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Buoyant Force on a Helium Balloon 
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#1
Jun1707, 01:23 AM

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1. The problem statement, all variables and given/known data
A helium balloon has volume V_{0} and temperature T_{0} at sea level where the pressure is P_{0} and the air density is [itex]\rho_0[/itex]. The balloon is allowed to float up in the air to altitude y where the temperature is T. (a) Show that the volume occupied by the balloon is then V = V_{0}(T/T_{0})e^{cy} where [itex]c = \rho_0g/P_0[/itex]. (b) Show that the buoyant force does not depend on altitude y. Assume that the skin of the balloon maintains the helium pressure at a constant factore of 1.05 times greater than the outside pressure. [Hint: Assume that the pressure change with altitude is P = P_{0}e^{cy}]. 2. Relevant equations The ideal gas law and the buoyant force equation. 3. The attempt at a solution (a) is pretty simple since the hint gives it away. What concerns me is (b). Since the buoyant force on the balloon is equal to the weight of the volume of air displaced by the balloon, and since the volume depends on altitude, then it seems logical that the buoyant force depends on altitude. What gives? 


#2
Jun1707, 07:11 AM

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But the density of the air decreases with altitude. What matters is how the product density*volumethe weight of the displaced airvaries with altitude.



#3
Jun1707, 12:51 PM

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The mass of air inside the balloon remains the same regardless of its change in volume. Thus, the buoyant force is constant right? (Actually, it wouldn't since the force of gravity decreases with altitude.)



#4
Jun1707, 02:22 PM

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Buoyant Force on a Helium Balloon



#5
Jun1707, 03:49 PM

P: 1,367

Ah, OK. I was getting confused. Let [itex]\rho = m/V[/itex] be the density of air at altitude y; m is the mass of air displaced by the balloon whose volume V is as given in the problem statement. The buoyant force is [itex]\rho V g = mg[/itex]. Hmm...so now how do I know that m is always the same?



#6
Jun1707, 06:42 PM

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How does the density of the air depend on pressure and temperature?



#7
Jun1807, 09:37 AM

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Applying the ideal gas law to air, then
[tex]PV = P \, \frac{m}{\rho} = nRT [/tex] Rearranging yields [tex]\rho = P \, \frac{m}{nRT}[/tex] The fraction m/n is surely constant as it equals air's molar mass. Now what about P/T? If they both decrease at the same rate, then it will definitely be constant. I don't have an expression for T though. 


#8
Jun1807, 09:52 AM

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Keep going. Now find the buoyant force, which is:
[tex]\rho g V[/tex] 


#9
Jun1807, 12:45 PM

P: 1,367

If I substitute and simplify, I get that the buoyant force is mg which is what I had derived in post #5. Again, I don't know if m is constant though.



#10
Jun1807, 02:07 PM

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What matters is how the density of the air varies. From the ideal gas law, the density of air will be proportional to:
[tex]\frac{nR}{V} = \frac{P}{T}[/tex] Plug that into the expression for buoyant force. 


#11
Jun1807, 05:21 PM

P: 1,367

I see. You're suggesting that the buoyant force is proportional to nRg right? However, like m, doesn't n depend on altitude as well?



#12
Jun1907, 08:04 AM

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All we are concerned with at this point in the solution (as in post #10) is what the air density depends on. I'm not concerned with any particular mass of air or number of moles.
It will turn out that the mass of air displaced by the balloon is independent of altitude, but no need to assume that at this point. 


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