What happens to the pressure change inside a rising balloon?

AI Thread Summary
As a hydrogen balloon rises, the external air pressure decreases, leading to an increase in the balloon's volume. According to Boyle's Law, this increase in volume results in a decrease in pressure inside the balloon. The balloon expands because the pressure difference between the inside and outside decreases, allowing it to rise. If the balloon's envelope is too tight, it may restrict the gas expansion, preventing it from rising effectively. Understanding these principles is crucial for the design of balloons, especially those used in meteorological applications.
axer
Messages
35
Reaction score
4
Hello, so i have a question regarding the pressure change inside a rising hydrogen balloon,

I know that the pressure change of air outside is decreasing(as we go up = less pressure)

And as the balloon is rising it's increasing its volume.

but what happens do the pressure change inside the balloon as it rises? decreses or increases?
 
Physics news on Phys.org
axer said:
Hello, so i have a question regarding the pressure change inside a rising hydrogen balloon,

I know that the pressure change of air outside is decreasing(as we go up = less pressure)

And as the balloon is rising it's increasing its volume.

but what happens do the pressure change inside the balloon as it rises? decreses or increases?
Constant mass, increasing volume means what?
 
russ_watters said:
Constant mass, increasing volume means what?
according to Boyle's law increasing volume means deceasing pressure,..

But can it be possible to have a decrease in the change in pressure both inside and outside?
 
axer said:
according to Boyle's law increasing volume means deceasing pressure,..

But can it be possible to have a decrease in the change in pressure both inside and outside?
Why not? What actually causes the change of pressure/volume inside if the envelope is loose?
 
russ_watters said:
What actually causes the change of pressure/volume inside if the envelope is loose?

I didn't quite understand what you meant. P = 1/V is based on a constant T..
 
axer said:
I didn't quite understand what you meant. P = 1/V is based on a constant T..
I don't mean equations, I mean objects/concepts. You have a balloon. It expands. Why?

If this were a party balloon at home on your birthday, you would answer "the balloon expands because I am blowing air into it." Does that happen here? If not, what physically is happening?
 
russ_watters said:
I don't mean equations, I mean objects/concepts. You have a balloon. It expands. Why?

If this were a party balloon at home on your birthday, you would answer "the balloon expands because I am blowing air into it." Does that happen here? If not, what physically is happening?
my explanation is that because the difference in pressure above is low therefore the volume of the balloon is increasing.

Other explanation is that because the volume of the air above is increasing, therefore the volume of the balloon is.. V(object)=V(air)

I don't know.. please tell me in your understandings
 
axer said:
my explanation is that because the difference in pressure above is low therefore the volume of the balloon is increasing.

Other explanation is that because the volume of the air above is increasing, therefore the volume of the balloon is.. V(object)=V(air)

I don't know.. please tell me in your understandings
There might be a language issue here, so fist a minor edit/correction of your wording:
..."the difference in pressure above is low therefore the volume of the balloon is increasing."

That has extra words in it. You should simply say that:
"as the balloon rises, the pressure of the air around it decreases."

Is that what you meant?
 
russ_watters said:
There might be a language issue here, so fist a minor edit/correction of your wording:
..."the difference in pressure above is low therefore the volume of the balloon is increasing."

That has extra words in it. You should simply say that:
"as the balloon rises, the pressure of the air around it decreases."

Is that what you meant?
Yes, i meant this.
 
  • #10
axer said:
Yes, i meant this.
...and if the pressure of the air around the balloon is decreasing, then the balloon will have a higher pressure than the air around it...unless it too expands.
 
  • #11
russ_watters said:
...and if the pressure of the air around the balloon is decreasing, then the balloon will have a higher pressure than the air around it...unless it too expands.
ooh, ok i think I got it, since the balloon is increasing in volume, therefore the pressure is decreasing. Ok thanks!
 
  • Like
Likes davenn and russ_watters
  • #12
axer said:
ooh, ok i think I got it, since the balloon is increasing in volume, therefore the pressure is decreasing. Ok thanks!
You have to remember that any 'serious' balloon (like a meteorological ballon) has a vast envelope that's only partially filled at sea level. By the time it is at a design height of several tens of km, the Helium inside has expanded and the envelope is more or less spherical. It's not a good idea to use a balloon with an elastic membrane that's under tension at low level because it will stop rising if the gas inside is constrained by the envelope - no lifting power.
 
Back
Top