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GAS MATH
Init 2/27/2008 by Daniel R. Barnes
2
22
1
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T
VP
T
VP
. . . 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
by clicking the link below.
“kPa” = “kilopascal” = a unit of pressure
P1 = 200 kPa
“mL” = “milliliter” = a unit of volume
V1 = 400 mL
V2 = ?P2 = 18,000 kPa
1. 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.
T1 = T2
Congratulations. You’ve justTRANSLATED ENGLISH INTO MATH
Now you can figure out which FORMULA from the reference sheet to use Skip to
answer
1. 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.
P1 = 200 kPa V1 = 400 mL
V2 = ?P2 = 18,000 kPa T1 = T2
2
22
1
11
T
VP
T
VP
In gas math problems that use the combined gas law, whatever remains constant can be crossed out of the equation.
P1V1 = P2V2
If you cross the temperatures out of the combined gas law, you get Boyle’s law.
Skip to answer
P1V1 = P2V2
1. 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.
P1 = 200 kPa V1 = 400 mL
V2 = ?P2 = 18,000 kPa T1 = T2
2
22
1
11
T
VP
T
VP
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 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
P1V1 = P2V2
P1 = 200 kPa V1 = 400 mL
V2 = ?P2 = 18,000 kPa T1 = T2
2
22
1
11
T
VP
T
VP
P2 P2
V2 = P1V1
P2
Now that we have V2 all by itself, it’s time to PLUG IN THE NUMBERS
V2 = (200 kPa) (400 mL) (18,000 kPa)
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.
Skip to answer
P1V1 = P2V2
P1 = 200 kPa V1 = 400 mL
V2 = ?P2 = 18,000 kPa T1 = T2
2
22
1
11
T
VP
T
VP
P2 P2
V2 = P1V1
P2
V2 = (200 kPa) (400 mL) (18,000 kPa)
= 4.4 mL = V2
1. 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.
Skip to answer
P1V1 = P2V2
P1 = 200 kPa V1 = 400 mL
V2 = ?P2 = 18,000 kPa T1 = T2
2
22
1
11
T
VP
T
VP
P2 P2
V2 = P1V1
P2
V2 = (200 kPa) (400 mL) (18,000 kPa)
= 4.4 mL = V2
1. 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.
GAS MATH INSTRUCTION MANUAL1. Translate English into math. (Declare your variables.)
2. Write the correct formula from the reference sheet/memory.
3. 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.
i. Determine what kind of problem you’re looking at
ii. Visualize the story
Steps before the steps:
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.
P1 = 100 kPa V1 = 800 mL
V2 = ? mLP2 = 60 kPa
T1 = T2
P1V1
T1
=P2V2
T2
P1V1 = P2V2
P2 P2
V2 = P1V1
P2
V2 = (100 kPa) (800 mL)
(60 kPa)=
8000 mL
6= 1333.3 mL = V2
Skip to answer
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.
P1 = 100 kPa V1 = 800 mL
V2 = ? mLP2 = 60 kPa
T1 = T2
P1V1
T1
=P2V2
T2
P1V1 = P2V2
P2 P2
V2 = P1V1
P2
V2 = (100 kPa) (800 mL)
(60 kPa)=
8000 mL
6= 1333.3 mL = V2
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
V2 = 30 mL
P1 = 100 kPa
P2 = ? kPa
T1 = T2
P1V1
T1
=P2V2
T2
P1V1 = P2V2
V2 V2
P2 = P1V1
V2
P2 = (100 kPa) (600 mL)
(30 mL)=
6000 kPa
3= 2000 kPa = P2
Skip to answer
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
V2 = 30 mL
P1 = 100 kPa
P2 = ? kPa
T1 = T2
P1V1
T1
=P2V2
T2
P1V1 = P2V2
V2 V2
P2 = P1V1
V2
P2 = (100 kPa) (600 mL)
(30 mL)=
6000 kPa
3= 2000 kPa = P2
P1V1 = P2V2
420 mmHg
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
V2 = 4000 mL
P1 = 700 mmHg
P2 = ? mmHg
T1 = T2
P1V1
T1
=P2V2
T2 V2 V2
P2 = P1V1
V2
P2 = (700 mmHg)(600 mL)
(4000 mL)=
4= 105 mmHg = P2
Skip to answer
P1V1 = P2V2
420 mmHg
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
V2 = 4000 mL
P1 = 700 mmHg
P2 = ? kPa
T1 = T2
P1V1
T1
=P2V2
T2 V2 V2
P2 = P1V1
V2
P2 = (700 mmHg)(600 mL)
(4000 mL)=
4= 105 mmHg = P2
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.
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
T1 = 127oC
T2 = ? oC
P1 = P2
= (127 + 273)K = 400 K
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.
To convert Celsius to kelvins, add 273.
Skip to answer
V1
T1
V2
T2
= V2
2400 K
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
T1 = 127oC
T2 = ? oC
P1 = P2
P1V1
T1
=P2V2
T2
T2 =
(60 mL)(400K)(480 mL)
=
48= 50 K
= (127 + 273)K = 400 K
V2
V1 1
T2
= V2 T1
V1 1
T2=
V2 T1
V1
T2 = V2 T1
= (50 – 273)oC = -223oC = T2
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.
Skip to answer
2400 K
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
T1 = 127oC
T2 = ? oC
P1 = P2
P1V1
T1
=P2V2
T2
T2 =
(60 mL)(400K)(480 mL)
=
48= 50 K
V1
T1
V2
T2
=
= (127 + 273)K = 400 K
V2V2
V1 1
T2
= V2 T1
V1 1
T2=
V2 T1
V1
T2 = V2 T1
= (50 – 273)oC = -223oC = T2
Now, remember . . .
P1V1
T1
=P2V2
T2
P1V1
T1
=P2V2
T2
When T2 is the unknown , there is danger . . .
P2V2 P2V2
But be careful! It’s not T2 on the right, it’s 1/T2 -- Just the opposite!
P1V1
T1
=1
T2P2V2 P1V1
T1=1
T2 P2V2
Wouldn’t it be ironic if a crab got cancer?
That would be like a wolf getting lupus.
V2
V1
T1
V2
T2
=
36,000 K
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.
V1 = 30 L
V2 = 1200 L
T1 = 27oC
T2 = ? oC
P1 = P2
P1V1
T1
=P2V2
T2
T2 =
(1200 L)(300K)(30 L)
= 3
12,000 K
= (27 + 273)K = 300 K
V2
V1 1
T2
= V2 T1
V1 1
T2=
V2 T1
V1
T2 = V2 T1
= (12,000 – 273)oC = 11,727oC = T2
=
Skip to answer
36,000 K
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.
V1 = 30 L
V2 = 1200 L
T1 = 27oC
T2 = ? oC
P1 = P2
P1V1
T1
=P2V2
T2
T2 =
(1200 L)(300K)(30 L)
= 3
12,000 K
V1
T1
V2
T2
=
= (27 + 273)K = 300 K
V2V2
V1 1
T2
= V2 T1
V1 1
T2=
V2 T1
V1
T2 = V2 T1
= (12,000 – 273)oC = 11,727oC = T2
=
You work hard.
You get tiny boxing bunny.
T2V1
T1
V2
T2
= P1V1
T1
=P2V2
T2
2100 cc
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.
V1 = 70 cc
V2 = ? cc
T1 = 327oC P1 = P2
(3000 K)(70 cc)(600 K) 6
= 350 cc = V2
= (327 + 273)K = 600 K
T2
T1
V2 = T2 V1
V2 = =
T2 = 2727oC = (2727 + 273)K = 3000 K
Skip to answer
V1
T1
V2
T2
=
2100 cc
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.
V1 = 70 cc
V2 = ? L
T1 = 327oC P1 = P2
P1V1
T1
=P2V2
T2
(3000 K)(70 cc)(600 K) 6
= 350 cc = V2
= (327 + 273)K = 600 K
T2T2
T1
V2 = T2 V1
V2 = =
T2 = 2727oC = (2727 + 273)K = 3000 K
T2V1
T1
V2
T2
=
3 mL
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.
V1 = 0.6 mL
V2 = ? mL
T1 = 27oC P1 = P2
P1V1
T1
=P2V2
T2
(50 K) (0.6mL)(300 K) 30
= 0.1 mL = V2
= (27 + 273)K = 300 K
T2
T1
V2 = T2 V1
V2 = =
T2 = -223oC = (-223 + 273)K = 50 K
Skip to answer
V1
T1
V2
T2
=
3 mL
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.
V1 = 0.6 mL
V2 = ? L
T1 = 27oC P1 = P2
P1V1
T1
=P2V2
T2
(50 K) (0.6mL)(300 K) 30
= 0.1 mL = V2
= (27 + 273)K = 300 K
T2T2
T1
V2 = T2 V1
V2 = =
T2 = -223oC = (-223 + 273)K = 50 K
2700 K
P2
P1
T1
P2
T2
=
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.
P1 = 30 lbs/in2
P2 = 90 lbs/in2
T1 = 27oC V1 = V2
P1V1
T1
=P2V2
T2
(90 lbs/in2)(300K)(30 lbs/in2)
= 900 K
= (27 + 273) K = 300 K
P2 P1
T2 = P2 T1
T2 = =
T2 = ? oC
P1
T2
= P2 T1
1
3
T2 = (900 – 273)oC = 627 oC = T2
Skip to answer
2700 K
P2
P1
T1
P2
T2
=
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.
P1 = 30 lbs/in2
P2 = 90 lbs/in2
T1 = 27oC V1 = V2
P1V1
T1
=P2V2
T2
(90 lbs/in2)(300K)(30 lbs/in2)
= 900 K
= (27 + 273) K = 300 K
P2 P1
T2 = P2 T1
T2 = =
T2 = ? oC
P1
T2
= P2 T1
1
3
T2 = (900 – 273)oC = 627 oC = T2
240 K
P2
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.
P1 = 20 lb/in2 T1 = 27oC = (27 + 273)K = 300 K
P2 = 8 lb/in2 T2 = ? oC V1 = V2
P1V1
T1
=P2V2
T2
P1
T1
P2
T2
= P2 P1
T2 = P2 T1P1
T2
= P2 T1
1
(8 lb/in2)(300K)(20 lb/in2)
= 120 KT2 = = 2
= (120 – 273) oC = -153 oC = T2
Skip to answer
240 K
P2
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.
P1 = 20 lb/in2 T1 = 27oC = (27 + 273)K = 300 K
P2 = 8 lb/in2 T2 = ? oC V1 = V2
P1V1
T1
=P2V2
T2
P1
T1
P2
T2
= P2 P1
T2 = P2 T1P1
T2
= P2 T1
1
(8 lb/in2)(300K)(20 lb/in2)
= 120 KT2 = = 2
= (120 – 273) oC = -153 oC = T2
(700 mmHg)
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.
T1 = 27oC = (27 + 273)K = 300 K
T2 = 2727oC = (2727 + 273)K = 3000 K
P1 = 700 mmHg
P2 = ? mmHg
P1V1
T1
=P2V2
T2
P1
T1
P2
T2
= T2 T2
V1 = V2
T1
P2 = T2 P1 P2 =
(3000 K)
(300 K)
21,000 mmHg=
3
P2 = 7000 mmHg
Skip to answer
(700 mmHg)
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.
T1 = 27oC = (27 + 273)K = 300 K
T2 = 2727oC = (2727 + 273)K = 3000 K
P1 = 700 mmHg
P2 = ? mmHg
P1V1
T1
=P2V2
T2
P1
T1
P2
T2
= T2 T2
V1 = V2
T1
P2 = T2 P1 P2 =
(3000 K)
(300 K)
21,000 mmHg=
3
P2 = 7000 mmHg
(900 mmHg)
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.
T1 = 27oC = (27 + 273)K = 300 K
T2 = -183oC = (-183 + 273)K = 90 K
P1 = 900 mmHg
P2 = ? mmHg
P1V1
T1
=P2V2
T2
P1
T1
P2
T2
= T2 T2
V1 = V2
T1
P2 = T2 P1 P2 =
(90 K)
(300 K)
810 mmHg=
3= 270 mmHg = P2
Skip to answer
P2
(900 mmHg)
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.
T1 = 27oC = (27 + 273)K = 300 K
T2 = -183oC = (-183 + 273)K = 90 K
P1 = 900 mmHg
P2 = ? mmHg
P1V1
T1
=P2V2
T2
P1
T1
P2
T2
= T2 T2
V1 = V2
T1
P2 = T2 P1 P2 =
(90 K)
(300 K)
810 mmHg=
3= 270 mmHg = P2P2
(800 m3)
(300 K)
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?
T1 = 300 K
T2 = 100 K
P1 = 15 psi
P2 = 10 psi
P1V1
T1
=P2V2
T2
V1 = 800 m3
P2T1
V2 =T2 P1V1
V2 = 400 m3
Skip to answer
V2 = ? m3
P2
T2
P2
T2
(15 psi)V2 =
(100 K)
(10 psi)V2 =
1200 m3
3
(800 m3)
(300 K)
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?
T1 = 300 K
T2 = 100 K
P1 = 15 psi
P2 = 10 psi
P1V1
T1
=P2V2
T2
V1 = 800 m3
P2T1
V2 =T2 P1V1
V2 = ? m3
P2
T2
P2
T2
(15 psi)V2 =
(100 K)
(10 psi)
V2 = 400 m3
V2 =1200 m3
3
Okay, so in real life the balloon would expand as it rises. Sorry. 100K is unrealistically cold.
(275 K)
(5 mL)
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?
T1 = 275 KT2 = 300 K
P1 = ?
P2 = 1 atm
P1V1
T1
=P2V2
T2
V1 = 5 mL
T2V1
P1 =P2V2T1
P1 = 33 atm
Skip to answer
V2 = 180 mL
V1
T1
(180 mL)P1 =
(1 atm)
(300K)P1 =
4950 atm
150
V1
T1
(275 K)
(5 mL)
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?
T1 = 275 KT2 = 300 K
P1 = ?
P2 = 1 atm
P1V1
T1
=P2V2
T2
V1 = 5 mL
T2V1
P1 =P2V2T1
P1 = 33 atm
V2 = 180 mL
V1
T1
(180 mL)P1 =
(1 atm)
(300K)P1 =
4950 atm
150
V1
T1
CH4
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?
Skip to answer
T1 = ?
P1
P2
T2
V2
V1
P1 = 21 atmV1 = 300 L
P2 = 3 atmT2 = 200 K
V2 = 200 L
T1 = ?
P1 = 21 atmV1 = 300 L
P2 = 3 atmT2 = 200 K
V2 = 200 L
T1 = ?
(300 L)
(200 L)
P1V1
T1
=P2V2
T2 P2V2
T1 =T2P1V1
T1 = 2100 K(21 atm)T1 =
(200 K)
(3 atm)T1 =
12600 K
6
P1V1
P1 = 21 atmV1 = 300 L
P2 = 3 atm T2 = 200 KV2 = 200 L
T1 = ?
P1V1
1
T2P1V1T1
P2V2=
Skip to answer
(300 L)
(200 L)
P1V1
T1
=P2V2
T2 P2V2
T1 =T2P1V1
T1 = 2100 K(21 atm)T1 =
(200 K)
(3 atm)T1 =
12600 K
6
P1V1
P1 = 21 atmV1 = 300 L
P2 = 3 atm T2 = 200 KV2 = 200 L
T1 = ?
P1V1
1
T2P1V1T1
P2V2=
(Q#15. Space Marine Mario)
