Boyle's Law and hot air balloon

Question

A bit dumb question because it is really difficult to imagine it. :-

Pressure is force per area. Talking about gases, the pressure is said to be the force molecules exert on walls of let's say a balloon.

Usually in examples of Boyle's law, our teachers mention hot air balloon, that the size of balloon increases as pressure decreases.

But at as height increases pressure decreases because there are less molecules above us.

How will the pressure of molecules outside the balloon effect the pressure of molecule inside the balloon.

Answer 1

Like you said, pressure arises from the force exerted by the molecules on stuff. In case of the balloon whose pressure inside is higher than the pressure outside, the number of collisions per second of the molecules outside the balloon is lower than that of the ones inside. Thus the balloon feels a net force that is outward. So the balloon expands until the net force becomes zero. Thus happens when the rate of collision outside equals that of inside. This is nothing but the condition where the pressure is equalised.

Answer 2

For the system to be in equilibrium, the pressure outside must be equal to the pressure inside, otherwise the inside will contract (if the pressure outside is higher) or expand (if lower). There is a minor contribution given by the elasticity of the balloon capable of handling some extra pressure but we can neglect that here.

So as the atmosphere gets thinner, the balloon expands a bit.

Answer 3

There are three forces at play here caused by the internal pressure, external pressure and the elastic balloon. The internal pressure balances out the other two to maintain equilibrium. For ease of calculations, I am assuming that no air is able to diffuse through the rubber and escape. Thereby the internal pressure stays constant throughout.

Boyle's law is applicable for both internal and external pressure. Hence when external pressure decreases, the volume increases. The rubber then applies greater force to try to compensate for this loss of external pressure.