Why gas balloons keep rising with the increase of altitude?

Something was bothering me. The whole concept of the law of buoyancy is based of the fact that in order that the balloon rises it is necessary that the gas inside the balloon (in our case Helium) is lighter than the air. And the difference between the weight of the air and the weight of the gas is the weight that can be lifted by the balloon. The bigger the balloon, the bigger this difference, and the bigger weight can be lifted. The difference between the total weight that can be lifted and the weight of what is lifted (payload, balloon skin, strings) will define the rate of ascent. But we know that as the altitude rises, the weight of the air decreases, and even if the weight of Helium also decreases, it is intuitive thatthe total weight that could be lifted should decrease. If this was true it would mean that somewhere the balloon would cease rising and would just float. But is this so?

Well there's nothing like going back to the maths to clarify this issue and better understand how the variables relate to each other. To do this let's consider the weight of 1 litre of air and helium at 0km and 20km of altitude, or put in another way, the air density and helium density at 0km and 20km.

We have seen before that at Standard conditions for temperature and pressure (STP) - we can consider those as being the conditions at 0km altitude - we have:

Gas / Mixture	Weight of 1 litre (g) at 0km  <=> density at 0km	Weight of 1 litre (g) at 20km <=> density at 20km
Helium	0.176	?
Air	1,275	?

Now let's determine the density of Helium and Air at 20km of altitude. To do that we shall used our old friend, the ideal gas law, but first let's see what is the pressure and temperature at 20km, that like we have seen before, and according to the International Standard Atmosphere (ISA), is:

• Absolute pressure at 20 km: 5474.9 Pa= 0.054 atm
• Temperature at 20km: -56.5 ºC= 216.65 K

So equipped with these values let's do the maths.

Step 1. Determine the volume (in litres) occupied by 1 mole of the Helium or Air (at 20 km):

At 20 km we have:

T (Temperature)= 216.65 K
P (Absolute pressure)= 0.054 atm

Additionally we have the ideal gas constant R= 0.08205746(14) L atm K^-1 mol^-1 which we can approximate to R= 0.082 L atm K^-1 mol^-1

And n= 1 since we want to determine the volume of 1 mole of helium or air.

Now, by using the ideal gas law we just have to replace the variables by their correspondent values:

PV = nRT
V= (nRT) / P
V= (1 x 0.082 x 216.65) / 0.054
V= 17.765 / 0.054 = 328.98 litres

Notice that above in the ideal gas law we have user as units for n, T and P, the same units in which the R constant is expressed: mol, K, and atm respectively.

So we have just determined that both 1 mole of Helium as well as 1 mole of Air occupy 328.98 litres at 20 km of altitude.

Step 2. With the result in 1. determine to how much moles correspond 1 litre of Helium or Air (at 20 km):

Now we have:

1 mol of Helium/Air ____(corresponds)_____ 328.98 litres
X mol of Helium/Air____(corresponds)_____ 1 litres

And by applying the rule of three, which is a particular form of cross-multiplication:

X = (1 * 1) / 328.98 = 0.003 mol

So we have just determined that 1 litre of Helium or 1 litre of air contain 0.003 moles.

Step 3. Using the molar mass of Helium and of the Air ,which is expressed in g/mol, and the result in 2, get the weight in grams (at 20 km):

And now we have reached to the last step. As seen in a previous post, the molar mass (molecular weight) of Helium is equal to its atomic weight (since Helium has only 1 atom). In this way, and knowing that the atomic weight of Helium A_r(Helium)= 4.003 g/mol, it is easy to calculate the molar mass of Helium, M(Helium):

M(Helium)= A_r(Helium)= 4.003 g/mol

In this way, we can now use once again the rule of three:

4.003 g         ____(corresponds)_____ 1 mol
X g of Helium____(corresponds)_____ 0.003 mol

X = (0.003 * 4.003) / 1 = 0.012g

Which means that the weight of 1 litre of Helium at 20 km of altitude is 0.012g, which is equivalent to:
Helium density (ρ) = 0.012g/l = 0.012 Kg/m³

Regarding the Air, we know by a previous post, that the molar mass (molecular weight) of Air is equal to:

M(Air)= 28.972 g/mol

In this way, we can now use once again the rule of three:

28.972 g         ____(corresponds)_____ 1 mol
X g of Air ____(corresponds)_____ 0.003 mol

X = (0.003 * 28.972) / 1 = 0.087g

Which means that the weight of 1 litre of Air at 20 km of altitude is 0.087g, which is equivalent to:
Air density (ρ) = 0.087g/l = 0.087 Kg/m³

And now we can complete the previous table containing the Helium and Air density at 0km and 20km as follows:

Gas / Mixture	Weight of 1 litre (g) at 0km  <=> density at 0km	Weight of 1 litre (g) at 20km <=> density at 20km
Helium	0.176	0,012
Air	1,275	0.087

Now, and after revising the post on how much weight can a Helium balloon lift, we know that:

• At 0 Km of altitude, 1 litre of Air weights 1,275g, whereas 1 litre of Helium weights 0,176g. This means that a 1 litre Helium balloon (plus its payload) shall rise if the weight of the balloon (without gas) plus the payload is less than 1,275-0,176g= 1,099g (if the weight is the same the balloon will float and not rise);
• At 20 Km of altitude, 1 litre of Air weights 0,087g, whereas 1 litre of Helium weights 0,012g. This means that a 1 litre Helium balloon (plus its payload) shall rise if the weight of the balloon (without gas) plus the payload is less than 0,087-0,012g= 0,075g (if the weight is the same the balloon will float and not rise).

This result is surprising, or not...

At 0 Km if you fill-in a balloon with, for instance, 500 litres of Helium,  the balloon (plus its payload) shall rise if the weight of the balloon (without gas) plus the payload that it may carry is less than 1,099 * 500 = 549,5g. This means that if we had a balloon that without gas weights 300g, we could carry a payload of a bit less than 549,5-300= 249,5g, for instance 240g.

But applying the same maths, the same balloon at 20km of altitude,  would only rise if the weight of the balloon (without gas) plus the payload that it carries is less than 0,075 * 500 = 37,5g. This means that with the same balloon weighting 300g, the balloon would already been descending much time before, meaning that would never reach the 20km of altitude. But we know that at this altitude the ballon is still rising. So what is failing in all this reasoning?

Well the problem is that we were assuming that the balloon filled with a volume of 500 litres of Helium at 0 km, would still have 500 litres of volume at 20 km of altitude. And that is wrong as we already have seen in our previous Post. So, let's determine the volume of the balloon at 20Km of altitude. To this let's use the general formula bellow:

$$V_{2}=V_{1}\frac{P_{1}T_{2}}{P_{2}T_{1}}$$
So the above formula is a general formula that allow us to determine the volume of the balloon in state 2 (S₂) given the balloon conditions in state 1 (S₁) and given the temperature and pressure of the balloon in state 2 (S₂).
Regarding the state 1, let's consider the that balloon at 0 km of altitude is filled in with 500 litres of Helium and that the pressure and temperature follow the STP values that we have indicated in this previous Post.

State 1 (S₁)

• Altitude= 0 meters (sea-level)
• Pressure= P1= 100000 Pa (at STP)
• Temperature= T1= 273.15 k (at STP)
• Volume= V1=  500 L=  0.5 m³

Then for state 2 (S₂) of the balloon, we would like to know the volume of the same balloon when at an altitude of 20000 meters. So  let's consider that for the absolute pressure at 20 km we have  5474.9 Pa  and the temperature is  -56.5 ºC, as indicated by the International Standard Atmosphere (ISA):

State 2 (S₂)

• Altitude=  20000 meters (20km)
• Pressure= P2= 5474.9 Pa
• Temperature= T2= -56.5 ºC= 216.65 k
• Volume= V2=  ?

Now it is just a question of using the general formula presented above as follows:

$$V_{2}=V_{1}\frac{P_{1}T_{2}}{P_{2}T_{1}} <=> V_{2}=500 \times \frac{100000 \times 216.65}{5474.9 \times 273.15} <=> V_{2}=500 \times \frac{21665000}{1495468.935} <=>$$

$$<=> V_{2}=500 \times 14.49 <=> V_{2}=7245L= 7.25m^{3}$$.

So now that we know that the balloon at 20km occupies a volume of 7245 litres (instead of the 500 litres at 0 Km of altitude), we also know that the balloon at 20km of altitude would rise if the weight of the balloon (without gas) plus the payload that it carries is less than 0,075 * 7245 = 543.38g. This is a value very similar to the 549.5g that the balloon is able to carry at 0 Km of altitude, corresponding to a drop of approximately 1 % in the weight that can be lifted.

This is just fantastic. As the balloon rises from 0km to 20km of altitude, the difference between the weight of the Helium and the Air around it drops substantially (14.65 times) what would makes us imagine that the lifting capability of the balloon would also drop 14.65 times. But in fact that is not the case because the volume occupied by the Helium rises also substantially (14.49 times) as the balloon rises up to 20km of altitude. This seems like magic, and it is... it is the magic of Nature and of the Physical world.

Just by curiosity let's see what would be the situation for the balloon in a third state at 30 km of altitude.

Step 1. Determine the volume (in litres) occupied by 1 mole of the Helium or Air (at 30 km):

At 30 km we have:

T (Temperature)=  226.5 K
P (Absolute pressure)= 1197 Pa= 0.012 atm

Additionally we have the ideal gas constant R= 0.08205746(14) L atm K^-1 mol^-1 which we can approximate to R= 0.082 L atm K^-1 mol^-1

And n= 1 since we want to determine the volume of 1 mole of helium or air.

Now, by using the ideal gas law we just have to replace the variables by their correspondent values:

PV = nRT
V= (nRT) / P
V= (1 x 0.082 x 226.5) / 0.012
V= 18.573 / 0.012 = 1547.75 litres

Notice that above in the ideal gas law we have user as units for n, T and P, the same units in which the R constant is expressed: mol, K, and atm respectively.

So we have just determined that both 1 mole of Helium as well as 1 mole of Air occupy 1547.75 litres at 30 km of altitude.

Step 2. With the result in 1. determine to how much moles correspond 1 litre of Helium or Air (at 20 km):

Now we have:

1 mol of Helium/Air ____(corresponds)_____ 1547.75 litres
X mol of Helium/Air____(corresponds)_____ 1 litres

And by applying the rule of three, which is a particular form of cross-multiplication:

X = (1 * 1) / 1547.75 = 0.00065 mol

So we have just determined that 1 litre of Helium or 1 litre of air contain 0.00065 moles.

Step 3. Using the molar mass of Helium and of the Air ,which is expressed in g/mol, and the result in 2, get the weight in grams (at 30 km):

And now we have reached to the last step. As seen in a previous post, the molar mass (molecular weight) of Helium is equal to its atomic weight (since Helium has only 1 atom). In this way, and knowing that the atomic weight of Helium A_r(Helium)= 4.003 g/mol, it is easy to calculate the molar mass of Helium, M(Helium):

M(Helium)= A_r(Helium)= 4.003 g/mol

In this way, we can now use once again the rule of three:

4.003 g         ____(corresponds)_____ 1 mol
X g of Helium____(corresponds)_____ 0.00065 mol

X = (0.00065 * 4.003) / 1 = 0.0026g

Which means that the weight of 1 litre of Helium at 30 km of altitude is 0.0026g, which is equivalent to:
Helium density (ρ) = 0.0026g/l = 0.0026 Kg/m³

Regarding the Air, we know by a previous post, that the molar mass (molecular weight) of Air is equal to:

M(Air)= 28.972 g/mol

In this way, we can now use once again the rule of three:

28.972 g         ____(corresponds)_____ 1 mol
X g of Air ____(corresponds)_____ 0.00065 mol

X = (0.00065 * 28.972) / 1 = 0.019g

Which means that the weight of 1 litre of Air at 30 km of altitude is 0.019g, which is equivalent to:
Air density (ρ) = 0.019g/l = 0.019 Kg/m³

So, at 30 Km of altitude, 1 litre of Air weights 0,019g, whereas 1 litre of Helium weights 0,0026g. This means that a 1 litre Helium balloon (plus its payload) shall rise if the weight of the balloon (without gas) plus the payload is less than 0,019-0,0026g= 0,0164g (if the weight is the same the balloon will float and not rise).

Additionally, like we have seen before at 30 Km:


State 3 (S₃)

• Altitude=  30000 meters (30km)
• Pressure= P3= 1197 Pa
• Temperature= T3= 226.5 k
• Volume= V3=  ?

Using again the above formula:
$$V_{3}=V_{1}\frac{P_{1}T_{3}}{P_{3}T_{1}} <=> V_{3}=500 \times \frac{100000 \times 226.5}{1197 \times 273.15} <=> V_{3}=500 \times \frac{22650000}{326960.55} <=>$$

$$<=> V_{3}=500 \times 69.27 <=> V_{3}=34635L= 34.64m^{3}$$.

So now that we know that the balloon at 30km occupies a volume of 34635 litres (instead of the 500 litres at 0 Km of altitude), we also know that the balloon at 30km of altitude would rise if the weight of the balloon (without gas) plus the payload that it carries is less than 0,0164 * 34635 = 568.014g. This is a value even slightly bigger similar to the 549.5g that the balloon is able to carry at 0 Km of altitude, corresponding to an increase of 3% in the weight that can be lifted.

This also shows that the factor that will determine the return of the balloon to the Earth surface is not its capability to lift the weight that it carries, but the extensibility of the balloon material, meaning its capacity to be able to increase in volume without bursting - and in this post we have seen that the balloon occupies a volume of 500 litres at the Earth's surface and then 7245 at 20km and 34635 at 30km.