Question

assuming that the air temperature is constant, what happens to a hot air balloon if it goes too far up in the atmosphere , where the outside air pressure is much less than the inside are pressure of the balloon?

a. it bursts

b. it loses air

c. it burns up

d. gravity begins to pull it back down to earth

Answer 1

A.. Because as the pressure builds up inside, it pushes outward and causes it to burst. If the pressure was less on the inside and more out side it would shrink.

Answer 2

This is because the air pressure is trying to find equilibrium between the air inside of the balloon and outside of the balloon. just try visualizing where the air wants to go sort of like osmosis.

Answer 3

None of these is correct. A hot-air balloon is open at the bottom, so the pressure is always identical inside and outside the balloon. The balloon rises because the balloon and enclosed hot air weigh less than the volume of (colder) outside air they displace -- Archimedes' Principle. This relies on the fact that hot air is less dense than cold air, which you can see by re-arranging the ideal gas law: d = P/RT, where d = n/V is the density. The higher T, the lower d. You can also see that the ratio of the densities of air in and out of the ballon is equal to the ratio of the temperatures only:

d_in/d_out = T_out/T_in.

All that matters for the ballon rising is that T_in > T_out, because that means d_in < d_out.

Of course, you can only raise the temperature of the air inside the balloon so far, because heat added will escape through the envelope of the balloon, and need to be replaced by the burners -- which have a limited ultimate capacity of supplying heat. (Only so many joules per hour, that is.)

Since heat leakage rate is generally proportional to the temperature differential dT = T_in - T_out (cf. Fourier's Law) you would expect, roughly, that the burners have an ultimate maximum temperature differential they can maintain. At that maximum dT heat leaks out of the balloon as fast as the burners can supply it.

As the balloon rises, the temperature outside falls, and if dT is fixed, than the ratio T_out/T_in = T_out/(T_out + dT) rises toward 1. That in turn implies that d_in/d_out also rises towards 1, and will eventually reach the highest value that allows the ballon to float. At that point the balloon will stop rising, with the burners going at max.

But it is very likely the person wh posed this question problem meant to specify something like a helium balloon, which is closed, and relies on the difference in densities of helium and nitrogen/oxygen at any pressure and temperature. When such a balloon rises, the volume of helium inside the balloon rises, putting pressure on the elastic envelope. Eventually the volume rises too much, and the balloon pops. This is in fact what happens to weather balloons. But not hot-air balloons.

Answer 4

Oops, now that I look at this question again, I think you can consider (b) as correct. As the balloon rises, the air inside the balloon expands. Since the balloon is open at the bottom, this just pushes some air out of the bottom. But I'm not sure your questioner wasn't thinking of a helium balloon, and therefore looking for (a).