Avogadro’s Law Made Easy

Avogadro’s Law Core Concepts

In this article, you will learn how quantity of moles relates to volume, and how to use Avogadro’s Law to determine the volume relation.

Topics Covered in Other Articles

Important Things to Consider

The gas law described in this article only applies to ideal gases, which you can read about on our article, The Ideal Gas Law.

Pressure and Moles

Consider a piston filled with 1 mole of oxygen. The piston has a pressure of 1 atm. As we know from our article What is Pressure, the pressure of the gas is dependent on the sum of collisions of the gas particles with the walls.

Now, consider what would happen if we added another mole of oxygen to the tank. Now, the amount of particles is doubled, and so the sum of the collisions is also doubled.

From this double in amount of collisions, the force exerted doubles and therefore the pressure also doubles. Since the piston can change in volume to equalize the pressure, the piston expands, according to Boyle’s Law, and the volume increases.

From this, we can start to see the relation between moles and volume.

Avogadro’s Law Relationship

Avogadro’s Law describes this relationship, which states that the amount of moles and the volume are proportional. This can be described by the equation: V\propto n.

From Boyle’s Law, we know that V\propto\frac{1}{P}. From this relation, we can rewrite Avogadro’s Law as P\propto n.

Constant Temperature Relation

Because the amount of moles is proportional to the volume, and the amount of moles of a gas in a container isn’t dependent on the identity of the gas, the volume of a certain amount of moles is always the same for a given temperature and pressure.

At standard conditions, for an ideal gas, this volume is 22.4 L. This means that any 1 mole of gas at 1 atm and 273 Celsius will occupy 22.4 L.

Avogadro’s Law Example Problem

Here’s an example of how you can use Avogadro’s Law to solve problems.

A weather balloon has been filled with 25 moles of helium gas, and occupies a volume of 600 L. The balloon is overinflated, and therefore the weather engineers designing it have decided to take some helium gas out. The standard size of a weather balloon is 521 L. How many moles of helium must be removed to decrease the volume of the balloon to a safe size?


     \begin{align*} n&\propto V\\ \implies \frac{V}{n}&=c\\ \implies \frac{V_1}{n_1}&=\frac{V_2}{n_2}\\ \therefore \frac{600\text{ L}}{25\text{ mol}}&=\frac{521\text{ L}}{x}\\ x&=\frac{25\text{ mol}}{600\text{ L}}\cdot\521\text{ L}\\ x&=21.7\text{ mol}\\ N_{f}&=N_{i}+\Delta N\\ 21.7&=25+\Delta N\\ \implies\Delta N&=\boxed{-3.29\text{ mol}} \end{align*}

Therefore, the engineers must take out 3.29 moles of helium gas.

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