Question

A hot air balloon has a volume of $2800m^3$ at $99^{\circ}C$. What is the volume if the air cools to $80^{\circ}C$?

Answer 1

We know that a hot air balloon uses the pressure built by the air to move or rise upwards above the ground. We know that gas molecules have the ability to modify or change their volume and/or shape in the presence of heat. Using the above knowledge, we can solve the question as discussed below.

Formula used:
Charles’ law : $\dfrac{V}{T}=k$, where $V$ is the volume of the given gas at temperature $T$ often taken in kelvin and $k$ is the non-zero constant.

Complete step by step solution:
Let us assume that the air in the hot air balloon is ideal. Let us also assume that the number of molecules of the air particles does not vary with the increase in temperature and the pressure in the balloon is constant.
Then from Charles’s law which gives the relationship between the volume $V$ and temperature $T$ at constant pressure as $V\propto T$or $\dfrac{V}{T}=k$
Given that the volume of $2800m^3$ at $99^{\circ}C$, let the volume at $80^{\circ}C$ will be $v$, then from Charles’ law $\dfrac{2800}{99}=K$ and $\dfrac{v}{80}=K$
Since the air is the same , we can say that the non-zero constant for both the cases are also the same, thus equating the two we have
$\implies \dfrac{2800}{99}=\dfrac{v}{80}$
$\implies v=\dfrac{2800\times 80}{99}$
$\therefore v=2262.62 \;m^3$
Hence the volume of the air at $80^{\circ}C$ will be $2262.62 \; m^3$

Note: The ideal gas law gives the relationship between the volume, pressure and temperature of 1 molecule of the given gas. It if formed by combining the following laws, namely the Boyle’s law, Charles’ law , Gay-Lussac’s law and Avogadro’s law and is also called the combined ideal gas law.