Mixing
As I inflate a balloon at a constant temperature does the pressure of gases inside it on the balloon increase?
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I know it's a dumb question but I am having a little misunderstanding with Boyle's Law.
Shouldn't pressure be inversely proportional with Volume , or this is meant for the pressure outside of gases outside the balloon,can you use another example for this relation .
pressure ideal-gas
asked 3 hours ago
user597368
132
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user597368 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
2
down vote
favorite
I know it's a dumb question but I am having a little misunderstanding with Boyle's Law.
Shouldn't pressure be inversely proportional with Volume , or this is meant for the pressure outside of gases outside the balloon,can you use another example for this relation .
pressure ideal-gas
asked 3 hours ago
user597368
132
New contributor
user597368 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Boyle's law is assuming that the quantity of gas (in moles or atoms) remains constant. Is this precondition met in the case you are considering?
– dmckee♦
3 hours ago
Oh surely its not ,Sorry I forgot about this
– user597368
3 hours ago
Is there anyway to delete this question now?
– user597368
3 hours ago
Please don't delete the question, I think it is an interesting case to look at!
– bRost03
2 hours ago
1
Agreed with @bRost03 and welcome to physics stack exchange! Questions are asked not only for their answers to the OP but to the benefit to the entire stack exchange community. Because you've asked it, now anyone that ever gets confused will have somewhere to turn to. Feel good that you've helped better the understanding of the world as a whole :)
– Joshua Ronis
2 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I know it's a dumb question but I am having a little misunderstanding with Boyle's Law.
Shouldn't pressure be inversely proportional with Volume , or this is meant for the pressure outside of gases outside the balloon,can you use another example for this relation .
pressure ideal-gas
asked 3 hours ago
user597368
132
New contributor
user597368 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I know it's a dumb question but I am having a little misunderstanding with Boyle's Law.
Shouldn't pressure be inversely proportional with Volume , or this is meant for the pressure outside of gases outside the balloon,can you use another example for this relation .
pressure ideal-gas
pressure ideal-gas
asked 3 hours ago
user597368
132
New contributor
user597368 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
user597368
132
New contributor
user597368 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
user597368
132
New contributor
user597368 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
user597368
132
asked 3 hours ago
user597368
132
132
New contributor
user597368 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user597368 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user597368 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Boyle's law is assuming that the quantity of gas (in moles or atoms) remains constant. Is this precondition met in the case you are considering?
– dmckee♦
3 hours ago
Oh surely its not ,Sorry I forgot about this
– user597368
3 hours ago
Is there anyway to delete this question now?
– user597368
3 hours ago
Please don't delete the question, I think it is an interesting case to look at!
– bRost03
2 hours ago
1
Agreed with @bRost03 and welcome to physics stack exchange! Questions are asked not only for their answers to the OP but to the benefit to the entire stack exchange community. Because you've asked it, now anyone that ever gets confused will have somewhere to turn to. Feel good that you've helped better the understanding of the world as a whole :)
– Joshua Ronis
2 hours ago
add a comment |Â
1
Boyle's law is assuming that the quantity of gas (in moles or atoms) remains constant. Is this precondition met in the case you are considering?
– dmckee♦
3 hours ago
Oh surely its not ,Sorry I forgot about this
– user597368
3 hours ago
Is there anyway to delete this question now?
– user597368
3 hours ago
Please don't delete the question, I think it is an interesting case to look at!
– bRost03
2 hours ago
1
Agreed with @bRost03 and welcome to physics stack exchange! Questions are asked not only for their answers to the OP but to the benefit to the entire stack exchange community. Because you've asked it, now anyone that ever gets confused will have somewhere to turn to. Feel good that you've helped better the understanding of the world as a whole :)
– Joshua Ronis
2 hours ago
1
1
Boyle's law is assuming that the quantity of gas (in moles or atoms) remains constant. Is this precondition met in the case you are considering?
– dmckee♦
3 hours ago
Boyle's law is assuming that the quantity of gas (in moles or atoms) remains constant. Is this precondition met in the case you are considering?
– dmckee♦
3 hours ago
Oh surely its not ,Sorry I forgot about this
– user597368
3 hours ago
Oh surely its not ,Sorry I forgot about this
– user597368
3 hours ago
Is there anyway to delete this question now?
– user597368
3 hours ago
Is there anyway to delete this question now?
– user597368
3 hours ago
Please don't delete the question, I think it is an interesting case to look at!
– bRost03
2 hours ago
Please don't delete the question, I think it is an interesting case to look at!
– bRost03
2 hours ago
1
1
Agreed with @bRost03 and welcome to physics stack exchange! Questions are asked not only for their answers to the OP but to the benefit to the entire stack exchange community. Because you've asked it, now anyone that ever gets confused will have somewhere to turn to. Feel good that you've helped better the understanding of the world as a whole :)
– Joshua Ronis
2 hours ago
Agreed with @bRost03 and welcome to physics stack exchange! Questions are asked not only for their answers to the OP but to the benefit to the entire stack exchange community. Because you've asked it, now anyone that ever gets confused will have somewhere to turn to. Feel good that you've helped better the understanding of the world as a whole :)
– Joshua Ronis
2 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
The general relation for an ideal gas is $$PV=nRT$$ where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles of the gas particles, $Rapprox 8.314 mbox J/ molcdotmboxK$ is the gas constant and $T$ is the temperature in Kelvin.
As you inflate a balloon at constant temperature, $T$ remains constant while $n$ increases because you are adding gas with your lungs, and $V$ increases because it is inflating. Your question is what happens to the pressure $P$?
The answer is... the gas equations doesn't give you enough information to figure this out. Since $RT$ is constant you have $PV/n=mboxconst.$. But both $P$ and $V$ are allowed to vary so you need more information to answer this than just the gas law alone.
However, you know that once you've blown into the balloon, the system will come to equilibrium, which means that forces on the balloon are balanced. You have pressure inside the balloon pushing out, and this must be balanced by the outside air pressure pushing in PLUS the force of the elastic which is also pushing in. The more the balloon is inflated, the more force the elastic will apply (think Hooke's law). Outside air pressure can be taken as constant. Therefore pressure will increase as you inflate the balloon, because the force from the elastic grows as the balloon is inflated.
No need to invoke the gas law at all. In fact, the gas law can't save you here!
edited 58 mins ago
sammy gerbil
21.5k42355
answered 2 hours ago
bRost03
1137
Yes I get it now. this case is dependent of numbers to know exactly what will happen, so a real example for Boyle's would be that the balloon already has a certain amount of gas and I increase outer pressure then its volume will decrease but There is a question if I do increase the pressure of gases inside the balloon somehow its volume will increase ,wouldn't it?
– user597368
2 hours ago
It's a little hard to increase the pressure of a system without changing anything else (the gas law says that if $V$, $n$, and $T$ all remain constant, then so does $P$). But either way, $P$ and $V$ go like inverses so a rise in pressure (assuming $T$ and $n$ are fixed) corresponds to a drop in volume. This makes perfect sense, if I squeeze my balloon (shrink the volume) - the pressure will go up.
– bRost03
2 hours ago
Perhaps you are thinking about when the balloon is out of equilibrium, so the pressure in the balloon is "too high". Like if we let go after squeezing the. When we let go, the pressure in the balloon is "too high" and the volume will increase. But when $V$ increases, $P$ will decrease - precisely because the pressure was "too high". It's wants to get to lower pressure, and it can do this by increasing it's volume. Or if you like, it could also do it by lowering the temperature (this is the basis for refrigerators), or ditching some molecules (like when you crack open a soda).
– bRost03
2 hours ago
1
Sorry I cannot upvote as I have less than 15 reputation ,cheers
– user597368
2 hours ago
1
Blowing into a balloon is hardest at the start. And after that it gets easier and easier. It must be related to decreasing thickness of material as it streches.
– physicsguy19
2 hours ago
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The general relation for an ideal gas is $$PV=nRT$$ where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles of the gas particles, $Rapprox 8.314 mbox J/ molcdotmboxK$ is the gas constant and $T$ is the temperature in Kelvin.
As you inflate a balloon at constant temperature, $T$ remains constant while $n$ increases because you are adding gas with your lungs, and $V$ increases because it is inflating. Your question is what happens to the pressure $P$?
The answer is... the gas equations doesn't give you enough information to figure this out. Since $RT$ is constant you have $PV/n=mboxconst.$. But both $P$ and $V$ are allowed to vary so you need more information to answer this than just the gas law alone.
However, you know that once you've blown into the balloon, the system will come to equilibrium, which means that forces on the balloon are balanced. You have pressure inside the balloon pushing out, and this must be balanced by the outside air pressure pushing in PLUS the force of the elastic which is also pushing in. The more the balloon is inflated, the more force the elastic will apply (think Hooke's law). Outside air pressure can be taken as constant. Therefore pressure will increase as you inflate the balloon, because the force from the elastic grows as the balloon is inflated.
No need to invoke the gas law at all. In fact, the gas law can't save you here!
edited 58 mins ago
sammy gerbil
21.5k42355
answered 2 hours ago
bRost03
1137
Yes I get it now. this case is dependent of numbers to know exactly what will happen, so a real example for Boyle's would be that the balloon already has a certain amount of gas and I increase outer pressure then its volume will decrease but There is a question if I do increase the pressure of gases inside the balloon somehow its volume will increase ,wouldn't it?
– user597368
2 hours ago
It's a little hard to increase the pressure of a system without changing anything else (the gas law says that if $V$, $n$, and $T$ all remain constant, then so does $P$). But either way, $P$ and $V$ go like inverses so a rise in pressure (assuming $T$ and $n$ are fixed) corresponds to a drop in volume. This makes perfect sense, if I squeeze my balloon (shrink the volume) - the pressure will go up.
– bRost03
2 hours ago
Perhaps you are thinking about when the balloon is out of equilibrium, so the pressure in the balloon is "too high". Like if we let go after squeezing the. When we let go, the pressure in the balloon is "too high" and the volume will increase. But when $V$ increases, $P$ will decrease - precisely because the pressure was "too high". It's wants to get to lower pressure, and it can do this by increasing it's volume. Or if you like, it could also do it by lowering the temperature (this is the basis for refrigerators), or ditching some molecules (like when you crack open a soda).
– bRost03
2 hours ago
1
Sorry I cannot upvote as I have less than 15 reputation ,cheers
– user597368
2 hours ago
1
Blowing into a balloon is hardest at the start. And after that it gets easier and easier. It must be related to decreasing thickness of material as it streches.
– physicsguy19
2 hours ago
 |Â
show 4 more comments
up vote
3
down vote
accepted
The general relation for an ideal gas is $$PV=nRT$$ where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles of the gas particles, $Rapprox 8.314 mbox J/ molcdotmboxK$ is the gas constant and $T$ is the temperature in Kelvin.
As you inflate a balloon at constant temperature, $T$ remains constant while $n$ increases because you are adding gas with your lungs, and $V$ increases because it is inflating. Your question is what happens to the pressure $P$?
The answer is... the gas equations doesn't give you enough information to figure this out. Since $RT$ is constant you have $PV/n=mboxconst.$. But both $P$ and $V$ are allowed to vary so you need more information to answer this than just the gas law alone.
However, you know that once you've blown into the balloon, the system will come to equilibrium, which means that forces on the balloon are balanced. You have pressure inside the balloon pushing out, and this must be balanced by the outside air pressure pushing in PLUS the force of the elastic which is also pushing in. The more the balloon is inflated, the more force the elastic will apply (think Hooke's law). Outside air pressure can be taken as constant. Therefore pressure will increase as you inflate the balloon, because the force from the elastic grows as the balloon is inflated.
No need to invoke the gas law at all. In fact, the gas law can't save you here!
edited 58 mins ago
sammy gerbil
21.5k42355
answered 2 hours ago
bRost03
1137
Yes I get it now. this case is dependent of numbers to know exactly what will happen, so a real example for Boyle's would be that the balloon already has a certain amount of gas and I increase outer pressure then its volume will decrease but There is a question if I do increase the pressure of gases inside the balloon somehow its volume will increase ,wouldn't it?
– user597368
2 hours ago
It's a little hard to increase the pressure of a system without changing anything else (the gas law says that if $V$, $n$, and $T$ all remain constant, then so does $P$). But either way, $P$ and $V$ go like inverses so a rise in pressure (assuming $T$ and $n$ are fixed) corresponds to a drop in volume. This makes perfect sense, if I squeeze my balloon (shrink the volume) - the pressure will go up.
– bRost03
2 hours ago
Perhaps you are thinking about when the balloon is out of equilibrium, so the pressure in the balloon is "too high". Like if we let go after squeezing the. When we let go, the pressure in the balloon is "too high" and the volume will increase. But when $V$ increases, $P$ will decrease - precisely because the pressure was "too high". It's wants to get to lower pressure, and it can do this by increasing it's volume. Or if you like, it could also do it by lowering the temperature (this is the basis for refrigerators), or ditching some molecules (like when you crack open a soda).
– bRost03
2 hours ago
1
Sorry I cannot upvote as I have less than 15 reputation ,cheers
– user597368
2 hours ago
1
Blowing into a balloon is hardest at the start. And after that it gets easier and easier. It must be related to decreasing thickness of material as it streches.
– physicsguy19
2 hours ago
 |Â
show 4 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The general relation for an ideal gas is $$PV=nRT$$ where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles of the gas particles, $Rapprox 8.314 mbox J/ molcdotmboxK$ is the gas constant and $T$ is the temperature in Kelvin.
As you inflate a balloon at constant temperature, $T$ remains constant while $n$ increases because you are adding gas with your lungs, and $V$ increases because it is inflating. Your question is what happens to the pressure $P$?
The answer is... the gas equations doesn't give you enough information to figure this out. Since $RT$ is constant you have $PV/n=mboxconst.$. But both $P$ and $V$ are allowed to vary so you need more information to answer this than just the gas law alone.
However, you know that once you've blown into the balloon, the system will come to equilibrium, which means that forces on the balloon are balanced. You have pressure inside the balloon pushing out, and this must be balanced by the outside air pressure pushing in PLUS the force of the elastic which is also pushing in. The more the balloon is inflated, the more force the elastic will apply (think Hooke's law). Outside air pressure can be taken as constant. Therefore pressure will increase as you inflate the balloon, because the force from the elastic grows as the balloon is inflated.
No need to invoke the gas law at all. In fact, the gas law can't save you here!
edited 58 mins ago
sammy gerbil
21.5k42355
answered 2 hours ago
bRost03
1137
The general relation for an ideal gas is $$PV=nRT$$ where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles of the gas particles, $Rapprox 8.314 mbox J/ molcdotmboxK$ is the gas constant and $T$ is the temperature in Kelvin.
As you inflate a balloon at constant temperature, $T$ remains constant while $n$ increases because you are adding gas with your lungs, and $V$ increases because it is inflating. Your question is what happens to the pressure $P$?
The answer is... the gas equations doesn't give you enough information to figure this out. Since $RT$ is constant you have $PV/n=mboxconst.$. But both $P$ and $V$ are allowed to vary so you need more information to answer this than just the gas law alone.
However, you know that once you've blown into the balloon, the system will come to equilibrium, which means that forces on the balloon are balanced. You have pressure inside the balloon pushing out, and this must be balanced by the outside air pressure pushing in PLUS the force of the elastic which is also pushing in. The more the balloon is inflated, the more force the elastic will apply (think Hooke's law). Outside air pressure can be taken as constant. Therefore pressure will increase as you inflate the balloon, because the force from the elastic grows as the balloon is inflated.
No need to invoke the gas law at all. In fact, the gas law can't save you here!
edited 58 mins ago
sammy gerbil
21.5k42355
answered 2 hours ago
bRost03
1137
edited 58 mins ago
sammy gerbil
21.5k42355
edited 58 mins ago
sammy gerbil
21.5k42355
edited 58 mins ago
sammy gerbil
21.5k42355
21.5k42355
answered 2 hours ago
bRost03
1137
answered 2 hours ago
bRost03
1137
answered 2 hours ago
bRost03
1137
1137
Yes I get it now. this case is dependent of numbers to know exactly what will happen, so a real example for Boyle's would be that the balloon already has a certain amount of gas and I increase outer pressure then its volume will decrease but There is a question if I do increase the pressure of gases inside the balloon somehow its volume will increase ,wouldn't it?
– user597368
2 hours ago
It's a little hard to increase the pressure of a system without changing anything else (the gas law says that if $V$, $n$, and $T$ all remain constant, then so does $P$). But either way, $P$ and $V$ go like inverses so a rise in pressure (assuming $T$ and $n$ are fixed) corresponds to a drop in volume. This makes perfect sense, if I squeeze my balloon (shrink the volume) - the pressure will go up.
– bRost03
2 hours ago
Perhaps you are thinking about when the balloon is out of equilibrium, so the pressure in the balloon is "too high". Like if we let go after squeezing the. When we let go, the pressure in the balloon is "too high" and the volume will increase. But when $V$ increases, $P$ will decrease - precisely because the pressure was "too high". It's wants to get to lower pressure, and it can do this by increasing it's volume. Or if you like, it could also do it by lowering the temperature (this is the basis for refrigerators), or ditching some molecules (like when you crack open a soda).
– bRost03
2 hours ago
1
Sorry I cannot upvote as I have less than 15 reputation ,cheers
– user597368
2 hours ago
1
Blowing into a balloon is hardest at the start. And after that it gets easier and easier. It must be related to decreasing thickness of material as it streches.
– physicsguy19
2 hours ago
 |Â
show 4 more comments
Yes I get it now. this case is dependent of numbers to know exactly what will happen, so a real example for Boyle's would be that the balloon already has a certain amount of gas and I increase outer pressure then its volume will decrease but There is a question if I do increase the pressure of gases inside the balloon somehow its volume will increase ,wouldn't it?
– user597368
2 hours ago
It's a little hard to increase the pressure of a system without changing anything else (the gas law says that if $V$, $n$, and $T$ all remain constant, then so does $P$). But either way, $P$ and $V$ go like inverses so a rise in pressure (assuming $T$ and $n$ are fixed) corresponds to a drop in volume. This makes perfect sense, if I squeeze my balloon (shrink the volume) - the pressure will go up.
– bRost03
2 hours ago
Perhaps you are thinking about when the balloon is out of equilibrium, so the pressure in the balloon is "too high". Like if we let go after squeezing the. When we let go, the pressure in the balloon is "too high" and the volume will increase. But when $V$ increases, $P$ will decrease - precisely because the pressure was "too high". It's wants to get to lower pressure, and it can do this by increasing it's volume. Or if you like, it could also do it by lowering the temperature (this is the basis for refrigerators), or ditching some molecules (like when you crack open a soda).
– bRost03
2 hours ago
1
Sorry I cannot upvote as I have less than 15 reputation ,cheers
– user597368
2 hours ago
1
Blowing into a balloon is hardest at the start. And after that it gets easier and easier. It must be related to decreasing thickness of material as it streches.
– physicsguy19
2 hours ago
Yes I get it now. this case is dependent of numbers to know exactly what will happen, so a real example for Boyle's would be that the balloon already has a certain amount of gas and I increase outer pressure then its volume will decrease but There is a question if I do increase the pressure of gases inside the balloon somehow its volume will increase ,wouldn't it?
– user597368
2 hours ago
Yes I get it now. this case is dependent of numbers to know exactly what will happen, so a real example for Boyle's would be that the balloon already has a certain amount of gas and I increase outer pressure then its volume will decrease but There is a question if I do increase the pressure of gases inside the balloon somehow its volume will increase ,wouldn't it?
– user597368
2 hours ago
It's a little hard to increase the pressure of a system without changing anything else (the gas law says that if $V$, $n$, and $T$ all remain constant, then so does $P$). But either way, $P$ and $V$ go like inverses so a rise in pressure (assuming $T$ and $n$ are fixed) corresponds to a drop in volume. This makes perfect sense, if I squeeze my balloon (shrink the volume) - the pressure will go up.
– bRost03
2 hours ago
It's a little hard to increase the pressure of a system without changing anything else (the gas law says that if $V$, $n$, and $T$ all remain constant, then so does $P$). But either way, $P$ and $V$ go like inverses so a rise in pressure (assuming $T$ and $n$ are fixed) corresponds to a drop in volume. This makes perfect sense, if I squeeze my balloon (shrink the volume) - the pressure will go up.
– bRost03
2 hours ago
Perhaps you are thinking about when the balloon is out of equilibrium, so the pressure in the balloon is "too high". Like if we let go after squeezing the. When we let go, the pressure in the balloon is "too high" and the volume will increase. But when $V$ increases, $P$ will decrease - precisely because the pressure was "too high". It's wants to get to lower pressure, and it can do this by increasing it's volume. Or if you like, it could also do it by lowering the temperature (this is the basis for refrigerators), or ditching some molecules (like when you crack open a soda).
– bRost03
2 hours ago
Perhaps you are thinking about when the balloon is out of equilibrium, so the pressure in the balloon is "too high". Like if we let go after squeezing the. When we let go, the pressure in the balloon is "too high" and the volume will increase. But when $V$ increases, $P$ will decrease - precisely because the pressure was "too high". It's wants to get to lower pressure, and it can do this by increasing it's volume. Or if you like, it could also do it by lowering the temperature (this is the basis for refrigerators), or ditching some molecules (like when you crack open a soda).
– bRost03
2 hours ago
1
1
Sorry I cannot upvote as I have less than 15 reputation ,cheers
– user597368
2 hours ago
Sorry I cannot upvote as I have less than 15 reputation ,cheers
– user597368
2 hours ago
1
1
Blowing into a balloon is hardest at the start. And after that it gets easier and easier. It must be related to decreasing thickness of material as it streches.
– physicsguy19
2 hours ago
Blowing into a balloon is hardest at the start. And after that it gets easier and easier. It must be related to decreasing thickness of material as it streches.
– physicsguy19
2 hours ago
 |Â
show 4 more comments
user597368 is a new contributor. Be nice, and check out our Code of Conduct.
user597368 is a new contributor. Be nice, and check out our Code of Conduct.
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1
Boyle's law is assuming that the quantity of gas (in moles or atoms) remains constant. Is this precondition met in the case you are considering?
– dmckee♦
3 hours ago
Oh surely its not ,Sorry I forgot about this
– user597368
3 hours ago
Is there anyway to delete this question now?
– user597368
3 hours ago
Please don't delete the question, I think it is an interesting case to look at!
– bRost03
2 hours ago
1
Agreed with @bRost03 and welcome to physics stack exchange! Questions are asked not only for their answers to the OP but to the benefit to the entire stack exchange community. Because you've asked it, now anyone that ever gets confused will have somewhere to turn to. Feel good that you've helped better the understanding of the world as a whole :)
– Joshua Ronis
2 hours ago