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GAS MATH

GAS MATH. Init 2/27/2008 by Daniel R. Barnes. SWBAT. . . . solve math problems using the combined gas law. What follows are step-by-step, click-by-click work-outs of the problems on Mr. Barnes’ “Gas Math Problems Worksheet”. A digital copy of the worksheet should be accessible

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GAS MATH

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

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

  3. What follows are step-by-step, click-by-click work-outs of the problems on Mr. Barnes’ “Gas Math Problems Worksheet”. A digital copy of the worksheet should be accessible by clicking the link below.

  4. V1 = 400 mL P = 200 kPa 1 V2 = ? T1 = T2 P2 = 18,000 kPa “kPa” = “kilopascal” = a unit of pressure “mL” = “milliliter” 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. = 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

  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. 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

  6. V1 = 400 mL P = 200 kPa 1 V2 = ? T1 = T2 P2 = 18,000 kPa P1V1 = P2V2 P2 P2 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. 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

  7. 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

  8. V2 = P1V1 P2 V1 = 400 mL P = 200 kPa 1 V2 = ? T1 = T2 P2 = 18,000 kPa P1V1 = P2V2 P2 P2 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. (200 kPa) (400 mL) V2 = = 4.4 mL = V2 (18,000 kPa) Skip to answer

  9. V2 = P1V1 P2 V1 = 400 mL P = 200 kPa 1 V2 = ? T1 = T2 P2 = 18,000 kPa P1V1 = P2V2 P2 P2 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. (200 kPa) (400 mL) V2 = = 4.4 mL = V2 (18,000 kPa)

  10. GAS MATH INSTRUCTION MANUAL 1. Translate English into math. (Declare your variables.) 2. Write the correct formula from the reference sheet/memory. • 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!) Steps before the steps: 6. Cancel units and zeros. i. Determine what kind of problem you’re looking at 7. Do the arithmetic. 8. Box the answer. ii. Visualize the story

  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 Skip to answer

  12. 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

  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 Skip to answer

  14. 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

  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 = ? mmHg P1V1 = P2V2 V2 V2 (700 mmHg) (600 mL) 420 mmHg P2 = = = 105 mmHg = P2 (4000 mL) 4 Skip to answer

  16. 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

  17. 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.

  18. 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

  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. 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

  20. 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

  21. Now, remember . . .

  22. When T2 is the unknown , there is danger . . . P2V2 P1V1 P2V2 P1V1 = = P2V2 T2 P2V2 T1 T2 T1 But be careful! It’s not T2 on the right, it’s 1/T2 -- Just the opposite! 1 P1V1 T2 P2V2 T1 = = P2V2 T2 T1 1 P1V1

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

  24. 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

  25. 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

  26. You work hard. You get tiny boxing bunny.

  27. 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 = ? cc = (2727 + 273)K = 3000 K T2 T2 2100 cc (3000 K) (70 cc) = 350 cc = V2 = V2 = (600 K) 6 Skip to answer

  28. 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

  29. 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 = ? mL = (-223 + 273)K = 50 K T2 T2 3 mL (50 K) (0.6mL) = 0.1 mL = V2 = V2 = Skip to answer (300 K) 30

  30. 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

  31. 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) = = 900 K T2 = (30 lbs/in2) 3 Skip to answer T2 = (900 – 273)oC = 627 oC = T2

  32. 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) = = 900 K T2 = (30 lbs/in2) 3 T2 = (900 – 273)oC = 627 oC = T2

  33. 10. Space Cadet Katrina wants to keep some leaking poison gases from escaping from the science lab on her space station. She knows that if she can reduce the pressure of the air in the lab, it will keep bad air from flowing out into the rest of the station. She can not manipulate the pressure directly, but she can control the air conditioning remotely. The pressure in the lab is currently 20 lb/in2 and the temperature is 27oC. She won’t feel safe until she reduces the pressure to 8 lbs/in2. To what temperature must she cool the lab? Assume constant volume. P1V1 P2V2 P1 P2 1 P1 = = = T1 T2 P2 T1 T1 T2 T2 T2 P2 T1 = P1 Skip to answer P1 = 20 lb/in2 T1 = 27oC = (27 + 273)K = 300 K P2 = 8 lb/in2 T2 = ? oC V1 = V2 P2 P2 (300K) 240 K (8 lb/in2) = -153 oC = T2 = = 120 K = (120 – 273) oC T2 = (20 lb/in2) 2

  34. 10. Space Cadet Katrina wants to keep some leaking poison gases from escaping from the science lab on her space station. She knows that if she can reduce the pressure of the air in the lab, it will keep bad air from flowing out into the rest of the station. She can not manipulate the pressure directly, but she can control the air conditioning remotely. The pressure in the lab is currently 20 lb/in2 and the temperature is 27oC. She won’t feel safe until she reduces the pressure to 8 lbs/in2. To what temperature must she cool the lab? Assume constant volume. P1V1 P2V2 P1 P2 1 P1 = = = T1 T2 P2 T1 T1 T2 T2 T2 P2 T1 = P1 P1 = 20 lb/in2 T1 = 27oC = (27 + 273)K = 300 K P2 = 8 lb/in2 T2 = ? oC V1 = V2 P2 P2 (300K) 240 K (8 lb/in2) = -153 oC = T2 = = 120 K = (120 – 273) oC T2 = (20 lb/in2) 2

  35. 11. Forest Ranger Steve wants to cook some creamed corn, but he forgot his pots and pans. He puts his unopened can of creamed corn directly into the campfire. The heat of the fire increases the temperature in the can from 27oC to 2727oC, but it also increases the pressure in the can from 700 mmHg to a high enough pressure to make the can explode and shoot high into the air. At what pressure did this occur? Assume constant volume. P1V1 P2V2 P1 P2 = = T1 T2 T1 T2 21,000 mmHg T2 P1 P2 = = 3 T1 T1 = 27oC = (27 + 273)K = 300 K P1 = 700 mmHg T2 = 2727oC = (2727 + 273)K = 3000 K P2 = ? mmHg V1 = V2 T2 T2 Skip to answer (3000 K) (700 mmHg) P2 = (300 K) P2 = 7000 mmHg

  36. 11. Forest Ranger Steve wants to cook some creamed corn, but he forgot his pots and pans. He puts his unopened can of creamed corn directly into the campfire. The heat of the fire increases the temperature in the can from 27oC to 2727oC, but it also increases the pressure in the can from 700 mmHg to a high enough pressure to make the can explode and shoot high into the air. At what pressure did this occur? Assume constant volume. P1V1 P2V2 P1 P2 = = T1 T2 T1 T2 21,000 mmHg T2 P1 P2 = = 3 T1 T1 = 27oC = (27 + 273)K = 300 K P1 = 700 mmHg T2 = 2727oC = (2727 + 273)K = 3000 K P2 = ? mmHg V1 = V2 T2 T2 (3000 K) (700 mmHg) P2 = (300 K) P2 = 7000 mmHg

  37. 12. Maybelline Cousteau’s backup oxygen tank reads 900 mmHg while on her boat, where the temperature is 27oC. However, she knows that when she dives down to the bottom of an unexplored methane lake on a recently-discovered moon of Neptune, the temperature will drop down to -183oC. What will the pressure in her backup tank be at that temperature? Assume constant volume. P1V1 P2V2 P1 P2 = = T1 T2 T1 T2 810 mmHg T2 P1 P2 = = 3 T1 T1 = 27oC = (27 + 273)K = 300 K P1 = 900 mmHg T2 = -183oC = (-183 + 273)K = 90 K P2 = ? mmHg V1 = V2 T2 T2 Skip to answer (90 K) (900 mmHg) P2 = (300 K) = 270 mmHg = P2 P2

  38. 12. Maybelline Cousteau’s backup oxygen tank reads 900 mmHg while on her boat, where the temperature is 27oC. However, she knows that when she dives down to the bottom of an unexplored methane lake on a recently-discovered moon of Neptune, the temperature will drop down to -183oC. What will the pressure in her backup tank be at that temperature? Assume constant volume. P1V1 P2V2 P1 P2 = = T1 T2 T1 T2 810 mmHg T2 P1 P2 = = 3 T1 T1 = 27oC = (27 + 273)K = 300 K P1 = 900 mmHg T2 = -183oC = (-183 + 273)K = 90 K P2 = ? mmHg V1 = V2 T2 T2 (90 K) (900 mmHg) P2 = (300 K) = 270 mmHg = P2 P2

  39. 13. A partially-inflated weather balloon has a volume of 800 m3, an internal pressure of 15 psi, and a temperature of 300 K when on the ground. After the balloon rises to a very high altitude, the pressure is 10 psi and the temperature is 100 K. What will its volume be at that higher altitude? T2 P1V1 T2 T2 V2 = P2T1 P2 P2 P1V1 P2V2 = T1 T2 V1 = 800 m3 P1 = 15 psi T1 = 300 K Skip to answer V2 = ? m3 P2 = 10 psi T2 = 100 K (100 K) (15 psi) (800 m3) 1200 m3 V2 = V2 = (10 psi) (300 K) 3 V2 = 400 m3

  40. 13. A partially-inflated weather balloon has a volume of 800 m3, an internal pressure of 15 psi, and a temperature of 300 K when on the ground. After the balloon rises to a very high altitude, the pressure is 10 psi and the temperature is 100 K. What will its volume be at that higher altitude? T2 P1V1 T2 T2 V2 = P2T1 P2 P2 P1V1 P2V2 = T1 T2 V1 = 800 m3 P1 = 15 psi T1 = 300 K V2 = ? m3 P2 = 10 psi T2 = 100 K (100 K) (15 psi) (800 m3) 1200 m3 V2 = V2 = (10 psi) (300 K) 3 Okay, so in real life the balloon would expand as it rises. Sorry. 100K is unrealistically cold. V2 = 400 m3

  41. 14. A bubble of methane forms very slowly at the bottom of a swamp in winter and then rises to the surface. At the bottom of the swamp, where the temperature is 275 K, the bubble is 5 mL in size. When it reaches the surface of the water, it has expanded to a size of 180 mL, where the pressure is 1 atm and the temperature is 300 K. What was the original pressure of the gas inside the bubble, when it was on the bottom of the swamp? P2V2T1 T1 T1 P1 = T2V1 V1 V1 P1V1 P2V2 = T1 T2 V1 = 5 mL P1 = ? T1 = 275 K Skip to answer T2 = 300 K V2 = 180 mL P2 = 1 atm (1 atm) (180 mL) (275 K) 4950 atm P1 = P1 = (300K) (5 mL) 150 P1 = 33 atm

  42. 14. A bubble of methane forms very slowly at the bottom of a swamp in winter and then rises to the surface. At the bottom of the swamp, where the temperature is 275 K, the bubble is 5 mL in size. When it reaches the surface of the water, it has expanded to a size of 180 mL, where the pressure is 1 atm and the temperature is 300 K. What was the original pressure of the gas inside the bubble, when it was on the bottom of the swamp? P2V2T1 T1 T1 P1 = T2V1 V1 V1 P1V1 P2V2 = T1 T2 V1 = 5 mL P1 = ? T1 = 275 K T2 = 300 K V2 = 180 mL P2 = 1 atm (1 atm) (180 mL) (275 K) 4950 atm P1 = P1 = (300K) (5 mL) 150 P1 = 33 atm CH4

  43. 15. Space Marine Mario crash-lands his recon cruiser in a molten sulfur lake on uncharted planet PU187, and quickly sinks to the bottom of the lake. In his heavy battle armor, he can take the heat and has plenty of air to breathe, but he can not float to the surface to escape. He realizes that his inflatable life raft can lift him to the top of the lake. He grabs the life raft, blasts a hole in the bottom of the cruiser, and swims out once his vehcile has filled with molten sulfur. Standing on the bottom of the lake, he inflates the liferaft. At the bottom of the lake, he estimates the volume of the liferaft to be 300 L, and the pressure gauge on the raft reads 21 atm, but the liferaft pulls him up toward the surface of the lake before he can read the temperature. At the surface, the pressure gauge on the raft reads 3 atm, his suit’s computer tells him that the temperature of the molten sulfur is 200 K, and the liferaft appears to have a volume of 200 L. What was the temperature of the bottom of the lake? V1 P1 P1 = 21 atm V1 = 300 L P2 T2 P2 = 3 atm V2 T2 = 200 K V2 = 200 L T1 = ? T1 = ? Skip to answer

  44. P1 = 21 atm V1 = 300 L P2 = 3 atm T2 = 200 K V2 = 200 L T1 = ?

  45. T2P1V1 1 T1 = P2V2 = P2V2 T1 T2P1V1 P1V1 P2V2 = T1 T2 V1 = 300 L P1 = 21 atm T1 = ? V2 = 200 L P2 = 3 atm T2 = 200 K P1V1 P1V1 (200 K) (21 atm) (300 L) 12600 K T1 = 2100 K T1 = T1 = (3 atm) (200 L) 6 Skip to answer

  46. T2P1V1 1 T1 = P2V2 = P2V2 T1 T2P1V1 P1V1 P2V2 = T1 T2 V1 = 300 L P1 = 21 atm T1 = ? V2 = 200 L P2 = 3 atm T2 = 200 K P1V1 P1V1 (200 K) (21 atm) (300 L) 12600 K T1 = 2100 K T1 = T1 = (3 atm) (200 L) 6 (Q#15. Space Marine Mario)

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