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## Homework Statement

A hot-air balloon stays aﬂoat because hot air at atmospheric pressure is less dense than cooler air at the same pressure.If the volume of the balloon is 500 m

^{3}and the surrounding air is at 60◦F. What is the maximum load (including the weight of balloon, but excluding the weight of the hot air) the balloon can lift if the hot air is at 400◦F? The air density at 60◦F is 1.23 kg/m

^{3}.

## Homework Equations

F = ma

F

_{buoyant}= ρ

_{DF}× g × V

_{DF}

pV = nRT

## The Attempt at a Solution

[/B]

1) My first thought is to apply Newton's Second Law:

F = ma ⇒ F

_{B}- F_{g}= 0_{out}gV

_{B}= (m

_{balloon}+ m

_{load}+ ρ

_{in}V

_{HotAir})g

[Where ρ

3) From here I can find ρ_{out}= air density outside of balloon and ρ_{in}V_{HotAir}= mass of the air in the balloon]_{in}using the ideal gas law, V

_{HotAir}= V

_{B}, and I can isolate m

_{balloon}+ m

_{load}but what is throwing me for a loop is the fact that it does not state this is an ideal gas and the say exclude the weight of the hot air. Would I just exclude ρ

_{in}V

_{HotAir}from step two then?