Trying to understand weather balloons better (ideal gas law)

In summary, the conversation discusses the variables involved in understanding the behavior of a balloon filled with a lighter-than-air gas. These variables include starting pressure, starting volume, starting temperature, final pressure, final volume, and final temperature. The conversation also touches on the importance of considering the tension in the balloon's surface and the relationship between pressure and altitude. The formula (P1V1)/T1 = (P2V2)/T2 is mentioned as a way to calculate the behavior of the balloon, but additional factors such as the weight of the gas and the pressure inside the balloon must also be taken into account. Finally, the conversation raises questions about determining the final volume of the balloon at different altitudes and suggests using tables or formulas to
  • #1
amwest
Here's my understanding...
(P1V1)/T1 = (P2V2)/T2
P1 = starting pressure, typically sea level pressure which is 1atm.
V1 = Starting volume of the gas inside your balloon.
T1 = starting temperature, temperature at ground level.

P2 = final pressure, this should be the pressure at your flight altitude.
V2 = final volume, should be 100% of balloon capacity.
T2 = final temperature, temp at flight altitude.
Now I'm trying to understand this from a lighter than air gas standpoint, so...
Do i assume that:
P1, T1 are what i stated above, and do i assume that P2, T2 are proportional, if the gas isn't heated by a secondary source such as a hot air balloon?

Thanks in advance, will post further questions once these are validated or corrected!
 
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  • #2
amwest said:
Here's my understanding...
(P1V1)/T1 = (P2V2)/T2
P1 = starting pressure, typically sea level pressure which is 1atm.
V1 = Starting volume of the gas inside your balloon.
T1 = starting temperature, temperature at ground level.

P2 = final pressure, this should be the pressure at your flight altitude.
V2 = final volume, should be 100% of balloon capacity.
T2 = final temperature, temp at flight altitude.
Now I'm trying to understand this from a lighter than air gas standpoint, so...
Do i assume that:
P1, T1 are what i stated above, and do i assume that P2, T2 are proportional, if the gas isn't heated by a secondary source such as a hot air balloon?

Thanks in advance, will post further questions once these are validated or corrected!
Temperature will be the ambient atmospheric temperature. Pressure will be the pressure that the balloon fabric can support and will be at least ambient atmospheric pressure. To determine whether the balloon will have positive buoyancy at a given T, P and V, what other parameter will you have to take into account (hint: apply ideal gas law)?

AM
 
  • #3
The weight of the gas?
 
  • #4
Andrew Mason said:
Pressure will be the pressure that the balloon fabric can support and will be at least ambient atmospheric pressure.
AM

i understand that the pressure has to be low enough to not pop the balloon. What do you mean by at least ambient atm pressure?
 
  • #5
amwest said:
What do you mean by at least ambient atm pressure?
I think the point here is that the pressure inside the balloon will be slightly higher or at least equal to the ambient pressure outside the balloon, depending on tension in the balloon's surface.

If there wasn't a limit on how much the balloon could expand (about 100 to 1 volume wise), a hydrogen filled weather balloon could reach the outer layer of atmosphere, above low orbiting objects.
 
  • #6
rcgldr said:
I think the point here is that the pressure inside the balloon will be slightly higher or at least equal to the ambient pressure outside the balloon, depending on tension in the balloon's surface.

If there wasn't a limit on how much the balloon could expand (about 100 to 1 volume wise), a hydrogen filled weather balloon could reach the outer layer of atmosphere, above low orbiting objects.

ok that makes sense.

So basicly a balloon is limited by:
V1 amount of gas needed to take off
then
V2 gas volume is maxed out at whatever altitude.
 
  • #7
So if i had V1 = 1m^3 of hydrogen weight of 0.0899kg/m^3
how do i determine what its V2 would be at different heights?
 
  • #8
amwest said:
how do i determine what its V2 would be at different heights?
The unknown is how much the pressure inside the balloon is affected by the tension in the balloon. You could assume zero tension and then assume pressure inside and outside the balloon is the same. You'll need to find pressure versus altitude forumulas / tables (wiki might have this).
 
  • #9
rcgldr said:
The unknown is how much the pressure inside the balloon is affected by the tension in the balloon. You could assume zero tension and then assume pressure inside and outside the balloon is the same. You'll need to find pressure versus altitude forumulas / tables (wiki might have this).

ok I've looked and found a few tables but not very complete and I've read that the tables are exponential but haven't found a fomula.
 
  • #10
amwest said:
ok I've looked and found a few tables but not very complete and I've read that the tables are exponential but haven't found a fomula.
Try http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/barfor.html#c1" If you want to factor temperature into it, you will have to find how temperature decreases with altitude.

AM
 
Last edited by a moderator:

1. What is the ideal gas law and how does it relate to weather balloons?

The ideal gas law is a fundamental equation in thermodynamics that describes the relationship between pressure, volume, temperature, and number of moles of a gas. It is often used in meteorology to understand the behavior of gases in the atmosphere, including the gases inside weather balloons.

2. How do weather balloons work?

Weather balloons are filled with a gas, typically helium or hydrogen, and then released into the atmosphere. As the balloon rises, the air pressure decreases and the gas inside the balloon expands, causing it to increase in volume. This expansion causes the balloon to rise until it reaches a point where the air pressure is equal to the surrounding atmosphere. The height at which this occurs is recorded by instruments attached to the balloon, providing valuable data about temperature, humidity, and wind in the upper atmosphere.

3. What is the role of the ideal gas law in predicting weather?

The ideal gas law is used in weather forecasting to help understand the behavior of gases in the atmosphere. By measuring the pressure, volume, and temperature of the gases, meteorologists can make predictions about how these gases will move and interact with each other, leading to more accurate weather forecasts.

4. How do changes in temperature and pressure affect weather balloons?

Changes in temperature and pressure can have a significant impact on weather balloons. As the balloon rises, the temperature and pressure of the surrounding air decrease, causing the gas inside to expand and the balloon to rise higher. However, if the temperature or pressure changes too drastically, it can cause the balloon to burst.

5. What are some limitations of weather balloons and the ideal gas law in predicting weather?

While weather balloons and the ideal gas law are valuable tools in weather prediction, they do have limitations. For example, weather balloons can only provide data at a specific location in the atmosphere, and they may not accurately represent conditions in other areas. Additionally, the ideal gas law assumes that gases behave in a perfectly ideal manner, which may not always be the case in the real atmosphere.

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