Pressure inside a helium balloon

In summary, a spherical balloon of radius 1.95 m made from a material of mass 4.56 kg and filled with helium gas at temperature 289K experiences a pressure of 678775.745 Pa.
  • #1
rubenhero
42
2

Homework Statement


g) A spherical balloon of radius R = 1.95 m is made from a material of mass M = 4.56 kg and is filled with helium gas at temperature T = 289 K. Assume the thickness of the balloon is negligible compared to the radius of the balloon, and the balloon just floats on air, neither rising nor falling. If the density of the surrounding air is ρ = 1.13 kg/m3, find P, the absolute pressure of the helium inside the balloon.
ASSUME: The balloon material displaces a negligible amount of air, and therefore creates no measurable buoyancy.


Homework Equations


M g = ρ V g, PV = nRT, Volume of sphere = 4/3(∏r3), n = total mass/molar mass


The Attempt at a Solution


V = 4/3(∏r3)
V = 4/3(∏1.953)
V = 31.05935577m3

M g = ρ V g
MHe g = 1.13 kg/m3 * 31.05935577m3 * g
MHe = 35.09707202 Kg = 35097.07202g

n = total mass/molar mass = total mass He/molar mass He
n = 35097.07202g/ (4g/mol)
n = 8774.268005 mol

P= (nRT)/V
P = (8774.268005 mol * 8.314 J/mol-K * 289K) / 31.05935577m3
P = 678775.745 Pa

I thought all the steps I took were right, but the answer I calculated is wrong. Any help with figuring out what I did wrong is appreciated!
 
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  • #2
You ignored the mass of the balloon material, M(balloon)=4.56 kg.

ehild
 
  • #3
Wouldn't that mass be ignored if the balloon material displaces a negligible amount of air?
 
  • #4
The volume of the balloon material is ignored, but not its mass. It is given as 4.56 kg. The buoyant force is equal to the total weight: the sum of the (mass of balloon material + the mass of helium) times g.

ehild
 
  • #5
Thank you for your help ehild, that "ASSUME: The balloon material displaces a negligible amount of air, and therefore creates no measurable buoyancy. " part really confused me. I accounted for the mass of the balloon material and got an answer to be 590585.5565Pa, I put in 5.90e5Pa and it is the right answer.
 
  • #6
Splendid!

ehild
 

1. What causes the pressure inside a helium balloon to increase?

The pressure inside a helium balloon increases as the temperature of the gas inside the balloon increases. This is because the molecules of gas move faster and collide more frequently with the walls of the balloon, resulting in an increase in pressure.

2. Why does a helium balloon deflate over time?

A helium balloon will eventually deflate over time due to the process of diffusion. The helium gas molecules are smaller and lighter than the molecules of air outside the balloon, so over time they will escape through the rubber walls of the balloon, causing it to deflate.

3. Is the pressure inside a helium balloon the same as the pressure outside?

No, the pressure inside a helium balloon is slightly higher than the pressure outside. This is because the gas inside the balloon is at a higher temperature, causing the molecules to move faster and collide more frequently with the balloon walls, resulting in a higher pressure.

4. How does the size of a helium balloon affect its internal pressure?

The size of a helium balloon does not directly affect its internal pressure. However, as a balloon expands, the gas molecules inside will have more room to move and will collide with the walls of the balloon less frequently, resulting in a decrease in pressure. Similarly, when a balloon is squeezed, the gas molecules will have less room to move and will collide with the walls more frequently, resulting in an increase in pressure.

5. Can the pressure inside a helium balloon be changed?

Yes, the pressure inside a helium balloon can be changed by altering the temperature or the volume of the gas inside. Increasing the temperature will result in an increase in pressure, while decreasing the volume of the gas will also increase the pressure. Similarly, decreasing the temperature or increasing the volume of the gas will result in a decrease in pressure.

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