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If the pressure inside and outside a balloon balance, then why does air leave when it pops?
Which balloon has the higher pressure and why?mimium force so that given ballon will not burstThe relation between Internal Pressure and flow rate in a balloonHow can a small volume of gas balance the pressure exerted by the whole atmosphere?The Balloon Model for Human VentilationHow does liquid stay in a Pasteur pipette/eye dropper instead of dripping out?Why a balloon explodes by itself?Boyle's Law-$ PV= nRT.$ What equation should be used to find pressure if n is not constant, like in an elastic system?Can a balloon be inflated with electrons?Which balloon has the higher pressure and why?About a balloon in a balloon in a balloon in a balloon in a balloonA perfectly fitting pot and its lid often stick after cooking
$begingroup$
Sorry for the primitive question but when we inflate a rubber balloon and tie the end, its volume increases until its inner pressure equals atmospheric pressure.
But after that equality is obtained why does the air goes out when we pop the balloon? If there is pressure equality what causes the air flow?
pressure atmospheric-science surface-tension
edited Jun 11 at 6:35
CJ Dennis
452413
asked Jun 8 at 20:05
panic attackpanic attack
20413
$endgroup$
add a comment |
$begingroup$
Sorry for the primitive question but when we inflate a rubber balloon and tie the end, its volume increases until its inner pressure equals atmospheric pressure.
But after that equality is obtained why does the air goes out when we pop the balloon? If there is pressure equality what causes the air flow?
pressure atmospheric-science surface-tension
edited Jun 11 at 6:35
CJ Dennis
452413
asked Jun 8 at 20:05
panic attackpanic attack
20413
$endgroup$
7
$begingroup$
The inner pressure does not equal the outer pressure. Rather, they are close enough that the difference can be ignored for most purposes (such as computing the buoyancy of the balloon).
$endgroup$
– Hot Licks
Jun 9 at 2:59
3
$begingroup$
Surely you mean a rubber balloon, not a plastic one?
$endgroup$
– David Conrad
Jun 9 at 20:42
2
$begingroup$
@DavidConrad true the elasticity of the balloon makes the air blow out. The analogy of a bin bag is best to understand this, as mentioned by Bilkokuya in the comment of the most voted answer.
$endgroup$
– shabby
Jun 11 at 6:07
$begingroup$
Can you explain why you believe that a pressurized balloon has the same pressure inside and out? In particular, suppose you have an uninflated balloon and you tie off the end: is the pressure inside the balloon in that case equal to, less than, or greater than the air pressure, in your conception of how the world works?
$endgroup$
– Eric Lippert
Jun 11 at 20:37
add a comment |
$begingroup$
Sorry for the primitive question but when we inflate a rubber balloon and tie the end, its volume increases until its inner pressure equals atmospheric pressure.
But after that equality is obtained why does the air goes out when we pop the balloon? If there is pressure equality what causes the air flow?
pressure atmospheric-science surface-tension
edited Jun 11 at 6:35
CJ Dennis
452413
asked Jun 8 at 20:05
panic attackpanic attack
20413
$endgroup$
Sorry for the primitive question but when we inflate a rubber balloon and tie the end, its volume increases until its inner pressure equals atmospheric pressure.
But after that equality is obtained why does the air goes out when we pop the balloon? If there is pressure equality what causes the air flow?
pressure atmospheric-science surface-tension
pressure atmospheric-science surface-tension
edited Jun 11 at 6:35
CJ Dennis
452413
asked Jun 8 at 20:05
panic attackpanic attack
20413
edited Jun 11 at 6:35
CJ Dennis
452413
asked Jun 8 at 20:05
panic attackpanic attack
20413
edited Jun 11 at 6:35
CJ Dennis
452413
edited Jun 11 at 6:35
CJ Dennis
452413
edited Jun 11 at 6:35
CJ Dennis
452413
452413
asked Jun 8 at 20:05
panic attackpanic attack
20413
asked Jun 8 at 20:05
panic attackpanic attack
20413
asked Jun 8 at 20:05
panic attackpanic attack
20413
20413
7
$begingroup$
The inner pressure does not equal the outer pressure. Rather, they are close enough that the difference can be ignored for most purposes (such as computing the buoyancy of the balloon).
$endgroup$
– Hot Licks
Jun 9 at 2:59
3
$begingroup$
Surely you mean a rubber balloon, not a plastic one?
$endgroup$
– David Conrad
Jun 9 at 20:42
2
$begingroup$
@DavidConrad true the elasticity of the balloon makes the air blow out. The analogy of a bin bag is best to understand this, as mentioned by Bilkokuya in the comment of the most voted answer.
$endgroup$
– shabby
Jun 11 at 6:07
$begingroup$
Can you explain why you believe that a pressurized balloon has the same pressure inside and out? In particular, suppose you have an uninflated balloon and you tie off the end: is the pressure inside the balloon in that case equal to, less than, or greater than the air pressure, in your conception of how the world works?
$endgroup$
– Eric Lippert
Jun 11 at 20:37
add a comment |
7
$begingroup$
The inner pressure does not equal the outer pressure. Rather, they are close enough that the difference can be ignored for most purposes (such as computing the buoyancy of the balloon).
$endgroup$
– Hot Licks
Jun 9 at 2:59
3
$begingroup$
Surely you mean a rubber balloon, not a plastic one?
$endgroup$
– David Conrad
Jun 9 at 20:42
2
$begingroup$
@DavidConrad true the elasticity of the balloon makes the air blow out. The analogy of a bin bag is best to understand this, as mentioned by Bilkokuya in the comment of the most voted answer.
$endgroup$
– shabby
Jun 11 at 6:07
$begingroup$
Can you explain why you believe that a pressurized balloon has the same pressure inside and out? In particular, suppose you have an uninflated balloon and you tie off the end: is the pressure inside the balloon in that case equal to, less than, or greater than the air pressure, in your conception of how the world works?
$endgroup$
– Eric Lippert
Jun 11 at 20:37
7
7
$begingroup$
The inner pressure does not equal the outer pressure. Rather, they are close enough that the difference can be ignored for most purposes (such as computing the buoyancy of the balloon).
$endgroup$
– Hot Licks
Jun 9 at 2:59
$begingroup$
The inner pressure does not equal the outer pressure. Rather, they are close enough that the difference can be ignored for most purposes (such as computing the buoyancy of the balloon).
$endgroup$
– Hot Licks
Jun 9 at 2:59
3
3
$begingroup$
Surely you mean a rubber balloon, not a plastic one?
$endgroup$
– David Conrad
Jun 9 at 20:42
$begingroup$
Surely you mean a rubber balloon, not a plastic one?
$endgroup$
– David Conrad
Jun 9 at 20:42
2
2
$begingroup$
@DavidConrad true the elasticity of the balloon makes the air blow out. The analogy of a bin bag is best to understand this, as mentioned by Bilkokuya in the comment of the most voted answer.
$endgroup$
– shabby
Jun 11 at 6:07
$begingroup$
@DavidConrad true the elasticity of the balloon makes the air blow out. The analogy of a bin bag is best to understand this, as mentioned by Bilkokuya in the comment of the most voted answer.
$endgroup$
– shabby
Jun 11 at 6:07
$begingroup$
Can you explain why you believe that a pressurized balloon has the same pressure inside and out? In particular, suppose you have an uninflated balloon and you tie off the end: is the pressure inside the balloon in that case equal to, less than, or greater than the air pressure, in your conception of how the world works?
$endgroup$
– Eric Lippert
Jun 11 at 20:37
$begingroup$
Can you explain why you believe that a pressurized balloon has the same pressure inside and out? In particular, suppose you have an uninflated balloon and you tie off the end: is the pressure inside the balloon in that case equal to, less than, or greater than the air pressure, in your conception of how the world works?
$endgroup$
– Eric Lippert
Jun 11 at 20:37
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
For an inflated and tied balloon, the inner and outer pressures aren't equal. The inner pressure is higher by an amount $2 gamma |H|$, where $gamma$ is the inflated balloon's surface tension and $H$ is its mean curvature (which is $-1/R$ for a sphere). This is called the Young-Laplace equation.
After the balloon is untied and deflates, the pressures equalize and the surface tension becomes negligible.
answered Jun 8 at 20:20
tparkertparker
25.1k152134
$endgroup$
3
$begingroup$
Not seen many balloons that are spherical...
$endgroup$
– Solar Mike
Jun 8 at 21:35
6
$begingroup$
@SolarMike Weird - almost every balloon I've ever seen has been pretty close to spherical.
$endgroup$
– tparker
Jun 9 at 13:56
85
$begingroup$
They are certainly more spherical than any cow I've ever seen!
$endgroup$
– Cort Ammon
Jun 9 at 15:16
6
$begingroup$
For non-spherical balloons, the relationship between the mean curvature and the dimensions is different, but the main idea remains.
$endgroup$
– dmckee♦
Jun 9 at 17:48
7
$begingroup$
Easy way to prove this to yourself; get a bin bag and catch a bunch of air in it. Let the top of it open - notice that the air barely tries to escape, until you push the sides. The balloon is the same as this, but it's always pushing in from its own sides.
$endgroup$
– Bilkokuya
Jun 10 at 10:05
|
show 3 more comments
$begingroup$
But after that equality is obtained why does the air goes out when we penetrate the balloon? If there is pressure equality what causes the air flow?
You need to take into account that the elastic tension of the balloon skin pulls inwards. This makes the pressure in the balloon greater than its surroundings. Since there is a pressure difference the air blows out when you penetrate the skin defeating the elastic tension of the balloon skin.
Think about what happens when you blow up a balloon. At the end when the balloon gets taut it gets harder it to blow it up until it bursts. Clearly the outside pressure has not changed. The elastic tension of the balloon material has increased, like what happens when you stretch a rubber band just before it snaps.
Hope this helps.
answered Jun 8 at 20:18
Bob DBob D
8,3073728
$endgroup$
9
$begingroup$
Actually, most balloons I know are much harder to blow up while they are small, especially at the point when we go from just unwrinkoiing the skin to actually expanding it - but this effect is due to material properties of the rubber and not relevant to the qualitative effect.
$endgroup$
– Hagen von Eitzen
Jun 9 at 15:21
2
$begingroup$
@BobD No, please see equation in tparker's answer. The balloon at the start has got smaller radius, hence greater curvature which requires higher pressure at the beginning.
$endgroup$
– mpasko256
Jun 10 at 8:29
4
$begingroup$
As a balloon expands, it is actually easier to blow up under certain differences in size. See the two-balloon experiment for an example. Until I read about this experiment on SE, I had never considered how balloon pressure changes with radius. physics.stackexchange.com/questions/317032/…
$endgroup$
– JMac
Jun 10 at 13:27
2
$begingroup$
@JMac At the beginning that's true. I was thinking of what happens at the end when the balloon gets very taut. I will edit to clarify. Thanks for your input.
$endgroup$
– Bob D
Jun 10 at 13:37
1
$begingroup$
@HagenvonEitzen If you watch professionals (either people selling normal balloons, or people making balloon animals), you will see that they start by stretching the rubber to warm it up and make it more pliable, which makes the initial inflation easier. If you just start blowing into a 'cold' balloon, you are trying to do that initial "warmup" with your lungs instead - and the balloon will also be stiffer and more prone to bursting too.
$endgroup$
– Chronocidal
Jun 11 at 10:51
|
show 1 more comment
$begingroup$
until its inner pressure equals to the atmospheric pressure
The inference that the balloon is not growing (or shrinking) because the pressure is the same is not correct.
The balloon is not growing because the effective force pushing the balloon out from inside is the same as the effective force pushing the balloon in from outside.
The force pushing outwards is indeed due to the pressure of the air inside the balloon.
But the force trying to collapse the balloon is the pressure of the air on the outside (atmospheric pressure) plus the elastics potential of the balloon trying to return to its original size and shape.
So, to counteract this additional force the pressure inside the balloon has to be higher than the air pressure outside the balloon.
answered Jun 10 at 7:30
FogmeisterFogmeister
1948
$endgroup$
add a comment |
$begingroup$
Blowing into a balloon is harder than just blowing into the air, because it takes higher air pressure to stretch the rubber. once the balloon is tied the stretched rubber continues to squeeze the air inside, so inner air pressure stays higher than outer air pressure. Untie the balloon and the stretched rubber will squeeze the air out until it shrinks to its normal un-stretched size. Sticking the inflated balloon with a needle will create a flaw in the stretched rubber causing it to split open and release the inner air pressure very fast, pop.
answered Jun 9 at 2:18
Adrian HowardAdrian Howard
2446
$endgroup$
add a comment |
$begingroup$
The one thing you have not included is the tension force from the balloon membrane, that is what forces the air out.
answered Jun 8 at 20:17
Solar MikeSolar Mike
32817
$endgroup$
1
$begingroup$
to be more understandable to non physicists I used stretched rubber squeezing, instead of surface tension of membrane
$endgroup$
– Adrian Howard
Jun 9 at 8:24
$begingroup$
@AdrianHoward I kept my answer succinct as I don't have any reason for using 17 words when one will do...
$endgroup$
– Solar Mike
Jun 9 at 8:32
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For an inflated and tied balloon, the inner and outer pressures aren't equal. The inner pressure is higher by an amount $2 gamma |H|$, where $gamma$ is the inflated balloon's surface tension and $H$ is its mean curvature (which is $-1/R$ for a sphere). This is called the Young-Laplace equation.
After the balloon is untied and deflates, the pressures equalize and the surface tension becomes negligible.
answered Jun 8 at 20:20
tparkertparker
25.1k152134
$endgroup$
3
$begingroup$
Not seen many balloons that are spherical...
$endgroup$
– Solar Mike
Jun 8 at 21:35
6
$begingroup$
@SolarMike Weird - almost every balloon I've ever seen has been pretty close to spherical.
$endgroup$
– tparker
Jun 9 at 13:56
85
$begingroup$
They are certainly more spherical than any cow I've ever seen!
$endgroup$
– Cort Ammon
Jun 9 at 15:16
6
$begingroup$
For non-spherical balloons, the relationship between the mean curvature and the dimensions is different, but the main idea remains.
$endgroup$
– dmckee♦
Jun 9 at 17:48
7
$begingroup$
Easy way to prove this to yourself; get a bin bag and catch a bunch of air in it. Let the top of it open - notice that the air barely tries to escape, until you push the sides. The balloon is the same as this, but it's always pushing in from its own sides.
$endgroup$
– Bilkokuya
Jun 10 at 10:05
|
show 3 more comments
$begingroup$
For an inflated and tied balloon, the inner and outer pressures aren't equal. The inner pressure is higher by an amount $2 gamma |H|$, where $gamma$ is the inflated balloon's surface tension and $H$ is its mean curvature (which is $-1/R$ for a sphere). This is called the Young-Laplace equation.
After the balloon is untied and deflates, the pressures equalize and the surface tension becomes negligible.
answered Jun 8 at 20:20
tparkertparker
25.1k152134
$endgroup$
3
$begingroup$
Not seen many balloons that are spherical...
$endgroup$
– Solar Mike
Jun 8 at 21:35
6
$begingroup$
@SolarMike Weird - almost every balloon I've ever seen has been pretty close to spherical.
$endgroup$
– tparker
Jun 9 at 13:56
85
$begingroup$
They are certainly more spherical than any cow I've ever seen!
$endgroup$
– Cort Ammon
Jun 9 at 15:16
6
$begingroup$
For non-spherical balloons, the relationship between the mean curvature and the dimensions is different, but the main idea remains.
$endgroup$
– dmckee♦
Jun 9 at 17:48
7
$begingroup$
Easy way to prove this to yourself; get a bin bag and catch a bunch of air in it. Let the top of it open - notice that the air barely tries to escape, until you push the sides. The balloon is the same as this, but it's always pushing in from its own sides.
$endgroup$
– Bilkokuya
Jun 10 at 10:05
|
show 3 more comments
$begingroup$
For an inflated and tied balloon, the inner and outer pressures aren't equal. The inner pressure is higher by an amount $2 gamma |H|$, where $gamma$ is the inflated balloon's surface tension and $H$ is its mean curvature (which is $-1/R$ for a sphere). This is called the Young-Laplace equation.
After the balloon is untied and deflates, the pressures equalize and the surface tension becomes negligible.
answered Jun 8 at 20:20
tparkertparker
25.1k152134
$endgroup$
For an inflated and tied balloon, the inner and outer pressures aren't equal. The inner pressure is higher by an amount $2 gamma |H|$, where $gamma$ is the inflated balloon's surface tension and $H$ is its mean curvature (which is $-1/R$ for a sphere). This is called the Young-Laplace equation.
After the balloon is untied and deflates, the pressures equalize and the surface tension becomes negligible.
answered Jun 8 at 20:20
tparkertparker
25.1k152134
answered Jun 8 at 20:20
tparkertparker
25.1k152134
answered Jun 8 at 20:20
tparkertparker
25.1k152134
answered Jun 8 at 20:20
tparkertparker
25.1k152134
25.1k152134
3
$begingroup$
Not seen many balloons that are spherical...
$endgroup$
– Solar Mike
Jun 8 at 21:35
6
$begingroup$
@SolarMike Weird - almost every balloon I've ever seen has been pretty close to spherical.
$endgroup$
– tparker
Jun 9 at 13:56
85
$begingroup$
They are certainly more spherical than any cow I've ever seen!
$endgroup$
– Cort Ammon
Jun 9 at 15:16
6
$begingroup$
For non-spherical balloons, the relationship between the mean curvature and the dimensions is different, but the main idea remains.
$endgroup$
– dmckee♦
Jun 9 at 17:48
7
$begingroup$
Easy way to prove this to yourself; get a bin bag and catch a bunch of air in it. Let the top of it open - notice that the air barely tries to escape, until you push the sides. The balloon is the same as this, but it's always pushing in from its own sides.
$endgroup$
– Bilkokuya
Jun 10 at 10:05
|
show 3 more comments
3
$begingroup$
Not seen many balloons that are spherical...
$endgroup$
– Solar Mike
Jun 8 at 21:35
6
$begingroup$
@SolarMike Weird - almost every balloon I've ever seen has been pretty close to spherical.
$endgroup$
– tparker
Jun 9 at 13:56
85
$begingroup$
They are certainly more spherical than any cow I've ever seen!
$endgroup$
– Cort Ammon
Jun 9 at 15:16
6
$begingroup$
For non-spherical balloons, the relationship between the mean curvature and the dimensions is different, but the main idea remains.
$endgroup$
– dmckee♦
Jun 9 at 17:48
7
$begingroup$
Easy way to prove this to yourself; get a bin bag and catch a bunch of air in it. Let the top of it open - notice that the air barely tries to escape, until you push the sides. The balloon is the same as this, but it's always pushing in from its own sides.
$endgroup$
– Bilkokuya
Jun 10 at 10:05
3
3
$begingroup$
Not seen many balloons that are spherical...
$endgroup$
– Solar Mike
Jun 8 at 21:35
$begingroup$
Not seen many balloons that are spherical...
$endgroup$
– Solar Mike
Jun 8 at 21:35
6
6
$begingroup$
@SolarMike Weird - almost every balloon I've ever seen has been pretty close to spherical.
$endgroup$
– tparker
Jun 9 at 13:56
$begingroup$
@SolarMike Weird - almost every balloon I've ever seen has been pretty close to spherical.
$endgroup$
– tparker
Jun 9 at 13:56
85
85
$begingroup$
They are certainly more spherical than any cow I've ever seen!
$endgroup$
– Cort Ammon
Jun 9 at 15:16
$begingroup$
They are certainly more spherical than any cow I've ever seen!
$endgroup$
– Cort Ammon
Jun 9 at 15:16
6
6
$begingroup$
For non-spherical balloons, the relationship between the mean curvature and the dimensions is different, but the main idea remains.
$endgroup$
– dmckee♦
Jun 9 at 17:48
$begingroup$
For non-spherical balloons, the relationship between the mean curvature and the dimensions is different, but the main idea remains.
$endgroup$
– dmckee♦
Jun 9 at 17:48
7
7
$begingroup$
Easy way to prove this to yourself; get a bin bag and catch a bunch of air in it. Let the top of it open - notice that the air barely tries to escape, until you push the sides. The balloon is the same as this, but it's always pushing in from its own sides.
$endgroup$
– Bilkokuya
Jun 10 at 10:05
$begingroup$
Easy way to prove this to yourself; get a bin bag and catch a bunch of air in it. Let the top of it open - notice that the air barely tries to escape, until you push the sides. The balloon is the same as this, but it's always pushing in from its own sides.
$endgroup$
– Bilkokuya
Jun 10 at 10:05
|
show 3 more comments
$begingroup$
But after that equality is obtained why does the air goes out when we penetrate the balloon? If there is pressure equality what causes the air flow?
You need to take into account that the elastic tension of the balloon skin pulls inwards. This makes the pressure in the balloon greater than its surroundings. Since there is a pressure difference the air blows out when you penetrate the skin defeating the elastic tension of the balloon skin.
Think about what happens when you blow up a balloon. At the end when the balloon gets taut it gets harder it to blow it up until it bursts. Clearly the outside pressure has not changed. The elastic tension of the balloon material has increased, like what happens when you stretch a rubber band just before it snaps.
Hope this helps.
answered Jun 8 at 20:18
Bob DBob D
8,3073728
$endgroup$
9
$begingroup$
Actually, most balloons I know are much harder to blow up while they are small, especially at the point when we go from just unwrinkoiing the skin to actually expanding it - but this effect is due to material properties of the rubber and not relevant to the qualitative effect.
$endgroup$
– Hagen von Eitzen
Jun 9 at 15:21
2
$begingroup$
@BobD No, please see equation in tparker's answer. The balloon at the start has got smaller radius, hence greater curvature which requires higher pressure at the beginning.
$endgroup$
– mpasko256
Jun 10 at 8:29
4
$begingroup$
As a balloon expands, it is actually easier to blow up under certain differences in size. See the two-balloon experiment for an example. Until I read about this experiment on SE, I had never considered how balloon pressure changes with radius. physics.stackexchange.com/questions/317032/…
$endgroup$
– JMac
Jun 10 at 13:27
2
$begingroup$
@JMac At the beginning that's true. I was thinking of what happens at the end when the balloon gets very taut. I will edit to clarify. Thanks for your input.
$endgroup$
– Bob D
Jun 10 at 13:37
1
$begingroup$
@HagenvonEitzen If you watch professionals (either people selling normal balloons, or people making balloon animals), you will see that they start by stretching the rubber to warm it up and make it more pliable, which makes the initial inflation easier. If you just start blowing into a 'cold' balloon, you are trying to do that initial "warmup" with your lungs instead - and the balloon will also be stiffer and more prone to bursting too.
$endgroup$
– Chronocidal
Jun 11 at 10:51
|
show 1 more comment
$begingroup$
But after that equality is obtained why does the air goes out when we penetrate the balloon? If there is pressure equality what causes the air flow?
You need to take into account that the elastic tension of the balloon skin pulls inwards. This makes the pressure in the balloon greater than its surroundings. Since there is a pressure difference the air blows out when you penetrate the skin defeating the elastic tension of the balloon skin.
Think about what happens when you blow up a balloon. At the end when the balloon gets taut it gets harder it to blow it up until it bursts. Clearly the outside pressure has not changed. The elastic tension of the balloon material has increased, like what happens when you stretch a rubber band just before it snaps.
Hope this helps.
answered Jun 8 at 20:18
Bob DBob D
8,3073728
$endgroup$
9
$begingroup$
Actually, most balloons I know are much harder to blow up while they are small, especially at the point when we go from just unwrinkoiing the skin to actually expanding it - but this effect is due to material properties of the rubber and not relevant to the qualitative effect.
$endgroup$
– Hagen von Eitzen
Jun 9 at 15:21
2
$begingroup$
@BobD No, please see equation in tparker's answer. The balloon at the start has got smaller radius, hence greater curvature which requires higher pressure at the beginning.
$endgroup$
– mpasko256
Jun 10 at 8:29
4
$begingroup$
As a balloon expands, it is actually easier to blow up under certain differences in size. See the two-balloon experiment for an example. Until I read about this experiment on SE, I had never considered how balloon pressure changes with radius. physics.stackexchange.com/questions/317032/…
$endgroup$
– JMac
Jun 10 at 13:27
2
$begingroup$
@JMac At the beginning that's true. I was thinking of what happens at the end when the balloon gets very taut. I will edit to clarify. Thanks for your input.
$endgroup$
– Bob D
Jun 10 at 13:37
1
$begingroup$
@HagenvonEitzen If you watch professionals (either people selling normal balloons, or people making balloon animals), you will see that they start by stretching the rubber to warm it up and make it more pliable, which makes the initial inflation easier. If you just start blowing into a 'cold' balloon, you are trying to do that initial "warmup" with your lungs instead - and the balloon will also be stiffer and more prone to bursting too.
$endgroup$
– Chronocidal
Jun 11 at 10:51
|
show 1 more comment
$begingroup$
But after that equality is obtained why does the air goes out when we penetrate the balloon? If there is pressure equality what causes the air flow?
You need to take into account that the elastic tension of the balloon skin pulls inwards. This makes the pressure in the balloon greater than its surroundings. Since there is a pressure difference the air blows out when you penetrate the skin defeating the elastic tension of the balloon skin.
Think about what happens when you blow up a balloon. At the end when the balloon gets taut it gets harder it to blow it up until it bursts. Clearly the outside pressure has not changed. The elastic tension of the balloon material has increased, like what happens when you stretch a rubber band just before it snaps.
Hope this helps.
answered Jun 8 at 20:18
Bob DBob D
8,3073728
$endgroup$
But after that equality is obtained why does the air goes out when we penetrate the balloon? If there is pressure equality what causes the air flow?
You need to take into account that the elastic tension of the balloon skin pulls inwards. This makes the pressure in the balloon greater than its surroundings. Since there is a pressure difference the air blows out when you penetrate the skin defeating the elastic tension of the balloon skin.
Think about what happens when you blow up a balloon. At the end when the balloon gets taut it gets harder it to blow it up until it bursts. Clearly the outside pressure has not changed. The elastic tension of the balloon material has increased, like what happens when you stretch a rubber band just before it snaps.
Hope this helps.
answered Jun 8 at 20:18
Bob DBob D
8,3073728
edited Jun 10 at 13:40
answered Jun 8 at 20:18
Bob DBob D
8,3073728
answered Jun 8 at 20:18
Bob DBob D
8,3073728
answered Jun 8 at 20:18
Bob DBob D
8,3073728
8,3073728
9
$begingroup$
Actually, most balloons I know are much harder to blow up while they are small, especially at the point when we go from just unwrinkoiing the skin to actually expanding it - but this effect is due to material properties of the rubber and not relevant to the qualitative effect.
$endgroup$
– Hagen von Eitzen
Jun 9 at 15:21
2
$begingroup$
@BobD No, please see equation in tparker's answer. The balloon at the start has got smaller radius, hence greater curvature which requires higher pressure at the beginning.
$endgroup$
– mpasko256
Jun 10 at 8:29
4
$begingroup$
As a balloon expands, it is actually easier to blow up under certain differences in size. See the two-balloon experiment for an example. Until I read about this experiment on SE, I had never considered how balloon pressure changes with radius. physics.stackexchange.com/questions/317032/…
$endgroup$
– JMac
Jun 10 at 13:27
2
$begingroup$
@JMac At the beginning that's true. I was thinking of what happens at the end when the balloon gets very taut. I will edit to clarify. Thanks for your input.
$endgroup$
– Bob D
Jun 10 at 13:37
1
$begingroup$
@HagenvonEitzen If you watch professionals (either people selling normal balloons, or people making balloon animals), you will see that they start by stretching the rubber to warm it up and make it more pliable, which makes the initial inflation easier. If you just start blowing into a 'cold' balloon, you are trying to do that initial "warmup" with your lungs instead - and the balloon will also be stiffer and more prone to bursting too.
$endgroup$
– Chronocidal
Jun 11 at 10:51
|
show 1 more comment
9
$begingroup$
Actually, most balloons I know are much harder to blow up while they are small, especially at the point when we go from just unwrinkoiing the skin to actually expanding it - but this effect is due to material properties of the rubber and not relevant to the qualitative effect.
$endgroup$
– Hagen von Eitzen
Jun 9 at 15:21
2
$begingroup$
@BobD No, please see equation in tparker's answer. The balloon at the start has got smaller radius, hence greater curvature which requires higher pressure at the beginning.
$endgroup$
– mpasko256
Jun 10 at 8:29
4
$begingroup$
As a balloon expands, it is actually easier to blow up under certain differences in size. See the two-balloon experiment for an example. Until I read about this experiment on SE, I had never considered how balloon pressure changes with radius. physics.stackexchange.com/questions/317032/…
$endgroup$
– JMac
Jun 10 at 13:27
2
$begingroup$
@JMac At the beginning that's true. I was thinking of what happens at the end when the balloon gets very taut. I will edit to clarify. Thanks for your input.
$endgroup$
– Bob D
Jun 10 at 13:37
1
$begingroup$
@HagenvonEitzen If you watch professionals (either people selling normal balloons, or people making balloon animals), you will see that they start by stretching the rubber to warm it up and make it more pliable, which makes the initial inflation easier. If you just start blowing into a 'cold' balloon, you are trying to do that initial "warmup" with your lungs instead - and the balloon will also be stiffer and more prone to bursting too.
$endgroup$
– Chronocidal
Jun 11 at 10:51
9
9
$begingroup$
Actually, most balloons I know are much harder to blow up while they are small, especially at the point when we go from just unwrinkoiing the skin to actually expanding it - but this effect is due to material properties of the rubber and not relevant to the qualitative effect.
$endgroup$
– Hagen von Eitzen
Jun 9 at 15:21
$begingroup$
Actually, most balloons I know are much harder to blow up while they are small, especially at the point when we go from just unwrinkoiing the skin to actually expanding it - but this effect is due to material properties of the rubber and not relevant to the qualitative effect.
$endgroup$
– Hagen von Eitzen
Jun 9 at 15:21
2
2
$begingroup$
@BobD No, please see equation in tparker's answer. The balloon at the start has got smaller radius, hence greater curvature which requires higher pressure at the beginning.
$endgroup$
– mpasko256
Jun 10 at 8:29
$begingroup$
@BobD No, please see equation in tparker's answer. The balloon at the start has got smaller radius, hence greater curvature which requires higher pressure at the beginning.
$endgroup$
– mpasko256
Jun 10 at 8:29
4
4
$begingroup$
As a balloon expands, it is actually easier to blow up under certain differences in size. See the two-balloon experiment for an example. Until I read about this experiment on SE, I had never considered how balloon pressure changes with radius. physics.stackexchange.com/questions/317032/…
$endgroup$
– JMac
Jun 10 at 13:27
$begingroup$
As a balloon expands, it is actually easier to blow up under certain differences in size. See the two-balloon experiment for an example. Until I read about this experiment on SE, I had never considered how balloon pressure changes with radius. physics.stackexchange.com/questions/317032/…
$endgroup$
– JMac
Jun 10 at 13:27
2
2
$begingroup$
@JMac At the beginning that's true. I was thinking of what happens at the end when the balloon gets very taut. I will edit to clarify. Thanks for your input.
$endgroup$
– Bob D
Jun 10 at 13:37
$begingroup$
@JMac At the beginning that's true. I was thinking of what happens at the end when the balloon gets very taut. I will edit to clarify. Thanks for your input.
$endgroup$
– Bob D
Jun 10 at 13:37
1
1
$begingroup$
@HagenvonEitzen If you watch professionals (either people selling normal balloons, or people making balloon animals), you will see that they start by stretching the rubber to warm it up and make it more pliable, which makes the initial inflation easier. If you just start blowing into a 'cold' balloon, you are trying to do that initial "warmup" with your lungs instead - and the balloon will also be stiffer and more prone to bursting too.
$endgroup$
– Chronocidal
Jun 11 at 10:51
$begingroup$
@HagenvonEitzen If you watch professionals (either people selling normal balloons, or people making balloon animals), you will see that they start by stretching the rubber to warm it up and make it more pliable, which makes the initial inflation easier. If you just start blowing into a 'cold' balloon, you are trying to do that initial "warmup" with your lungs instead - and the balloon will also be stiffer and more prone to bursting too.
$endgroup$
– Chronocidal
Jun 11 at 10:51
|
show 1 more comment
$begingroup$
until its inner pressure equals to the atmospheric pressure
The inference that the balloon is not growing (or shrinking) because the pressure is the same is not correct.
The balloon is not growing because the effective force pushing the balloon out from inside is the same as the effective force pushing the balloon in from outside.
The force pushing outwards is indeed due to the pressure of the air inside the balloon.
But the force trying to collapse the balloon is the pressure of the air on the outside (atmospheric pressure) plus the elastics potential of the balloon trying to return to its original size and shape.
So, to counteract this additional force the pressure inside the balloon has to be higher than the air pressure outside the balloon.
answered Jun 10 at 7:30
FogmeisterFogmeister
1948
$endgroup$
add a comment |
$begingroup$
until its inner pressure equals to the atmospheric pressure
The inference that the balloon is not growing (or shrinking) because the pressure is the same is not correct.
The balloon is not growing because the effective force pushing the balloon out from inside is the same as the effective force pushing the balloon in from outside.
The force pushing outwards is indeed due to the pressure of the air inside the balloon.
But the force trying to collapse the balloon is the pressure of the air on the outside (atmospheric pressure) plus the elastics potential of the balloon trying to return to its original size and shape.
So, to counteract this additional force the pressure inside the balloon has to be higher than the air pressure outside the balloon.
answered Jun 10 at 7:30
FogmeisterFogmeister
1948
$endgroup$
add a comment |
$begingroup$
until its inner pressure equals to the atmospheric pressure
The inference that the balloon is not growing (or shrinking) because the pressure is the same is not correct.
The balloon is not growing because the effective force pushing the balloon out from inside is the same as the effective force pushing the balloon in from outside.
The force pushing outwards is indeed due to the pressure of the air inside the balloon.
But the force trying to collapse the balloon is the pressure of the air on the outside (atmospheric pressure) plus the elastics potential of the balloon trying to return to its original size and shape.
So, to counteract this additional force the pressure inside the balloon has to be higher than the air pressure outside the balloon.
answered Jun 10 at 7:30
FogmeisterFogmeister
1948
$endgroup$
until its inner pressure equals to the atmospheric pressure
The inference that the balloon is not growing (or shrinking) because the pressure is the same is not correct.
The balloon is not growing because the effective force pushing the balloon out from inside is the same as the effective force pushing the balloon in from outside.
The force pushing outwards is indeed due to the pressure of the air inside the balloon.
But the force trying to collapse the balloon is the pressure of the air on the outside (atmospheric pressure) plus the elastics potential of the balloon trying to return to its original size and shape.
So, to counteract this additional force the pressure inside the balloon has to be higher than the air pressure outside the balloon.
answered Jun 10 at 7:30
FogmeisterFogmeister
1948
answered Jun 10 at 7:30
FogmeisterFogmeister
1948
answered Jun 10 at 7:30
FogmeisterFogmeister
1948
answered Jun 10 at 7:30
FogmeisterFogmeister
1948
1948
add a comment |
add a comment |
$begingroup$
Blowing into a balloon is harder than just blowing into the air, because it takes higher air pressure to stretch the rubber. once the balloon is tied the stretched rubber continues to squeeze the air inside, so inner air pressure stays higher than outer air pressure. Untie the balloon and the stretched rubber will squeeze the air out until it shrinks to its normal un-stretched size. Sticking the inflated balloon with a needle will create a flaw in the stretched rubber causing it to split open and release the inner air pressure very fast, pop.
answered Jun 9 at 2:18
Adrian HowardAdrian Howard
2446
$endgroup$
add a comment |
$begingroup$
Blowing into a balloon is harder than just blowing into the air, because it takes higher air pressure to stretch the rubber. once the balloon is tied the stretched rubber continues to squeeze the air inside, so inner air pressure stays higher than outer air pressure. Untie the balloon and the stretched rubber will squeeze the air out until it shrinks to its normal un-stretched size. Sticking the inflated balloon with a needle will create a flaw in the stretched rubber causing it to split open and release the inner air pressure very fast, pop.
answered Jun 9 at 2:18
Adrian HowardAdrian Howard
2446
$endgroup$
add a comment |
$begingroup$
Blowing into a balloon is harder than just blowing into the air, because it takes higher air pressure to stretch the rubber. once the balloon is tied the stretched rubber continues to squeeze the air inside, so inner air pressure stays higher than outer air pressure. Untie the balloon and the stretched rubber will squeeze the air out until it shrinks to its normal un-stretched size. Sticking the inflated balloon with a needle will create a flaw in the stretched rubber causing it to split open and release the inner air pressure very fast, pop.
answered Jun 9 at 2:18
Adrian HowardAdrian Howard
2446
$endgroup$
Blowing into a balloon is harder than just blowing into the air, because it takes higher air pressure to stretch the rubber. once the balloon is tied the stretched rubber continues to squeeze the air inside, so inner air pressure stays higher than outer air pressure. Untie the balloon and the stretched rubber will squeeze the air out until it shrinks to its normal un-stretched size. Sticking the inflated balloon with a needle will create a flaw in the stretched rubber causing it to split open and release the inner air pressure very fast, pop.
answered Jun 9 at 2:18
Adrian HowardAdrian Howard
2446
answered Jun 9 at 2:18
Adrian HowardAdrian Howard
2446
answered Jun 9 at 2:18
Adrian HowardAdrian Howard
2446
answered Jun 9 at 2:18
Adrian HowardAdrian Howard
2446
2446
add a comment |
add a comment |
$begingroup$
The one thing you have not included is the tension force from the balloon membrane, that is what forces the air out.
answered Jun 8 at 20:17
Solar MikeSolar Mike
32817
$endgroup$
1
$begingroup$
to be more understandable to non physicists I used stretched rubber squeezing, instead of surface tension of membrane
$endgroup$
– Adrian Howard
Jun 9 at 8:24
$begingroup$
@AdrianHoward I kept my answer succinct as I don't have any reason for using 17 words when one will do...
$endgroup$
– Solar Mike
Jun 9 at 8:32
add a comment |
$begingroup$
The one thing you have not included is the tension force from the balloon membrane, that is what forces the air out.
answered Jun 8 at 20:17
Solar MikeSolar Mike
32817
$endgroup$
1
$begingroup$
to be more understandable to non physicists I used stretched rubber squeezing, instead of surface tension of membrane
$endgroup$
– Adrian Howard
Jun 9 at 8:24
$begingroup$
@AdrianHoward I kept my answer succinct as I don't have any reason for using 17 words when one will do...
$endgroup$
– Solar Mike
Jun 9 at 8:32
add a comment |
$begingroup$
The one thing you have not included is the tension force from the balloon membrane, that is what forces the air out.
answered Jun 8 at 20:17
Solar MikeSolar Mike
32817
$endgroup$
The one thing you have not included is the tension force from the balloon membrane, that is what forces the air out.
answered Jun 8 at 20:17
Solar MikeSolar Mike
32817
answered Jun 8 at 20:17
Solar MikeSolar Mike
32817
answered Jun 8 at 20:17
Solar MikeSolar Mike
32817
answered Jun 8 at 20:17
Solar MikeSolar Mike
32817
32817
1
$begingroup$
to be more understandable to non physicists I used stretched rubber squeezing, instead of surface tension of membrane
$endgroup$
– Adrian Howard
Jun 9 at 8:24
$begingroup$
@AdrianHoward I kept my answer succinct as I don't have any reason for using 17 words when one will do...
$endgroup$
– Solar Mike
Jun 9 at 8:32
add a comment |
1
$begingroup$
to be more understandable to non physicists I used stretched rubber squeezing, instead of surface tension of membrane
$endgroup$
– Adrian Howard
Jun 9 at 8:24
$begingroup$
@AdrianHoward I kept my answer succinct as I don't have any reason for using 17 words when one will do...
$endgroup$
– Solar Mike
Jun 9 at 8:32
1
1
$begingroup$
to be more understandable to non physicists I used stretched rubber squeezing, instead of surface tension of membrane
$endgroup$
– Adrian Howard
Jun 9 at 8:24
$begingroup$
to be more understandable to non physicists I used stretched rubber squeezing, instead of surface tension of membrane
$endgroup$
– Adrian Howard
Jun 9 at 8:24
$begingroup$
@AdrianHoward I kept my answer succinct as I don't have any reason for using 17 words when one will do...
$endgroup$
– Solar Mike
Jun 9 at 8:32
$begingroup$
@AdrianHoward I kept my answer succinct as I don't have any reason for using 17 words when one will do...
$endgroup$
– Solar Mike
Jun 9 at 8:32
add a comment |
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7
$begingroup$
The inner pressure does not equal the outer pressure. Rather, they are close enough that the difference can be ignored for most purposes (such as computing the buoyancy of the balloon).
$endgroup$
– Hot Licks
Jun 9 at 2:59
3
$begingroup$
Surely you mean a rubber balloon, not a plastic one?
$endgroup$
– David Conrad
Jun 9 at 20:42
2
$begingroup$
@DavidConrad true the elasticity of the balloon makes the air blow out. The analogy of a bin bag is best to understand this, as mentioned by Bilkokuya in the comment of the most voted answer.
$endgroup$
– shabby
Jun 11 at 6:07
$begingroup$
Can you explain why you believe that a pressurized balloon has the same pressure inside and out? In particular, suppose you have an uninflated balloon and you tie off the end: is the pressure inside the balloon in that case equal to, less than, or greater than the air pressure, in your conception of how the world works?
$endgroup$
– Eric Lippert
Jun 11 at 20:37