Calculate Net Force on Hot-Air Balloon Using Ideal Gas Law | Homework Solution

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To calculate the net force on a hot-air balloon using the ideal gas law, the volume of the balloon is 1.5 m^3, with the air inside at 75°C and outside at 24°C. The pressure inside the balloon is assumed to be equal to the outside pressure of 1 atm. By applying the ideal gas law, the density of the air inside the balloon is found to be just over 1 kg/m^3, compared to 1.19 kg/m^3 for the outside air. The net force can then be determined by calculating the difference in buoyant force due to the density difference, confirming that the open bottom of the balloon allows for this assumption. This approach effectively utilizes Archimedes' principle to solve the problem.
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Homework Statement


A hot-air balloon has a volume of 1.5m^3 and is open at the bottom. If the air inside the balloon is at temperature of 75C, while the temperature of the air outside the balloon is 24C, at a pressure of about 1 atm, what is the net force on the balloon and its contents? (Neglect the Weight of the balloon itself).



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The Attempt at a Solution


I know that this question is about the ideal-gas law, but for some reason I just can't think how to do it.
 
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Force is pressure times area. This is all about the pressure of the gas inside the balloon.. That should be enough to get you going.
 
Consider Archimedes' principle. How does the density of the air inside the balloon compare to that outside?
 
So to find the density, I presume I first find the number of moles in the balloon using the ideal gas law. But to do so, can I assume that the pressure inside the balloon is the same as the pressure outside?

When I do this, I calculate 35 moles per m^3. This leads me to a density of just over 1kg/m^3. This is compared to the density of the air outside of 1.19kg/m^3

Oh.. ok I got it. So now I just figure out how much more gravity would pull down 1.5m^3 of the normal density air, and this is the force on the balloon.

Initially my problem was just that I wasn't sure if I could assume pressure inside to be equal to outside - because I thought the material of the balloon would add its own elasticity to the forces making the balloon smaller - but I guess because the bottom of the balloon is open this never becomes an issue.

Anyway, thanks for the help!
 
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