In 2009 Pixar Animation studios released the movie “Up”.

On it we see a house floating in the air, pulled up by many helium balloons.

**What volume of helium is needed to to make a house
airborne?**

(We assume a single balloon, to make it easy to answer)

- Estimate a range for the weight of the house.
- Calculate the weight of a ballon full of air and another full of Helium.

You can assume that the balloon has a volume of one cubic meter
(1m^{3}). This will give you how much weight can be lifted by
such balloon.

You can assume that air and helium are ideal gases.

The ideal gas law says that \(PV=nRT\) where

- \(P\) is the absolute pressure of the gas,
- \(V\) is the volume of the gas,
- \(n\) is the amount of substance of gas (number of moles),
- \(R\) is the gas constant, equal to \[8.31446261815324 \frac{m^3⋅Pa}{K⋅ mol}\]

The house is pulled down by its weight

- It is pulled up by the balloon

The ballon is pushed down by its weight

- It is pulled up by the air

The push-up force is given by Archimedes principle

*“The force up is equal to the weight of air displaced by the balloon”*

A quick Google Search suggests that such house weight between 120 and 180 tonnes

(a metric ton is 1000 Kg)

It is important that you check the source of these estimations

Use only *well recognized* sources

Include explicit references to all used sources

We can assume that the balloon has a volume of one cubic meter
(1m^{3})

To know the weight of 1m^{3} of helium, we need the number of
mols

Then we will multiply it by the atomic weight of helium

The ideal gas law says \(PV=nRT\), therefore \[n=\frac{PV}{RT}\]

It is reasonable to assume that \[\begin{aligned} P &= 1 \text{ atm} = 101325 \text{ Pa}\\ T &= 27 \text{°C} = 300\text{ K} \end{aligned}\] If you like, you can use intervals instead

According to WikiPedia, the International Union of Pure and Applied Chemistry (IUPAC) says that the atomic weight of helium is \[A_r°(He) = 4.002602±0.000002\text{ dalton}\]

This is also the weight in grams of a mol of helium

IUPAC also publishes abridged values, rounded to five significant figures.

\[A_{r, \text{abridged}}°(He) = 4.0026\text{ dalton}\]

The uncertainty is small, we can take it as a precise value

The number of moles of air is the same as for helium

Air is a mix of gases. According to WikiPedia:

“By number of molecules, dry air contains

- 78.08% nitrogen,
- 20.95% oxygen,
- 0.93% argon,
- 0.04% carbon dioxide,
- and small amounts of other gases”

Element | Range |
---|---|

N | [14.00643, 14.00728] |

O | [15.99903, 15.99977] |

Ar | [39.792, 39.963] |

C | [12.0096, 12.0116] |

Keep in mind that in air the molecules are N_{2},
O_{2}, Ar, and CO_{2}

Since the number of moles is the same, the only factor determining the buoyancy is \[\text{Molecular weight of Helium} - \text{Molecular weight of Air}\]

From there the rest is just calculation