# Why do helium balloons rise and fall?

I understand why a regular party balloon filled with helium falls over time due to leakage of the helium. However I've also noticed that recently filled helium balloons put outside rise and fall. At one point in the afternoon they were dropping but later in the evening it was fully upright again.

Why is this? Is it because of a change in atmospheric pressure? The weather did go from a little rainy and overcast to dry and slightly clearer skies later. It must be something to do with P = VxT.

• This is related: physics.stackexchange.com/q/2415 and deals with buoyancy and density differences between the helium inside the balloon and the atmosphere outside of it.
– user81619
Nov 7 '15 at 19:18
• I read that question before posting but I felt it didn't cover why they rise and fall or at least it wasn't clear to me from reading those answers. Nov 7 '15 at 19:22

The upthrust on the balloon is equal to the weight of air displaced, so we get:

$$F = V_b \rho g \tag{1}$$

where $V_b$ is the volume of the balloon and $\rho$ is the density of the air. Assuming air is approximately an ideal gas it obeys the equation of state:

$$PV = nRT$$

so the molar density is:

$$\rho_M = \frac{n}{V} = \frac{P}{RT}$$

where $n$ is the number of moles of air. The density in kg/m$^3$ is given by multiplying the molar density by the (average) molar mass of the air $M_{\text{air}}$, and substituting this in equation (1) we get:

$$F = V_b M_{\text{air}} g \frac{P}{RT} \tag{2}$$

Now let's consider what happens to the volume of the balloon. We'll take the two extreme cases where the rubber skin is infinitely rigid and where it's infinitely compliant.

First consider the case where the rubber skin is infinitely compliant i.e. it doesn't exert any force on the helium inside it. In that case the volume of the helium is (approximately) given by the ideal gas equation:

$$V_b = \frac{n_{\text{He}}RT}{P}$$

where $n_{\text{He}}$ is the number of moles of helium. Substituting this into equation (2) we get:

$$F = n_{\text{He}} M_{\text{air}} g$$

which is constant. So in this case we find that the bouyancy is unaffected as the pressure and temperature change.

Now consider what happens if the rubber skin is infinitely rigid, in which case the volume $V_b$ is constant. We end up with:

$$F \propto \frac{P}{T}$$

In this case the bouyancy is affected by the pressure and temperature. Assuming the pressure is approximately constant the bouyancy is inversely proportional to temperature so the balloon will rise when it gets cold and fall when it gets hot, which matches your observation.

I've taken the two extreme cases because I don't know the equation for the force produced by the rubber skin of the balloon. However it is presumably somewhere in between the two extremes I've discussed, so we expect the behaviour of the balloon to fall between those two extremes. That means we expect the bouyancy will increase as the temperature decreases and vice versa.

• Excellent, thank you very much for an extremely thorough answer. I'm new to the physics stack exchange so I can only accept the answer and not vote it up. Nov 8 '15 at 11:07

Why do helium balloons rise and fall? At one point in the afternoon they were dropping but later in the evening it was fully upright again. ... The weather did go from a little rainy and overcast to dry and slightly clearer skies later.

Likely many droplets of water adhered to the helium balloons, when the weather was rainy, and made them heavier. When the water evaporated the balloons became lighter and rose again.

• I hadn't thought of something as obvious as that and neither did the two other people who observed it. It is quite likely even a small amount of moisture was on the balloon. But what would the reason be assuming it was a perfectly dry day with no moisture on the balloon? Could a change in atmospheric pressure affect the buoyancy of the balloon? Nov 7 '15 at 19:34
• A change in the air density (cold weather - hot weather) affects the buoyancy of a helium balloon. Nov 7 '15 at 19:55
• Exactly. That's what I thought. Can you expand that into an answer please? What you're saying is along the lines of what I'm looking for. Nov 7 '15 at 19:57