# Helium balloons losing buoyancy in cold environment

I recently conducted an experiment in the mountains (approx 1700m above sea level) using a series of large tethered helium balloons to measure windspeed at different heights.

After only a few hours, the lower balloons started to lose buoyancy and sink, followed by the higher balloons later in the night. There was mist, which formed condensation on the balloon skins that later froze (see picture below).

The balloons were filled at a temperature of roughly 14°C to a free lift of approximately 0.25N, and the temperature dropped to a minimum of 2°C at night. Obviously the weight of the combined components (helium, balloon skin, tether and frost) overcame the lift force, causing the balloon to sink. I'm wondering if that is entirely due to the weight of the frost/condensation (which I had no way to measure at the time), or whether the temperature drop caused a decrease buoyancy.

Regarding the latter, my thinking went as follows: According to Archimedes Principle, the lift force is equal to the weight of the displaced air.

$$F_b = g V \rho_{air}$$ (The mass of the helium should be constant, so I am ignoring that for now)

The density of air and the balloon volume are both dependant on temperature, in opposite ways which should cancel each other out. Following the ideal gas law:

$$\rho_{air}= \frac{n M}{V} = \frac{M P_{atm}}{RT}$$ where M is the molar mass of air, and $$V_{balloon}= \frac{n_{He} RT}{P_{balloon}}$$ Where $$n_{He}$$ is the moles of Helium. Combining these with the first equation gives the following, with RT cancelling out: $$F_b = g \frac{n_{He}}{P_{balloon}} M P_{atm}$$ I assume this means that the lift is not directly dependent on temperature, but rather indirectly through the atmospheric pressure term. However, this should increase with colder temperature, so it doesn't explain the decrease in lift. The balloon pressure should stay constant as long as we assume that the elastic force of the skin increases linearly with radius according to Hooke's Law (meaning that pressure changes negligibly with volume).

This all leads me to think that it is just the weight of the frost that caused the balloons to sink. Please, someone check my logic and throw in any ideas you might have. Many thanks.

• Why do you assume that these gases are ideal? They are further from ideal, the colder they get. Why do you assume that they follow exactly the same path of volume change with temperature? There is no reason to that. Anyway, there is lots of frost of the balloon you showed. So measure how much it weighs and there you have it's gravity. If it is more than your 0.25 newtons, you are at home , Commented May 14 at 5:35
• Another issue, as any party store manager will warn you: helium balloons leak pretty fast. Commented May 14 at 13:42
• @RadekD it is unfortunately impossible to go back and measure the weight of the frost. The closes would be replicating the experiment under similar conditions, but I would rather eliminate other possible causes first. At least the helium leakage is something that can be tested in a controlled environment fairly easily. Commented May 22 at 23:21
• All right. To have 0.25N of water, you need roughly 2 tablespoons. There seems to be enough of this ice there. Commented May 24 at 16:44

Your math seems valid to me. Assuming ideal gas equations I would just write for the force on the balloon neglecting the weight of the non gas components: $$F_up = g \cdot V \cdot \rho_{air} - g \cdot V \cdot \rho_{He} = g V (\rho_{air} - \rho_{He}) = g V (n_{air} \cdot m_{air} - n_{He} \cdot m_{He})$$ with the particle density $$n$$ and the average particle mass $$m$$.

If we assumed that your balloon is not exerting significant pressure on the gas inside and is in thermal equilibrium with its outside, since we assumed ideal gas equations for both gases, we can write $$n_{air} = n_{He} = n$$ and $$n V = N$$ the particle number of an ideal gas in the volume of your ballon at the given temperature and pressure. N should be independent of the change of your temperature and pressure. So I get $$F_up = g N (m_{air} - m_{He})$$, which is consistent with your result in the case that $$P_{ballon} \approx P_{atm}$$. In the more general case, that your balloon is exerting significant pressure on your gas, which would seem like poor engineering to me, I believe your formula to be correct.

So yeah, I think the ice-explanation seems reasonable.

As pointed out in the comments, it might be interesting to do a calculation with the Van der Waals equation, to check how much of a difference it actually causes. While I am sufficiently sure, that the effect will be very small compared to the one of the ice, it would still do a significant qualitative difference to the ideal gas result in the aspect, that the ideal result does not predict any change with temperature what so ever.

• No, Radek is correct, it should be cold enough that deviation from ideal-ness should start to become appreciable. This is compounded by the fact that, as temperature decreases, the ability of the balloon to pull on the gas is decreased, leading to an opposite contribution than expected. This means that whatever is causing it to drop, needs to be a greater effect. I think ice explanation, and gas leakage, are reasonable. Commented May 14 at 7:52
• @naturallyInconsistent I could be wrong but I recall reading that for gases such as helium and air it is reasonable to assume ideal gas behavior in the vicinity of STP (0 C, 1 atm) Commented May 14 at 9:57
• @BobD Oh, it is definitely reasonable to assume ideal gas behaviour. The issue is that the ideal gas prediction is zero, and so any correction is fractionally infinitely larger. In cases like these, it is then kind of necessary to take these corrections into account if we want to describe those behaviours correctly. Commented May 14 at 10:31
• @naturallyInconsistent But I'm thinking the deviation from ideal gas behavior in this case is less of an influence than the increase in the density of the balloon due to the ice formation. But I haven't done any calculations to back it up. Commented May 14 at 10:42
• Regarding helium pressure under the condition of 2 degrees Celsius and van der Waals equation, this is the result: wolframalpha.com/…. So it is different a lot compared to perfect gas equation wolframalpha.com/… Commented May 24 at 19:06