Effects of Temperature on a Balloon's Buoyancy?

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SUMMARY

The discussion focuses on the effects of temperature on a balloon's buoyancy, specifically in the context of large helium-filled air bags used for high-altitude measurements. As the balloon rises, the surrounding air density decreases, but the buoyant force remains constant due to the ideal gas law (PV = nRT), which indicates that the weight of the displaced air does not change with altitude. The confusion arises regarding the relationship between atmospheric temperature and buoyancy, as the same weight of air is displaced regardless of altitude, leading to a consistent buoyant force.

PREREQUISITES
  • Understanding of the ideal gas law (PV = nRT)
  • Knowledge of buoyancy principles and Archimedes' principle
  • Familiarity with concepts of air density and atmospheric pressure
  • Basic physics of gases and their behavior under varying conditions
NEXT STEPS
  • Research the implications of the ideal gas law on buoyancy in different altitudes
  • Study the relationship between temperature, pressure, and density in gases
  • Explore the physics of buoyancy in varying atmospheric conditions
  • Investigate the design and function of high-altitude balloons and their applications
USEFUL FOR

Students studying physics, educators teaching buoyancy concepts, and engineers involved in the design of high-altitude measurement instruments.

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Homework Statement



Very large air bags are used to lift instruments for high altitude measurements as shown in the image below. At ground level the bag is only partially filled with helium, just buoyant enough to rise.

A) As the balloon rises, describe what happens to the magnitude of the buoyant force on the balloon.

B) Given that air temperature steadily decreases at higher altitudes, explain the effect of atmospheric temperature on the net buoyant force of this system?

Homework Equations



No equations were given.

The Attempt at a Solution



A) As the balloon rises from sea level, the surrounding air becomes less dense, causing the volume of the helium inside the balloon to increase. However, due to the ideal gas law, the weight of the displaced air remains the same as it was at sea level. The ideal gas law is PV = nRT. P represents pressure and V volume. On the other side of the equation there is: n (the number of moles), the universal gas constant R, and absolute temperature T. From sea level, the balloon rises due to a buoyant force acting on it. This force results from the helium displacing a weight of air greater than itself. However, as mentioned, the same weight of air is displaced despite the air becoming less dense as the balloon rises. Thus, the magnitude of the buoyant force on the balloon remains the same as the balloon rises.

B) I have no idea how this is different from question A, which is where part of the confusion is. Any help is appreciated.
 
Physics news on Phys.org
Why is the same weight of air being displaced at any height?
How did you deduce that?
What are the assumptions?
Are there no limit to your deduction?
Could you go up to the moon with this balloon?
Don't hesitate to be analytical, that's the aim.
 

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