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Buoyancy force of a hot-air balloon

by smeiste
Tags: balloon, buoyancy, force, hotair
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smeiste
#1
Jan25-12, 06:36 PM
P: 36
1. The problem statement, all variables and given/known data

The envelope and basket of a hot-air balloon have combined mass of 331 kg. The spherical envelope of the balloon has a radius of 10 m when fully inflated. What is the maximum mass of the passengers the balloon can carry when the temperature of the gas is 108 o C? Use M(air) = 29 g/mol. Assume that the surrounding air is at 25 o C and is treated as an ideal gas.

2. Relevant equations

Fbuoyancy = ρVg (density times volume times acceleration due to gravity)

3. The attempt at a solution
Not entirely sure if a buoyant force is acting here, but it makes sense as that's what we're learning.
Attempt:
Volume at 25 degrees = 4/3πr^3 = 4188 m^3

(101330Pa)(4188m^3) = n (8.314)(298K)
n = 171317 moles

(171317mol - n hotair) = (171317mol)(298K)/(381K)
n hotair = 37321 mol

171317 - 37321 = 133995 mol

(133995mol/m^3)(0.029kg/mol) = 3885 kg/m^3

Now assuming that is the density, I was trying to plug that into the buoyancy formula and calculate the force. However my force ends up way too big. The correct answer in the end should be 751 kg. Any help is appreciated!
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ehild
#2
Jan26-12, 02:25 AM
HW Helper
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P: 10,375
Hi smeiste,
I can not follow your work...

Quote Quote by smeiste View Post
Volume at 25 degrees = 4/3πr^3 = 4188 m^3

(101330Pa)(4188m^3) = n (8.314)(298K)
n = 171317 moles

(171317mol - n hotair) = (171317mol)(298K)/(381K)
n hotair = 37321 mol

171317 - 37321 = 133995 mol

(133995mol/m^3)(0.029kg/mol) = 3885 kg/m^3
Could you please explain what you calculated?What are n and n hotair?

Quote Quote by smeiste View Post

133995mol/m^3)(0.029kg/mol) = 3885 kg/m^3
Now assuming that is the density, I was trying to plug that into the buoyancy formula and calculate the force. However my force ends up way too big. The correct answer in the end should be 751 kg. Any help is appreciated!
What density did you use when calculating the buoyant force?

ehild
smeiste
#3
Jan26-12, 08:09 AM
P: 36
umm n is the number of moles. n hot air is the number of moles in the balloon at 108 degrees Celsius. I'm not entirely sure what exactly I did haha. and I used the density I calculated 3885 kg/m^3. I probably did it all wrong

ehild
#4
Jan26-12, 08:19 AM
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P: 10,375
Buoyancy force of a hot-air balloon

It is not sure you did it wrong, try to explain.

Quote Quote by smeiste View Post
umm n is the number of moles.
Number of moles of what?


Quote Quote by smeiste View Post
and I used the density I calculated 3885 kg/m^3. I probably did it all wrong
Yes but the density of what? OK, air, but inside the balloon or outside? At what temperature?


ehild
smeiste
#5
Jan26-12, 10:20 AM
P: 36
i honestly have no idea what moles I calculated.. There was a similar question only it wanted the temperature inside the balloon and I tried to do a similar thing. I don't really understand what density I calculated either.
ehild
#6
Jan26-12, 11:34 AM
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P: 10,375
The balloon rises or at least it does not fall down so its weight must be balanced by the buoyant force: The weight of the whole balloon with the gas inside and with the load must be balanced by the weight of the air displaced.
The volume of the air inside the balloon is the same 4000pi/3 m3 as the volume of the air displaced, but the temperature of the air in the balloon is 381 K and that of the displaced air is 298 K. The same amount of air consist of less molecules inside the balloon than outside. You calculated the number of the displaced moles - that was n, and the number of the moles inside the balloon that was n(hotair). Multiplying n(hotair) with 0.029 kg/mol gives the mass of the gas inside the balloon; multiplying n by the same 0.029 kg/mol gives the mass of air displaced.
Write out the equation for the balance of the weights.

ehild
smeiste
#7
Jan26-12, 12:27 PM
P: 36
ohh i see. I just went too far with the problem haha. Thank you!


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