What changes about a helium-filled balloon on the surface of the moon? In terms of air pressure, I think that the pressure inside the balloon should be equal to the air pressure outside so that it does not burst. So how will a helium-filled balloon behave on the moon in comparison to earth? because here, the balloon does not burst unless it has been hit with enough force to burst it. This means that the air pressure inside must be equal to air pressure outside? Will this change because there is no pressure on the moon?
 A: You don't really need to have an equal pressure inside and outside of the balloon as long as the balloon can withstand some pressure gradient.
Without any data on balloon resistance to pressure it's hard to predict exactly what will happen, but I think it's a good guess to say that if you inflate an helium balloon on the Moon you will need a really small amount of gas in comparison to Earth to create the correct pressure gradient.
Moreover, since there is no atmosphere to give buoyancy to the helium balloon, it will not float, but rather drop to the floor as any other object.
A: The important thing to understand is that there is another force in addition to the forces exerted by the gases inside and outside the balloon: the force created in stretching the balloon material. The full equilibrium state must account for all  three of these forces.
On the moon there is no gas outside the balloon and so when it is inflated there are only two forces that come to equilibrium: the gas pressure inside the balloon, and the stretched balloon.
Relative to a near vacuum state of the moon's atmosphere, and the fact the balloon has mass will cause it to fall to the surface, not float away.
