volume Congratulations. You’ve just TRANSLATED ENGLISH INTO MATH Now you can figure out which FORMULA from the reference sheet to use Skip to answer

4. V1 = 400 mL P = 200 kPa 1 V2 = ? T1 = T2 P2 = 18,000 kPa P1V1 = P2V2 The air inside a tennis ball in a sinking cruise ship is initially at 200 kPa and occupies 400 mL. What will its volume be when the ship has sunk enough that the pressure has gone up to 18,000 kPa? Assume that the temperature remains constant. In gas math problems that use the combined gas law, whatever remains constant can be crossed out of the equation. If you cross the temperatures out of the combined gas law, you get Boyle’s law. Skip to answer

5. V1 = 400 mL P = 200 kPa 1 V2 = ? T1 = T2 P2 = 18,000 kPa P1V1 = P2V2 The air inside a tennis ball in a sinking cruise ship is initially at 200 kPa and occupies 400 mL. What will its volume be when the ship has sunk enough that the pressure has gone up to 18,000 kPa? Assume that the temperature remains constant. P2 P2 V2 is the unknown, so we need to isolate V2 so we can solve for it. That means we need to get V2 by itself. V2 has P2 stuck to it, so we need to get rid of P2. We do this by dividing both sides of the equation by P2. P2 is now on the top and the bottom of the right hand side of the equation, so we can cancel them out. Skip to answer

6. Now that we have V2 all by itself, it’s time to PLUG IN THE NUMBERS V2 = P1V1 P2 Don’t forget to include the units along with the numbers. We’ve got kPa on the top and on the bottom, so they can cancel each other out. V1 = 400 mL P = 200 kPa 1 V2 = ? T1 = T2 P2 = 18,000 kPa P1V1 = P2V2 P2 P2 (200 kPa) (400 mL) V2 = (18,000 kPa) Skip to answer

7. V2 = P1V1 P2 V1 = 400 mL P = 200 kPa 1 V2 = ? T1 = T2 P2 = 18,000 kPa P1V1 = P2V2 The air inside a tennis ball in a sinking cruise ship is initially at 200 kPa and occupies 400 mL. What will its volume be when the ship has sunk enough that the pressure has gone up to 18,000 kPa? Assume that the temperature remains constant. P2 P2 (200 kPa) (400 mL) V2 = = 4.4 mL = V2 (18,000 kPa) Skip to answer

8. V2 = P1V1 P2 V1 = 400 mL P = 200 kPa 1 V2 = ? T1 = T2 P2 = 18,000 kPa P1V1 = P2V2 The air inside a tennis ball in a sinking cruise ship is initially at 200 kPa and occupies 400 mL. What will its volume be when the ship has sunk enough that the pressure has gone up to 18,000 kPa? Assume that the temperature remains constant. P2 P2 (200 kPa) (400 mL) V2 = = 4.4 mL = V2 (18,000 kPa)

9. GAS MATH INSTRUCTION MANUAL 1. Translate English into math. (Declare your variables.) 2. Write the correct formula from the reference sheet. • Cross out anything that remains constant • (or anything that is not even mentioned). 4. Rearrange the formula to isolate the unknown. 5. Plug in the numbers. (Include the units!) 6. Cancel units and zeros. 7. Do the arithmetic. 8. Box the answer.

10. 2. Joannie is driving up into the mountains to have a picnic. The air inside an airtight bag of potato chips in her backpack is initially at 100 kPa and has a volume of 800 mL. What will its volume be when she has driven up to an altitude where the pressure is 60 kPa? Assume that the temperature remains constant. V2 = P1V1 P2 P1V1 P2V2 = T1 T2 P1 = 100 kPa V1 = 800 mL T1 = T2 P2 = 60 kPa V2 = ? mL P1V1 = P2V2 P2 P2 (100 kPa) (800 mL) 8000 mL V2 = = = 1333.3 mL = V2 (60 kPa) 6 Skip to answer

11. 2. Joannie is driving up into the mountains to have a picnic. The air inside an airtight bag of potato chips in her backpack is initially at 100 kPa and has a volume of 800 mL. What will its volume be when she has driven up to an altitude where the pressure is 60 kPa? Assume that the temperature remains constant. V2 = P1V1 P2 P1V1 P2V2 = T1 T2 P1 = 100 kPa V1 = 800 mL T1 = T2 P2 = 60 kPa V2 = ? mL P1V1 = P2V2 P2 P2 (100 kPa) (800 mL) 8000 mL V2 = = = 1333.3 mL = V2 (60 kPa) 6

12. P2 = P1V1 P1V1 P2V2 V2 = T1 T2 3. A man is pumping up the tires in his bicycle. With each stroke of the pump, he reduces the volume in the pump from 600 mL to 30 mL. Before he pushes down on the pump, the pressure is 100 kPa. What will the pressure be when he has pushed down on the pump and reduced the volume to the lower size? Assume that temperature remains constant. V1 = 600 mL P1 = 100 kPa T1 = T2 V2 = 30 mL P2 = ? kPa P1V1 = P2V2 V2 V2 (100 kPa) (600 mL) 6000 kPa P2 = = = 2000 kPa = P2 (30 mL) 3 Skip to answer

13. P2 = P1V1 P1V1 P2V2 V2 = T1 T2 3. A man is pumping up the tires in his bicycle. With each stroke of the pump, he reduces the volume in the pump from 600 mL to 30 mL. Before he pushes down on the pump, the pressure is 100 kPa. What will the pressure be when he has pushed down on the pump and reduced the volume to the lower size? Assume that temperature remains constant. V1 = 600 mL P1 = 100 kPa T1 = T2 V2 = 30 mL P2 = ? kPa P1V1 = P2V2 V2 V2 (100 kPa) (600 mL) 6000 kPa P2 = = = 2000 kPa = P2 (30 mL) 3

14. P2 = P1V1 P1V1 P2V2 V2 = T1 T2 4. A man in a hot air balloon has a gas bladder filled with 600 mL of air when on the ground, where the pressure is 700 mm Hg. He rises up in the sky, and the bladder swells up to a volume of 4000 mL. What is the pressure at the higher altitude? Assume that temperature remains constant. V1 = 600 mL P1 = 700 mmHg T1 = T2 V2 = 4000 mL P2 = ? kPa P1V1 = P2V2 V2 V2 (700 mmHg) (600 mL) 420 mmHg P2 = = = 105 mmHg = P2 (4000 mL) 4 Skip to answer

15. P2 = P1V1 P1V1 P2V2 V2 = T1 T2 4. A man in a hot air balloon has a gas bladder filled with 600 mL of air when on the ground, where the pressure is 700 mm Hg. He rises up in the sky, and the bladder swells up to a volume of 4000 mL. What is the pressure at the higher altitude? Assume that temperature remains constant. V1 = 600 mL P1 = 700 mmHg T1 = T2 V2 = 4000 mL P2 = ? kPa P1V1 = P2V2 V2 V2 (700 mmHg) (600 mL) 420 mmHg P2 = = = 105 mmHg = P2 (4000 mL) 4

16. So far, all the problems have invovled constant temperature The next problem invovles changing temperature, and introduces a few new tricky things that the previous problems didn’t have. Get ready for some new stuff.

17. 5. Professor Cryopithecus pours liquid nitrogen onto a balloon, causing it to shrivel from 480 mL to 60 mL. If the balloon was at 127oC before he poured the liquid nitrogen on the balloon, what temperature must the balloon have had after he poured the liquid nitrogen onto it? Assume constant pressure. V1 = 480 mL V2 = 60 mL P1 = P2 To convert Celsius to kelvins, add 273. T1 = 127oC = (127 + 273)K = 400 K T2 = ? oC Here’s new thing #1. You’re not allowed to plug Celsius temperatures into gas math formulas. T = KEave. Since KE = ½ mv2, KE can never be negative, so T shouldn’t ever really be negative either. Both the Celsius and Fahrenheit temperature scales can go negative, but Kelvin only goes down to zero, so Kelvin is the only scale appropriate for any math that depends upon the mechanics of kinetic theory. In the kelvin temperature scale, zero really means zero. At zero kelvins, absolute zero, molecules truly stop moving. Skip to answer

18. V1 V2 V1 1 = = P1V1 P2V2 T1 T2 V2 T1 T2 = T1 T2 V2 T1 T2 V2 T1 T2 = = V1 1 V1 5. Professor Cryopithecus pours liquid nitrogen onto a balloon, causing it to shrivel from 480 mL to 60 mL. If the balloon was at 127oC before he poured the liquid nitrogen on the balloon, what temperature must the balloon have had after he poured the liquid nitrogen onto it? Assume constant pressure. New thing #2: Notice that T2, our unknown, is on the bottom of the fraction. It can’t stay there. We need to flip the equation. New thing #3: Since T1 was originally given in oC, we’d better convert back from Kelvins to Celsius. V1 = 480 mL V2 = 60 mL P1 = P2 T1 = 127oC = (127 + 273)K = 400 K T2 = ? oC V2 V2 (60 mL) (400K) = (480 mL) 2400 K = 50 K = (50 – 273)oC T2 = = -223oC = T2 Skip to answer 48

19. V1 V2 V1 1 = = P1V1 P2V2 T1 T2 V2 T1 T2 = T1 T2 V2 T1 T2 V2 T1 T2 = = V1 1 V1 5. Professor Cryopithecus pours liquid nitrogen onto a balloon, causing it to shrivel from 480 mL to 60 mL. If the balloon was at 127oC before he poured the liquid nitrogen on the balloon, what temperature must the balloon have had after he poured the liquid nitrogen onto it? Assume constant pressure. V1 = 480 mL V2 = 60 mL P1 = P2 T1 = 127oC = (127 + 273)K = 400 K T2 = ? oC V2 V2 (60 mL) (400K) = (480 mL) 2400 K = 50 K = (50 – 273)oC T2 = = -223oC = T2 48

20. Wouldn’t it be ironic if a crab got cancer? That would be like a wolf getting lupus.

21. V1 V2 V1 1 = = P1V1 P2V2 T1 T2 V2 T1 T2 = T1 T2 V2 T1 T2 V2 T1 T2 = = V1 1 V1 6. Bernie, the village pyromaniac, fills a hefty bag with a mixture of methane and pure oxygen. Before he lights it on fire, its temperature is 27oC, and its volume is 30 L. Judging from the size of the fireball produced by the explosion, the gases expanded to a volume of 1200 L. What was the temperature of the gases in the fireball? Assume that the pressure in the fireball at its maximum size was equal to the pressure of the gas before it exploded. P1 = P2 T1 = 27oC = (27 + 273)K = 300 K V1 = 30 L T2 = ? oC V2 = 1200 L V2 V2 (1200 L) (300K) 36,000 K = = (30 L) 3 T2 = 12,000 K = (12,000 – 273)oC = 11,727oC = T2 Skip to answer

22. V1 V2 V1 1 = = P1V1 P2V2 T1 T2 V2 T1 T2 = T1 T2 V2 T1 T2 V2 T1 T2 = = V1 1 V1 6. Bernie, the village pyromaniac, fills a hefty bag with a mixture of methane and pure oxygen. Before he lights it on fire, its temperature is 27oC, and its volume is 30 L. Judging from the size of the fireball produced by the explosion, the gases expanded to a volume of 1200 L. What was the temperature of the gases in the fireball? Assume that the pressure in the fireball at its maximum size was equal to the pressure of the gas before it exploded. P1 = P2 T1 = 27oC = (27 + 273)K = 300 K V1 = 30 L T2 = ? oC V2 = 1200 L V2 V2 (1200 L) (300K) 36,000 K = = (30 L) 3 T2 = 12,000 K = (12,000 – 273)oC = 11,727oC = T2

23. V1 V2 T2 V1 V2 = = P1V1 P2V2 T1 T2 T1 = T1 T2 7. The gases in a cylinder of a methane-burning engine take up 70 cc and are at 327oC before combusting. After the spark plug fires, the gases explode and reach a temperature of 2727oC. To what volume should the gases expand? Assume constant pressure. T1 = 327oC V1 = 70 cc = (327 + 273)K = 600 K P1 = P2 T2 = 2727oC V2 = ? L = (2727 + 273)K = 3000 K T2 T2 2100 cc (3000 K) (70 cc) = 350 cc = V2 = V2 = (600 K) 6 Skip to answer

24. V1 V2 T2 V1 V2 = = P1V1 P2V2 T1 T2 T1 = T1 T2 7. The gases in a cylinder of a methane-burning engine take up 70 cc and are at 327oC before combusting. After the spark plug fires, the gases explode and reach a temperature of 2727oC. To what volume should the gases expand? Assume constant pressure. T1 = 327oC V1 = 70 cc = (327 + 273)K = 600 K P1 = P2 T2 = 2727oC V2 = ? L = (2727 + 273)K = 3000 K T2 T2 2100 cc (3000 K) (70 cc) = 350 cc = V2 = V2 = (600 K) 6

25. V1 V2 T2 V1 V2 = = P1V1 P2V2 T1 T2 T1 = T1 T2 8. Mike the maniac receives a porcelain elephant for his birthday. As he unpacks his gift, he decides to put the bubble wrap protecting the elephant in his liquid helium-cooled freezer. Before he puts the bubble wrap in the freezer, each bubble has a volume of 0.6 mL, at a temperature of 27oC. What will the volume of each bubble be if he lets the bubble wrap cool down to -223oC? Assume that pressure remains constant. T1 = 27oC V1 = 0.6 mL = (27 + 273)K = 300 K P1 = P2 T2 = -223oC V2 = ? L = (-223 + 273)K = 50 K T2 T2 3 mL (50 K) (0.6mL) = 0.1 mL = V2 = V2 = Skip to answer (300 K) 30

26. V1 V2 T2 V1 V2 = = P1V1 P2V2 T1 T2 T1 = T1 T2 8. Mike the maniac receives a porcelain elephant for his birthday. As he unpacks his gift, he decides to put the bubble wrap protecting the elephant in his liquid helium-cooled freezer. Before he puts the bubble wrap in the freezer, each bubble has a volume of 0.6 mL, at a temperature of 27oC. What will the volume of each bubble be if he lets the bubble wrap cool down to -223oC? Assume that pressure remains constant. T1 = 27oC V1 = 0.6 mL = (27 + 273)K = 300 K P1 = P2 T2 = -223oC V2 = ? L = (-223 + 273)K = 50 K T2 T2 3 mL (50 K) (0.6mL) = 0.1 mL = V2 = V2 = (300 K) 30

27. P1V1 P2V2 P1 P2 1 P1 = = = T1 T2 P2 T1 T1 T2 T2 T2 P2 T1 = P1 9. Professor F. Idiothead runs out of hair spray. The gases in the hair spray can are at a temperature of 27oC and a pressure of 30 lbs/in2. The professor knows that if he heats up the can, it will cause a pressure increase that will make the last bits of hair spray come out better. What he doesn’t know is that if the gases in the can reach a pressure of 90 lbs/in2, the can will explode, ruining his hair-do. To what temperature must the gases be raised for the can to explode? Assume constant volume. T1 = 27oC = (27 + 273)K = 300 K P1 = 30 lbs/in2 V1 = V2 T2 = ? oC P2 = 90 lbs/in2 P2 P2 (300K) 2700 K (90 lbs/in2) Sorry. No click-by-click animation yet. = = 900 K T2 = (30 lbs/in2) 3 T2 = (900 – 273)oC = 627 oC = T2