Discovered by Amedo Avogadro, of Avogadro's Hypothesis fame. The ChemTeam is not sure when, but probably sometime in the early 1800s.

Gives the relationship between volume and amount when pressure and temperature are held constant. Remember amount is measured in moles. Also, since volume is one of the variables, that means the container holding the gas is flexible in some way and can expand or contract.

If the amount of gas in a container is increased, the volume increases.

If the amount of gas in a container is decreased, the volume decreases.

Why?

Suppose the amount is increased. This means there are more gas molecules and this will increase the number of impacts on the container walls. This means the gas pressure inside the container will increase (for an instant), becoming greater than the pressure on the outside of the walls. This causes the walls to move outward. Since there is more wall space the impacts will lessen and the pressure will return to its original value.

The mathematical form of Avogadro's Law is: V ÷ n = k

This means that the volume-amount fraction will always be the same value if the pressure and temperature remain constant.

Let V_{1} and n_{1} be a volume-amount pair of data at the start of an experiment. If the amount is changed to a new value called n_{2}, then the volume will change to V_{2}.

We know this: V_{1} ÷ n_{1} = k

And we know this: V_{2} ÷ n_{2} = k

Since k = k, we can conclude that V_{1} ÷ n_{1} = V_{2} ÷ n_{2}.

This equation of V_{1} ÷ n_{1} = V_{2} ÷ n_{2} will be very helpful in solving Avogadro's Law problems.
Here is the Law done up in fractional form, something HTML isn't good at:

Avogadro's Law is a direct mathematical relationship.

You prove Avogadro's Law every time you blow up a balloon.

**Example #1:** 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)?

**Solution:**

I'll use V_{1}n_{2}= V_{2}n_{1}(5.00 L) (1.80 mol) = (x) (0.965 mol)

**Example #2:** A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 2.70 L? (The temperature was held constant.)

**Solution:**

1) Convert grams of He to moles:

2.00 g / 4.00 g/mol = 0.500 mol

2) Use Avogadro's Law:

V_{1}/n_{1}= V_{2}/n_{2}2.00 L / 0.500 mol = 2.70 L / x

x = 0.675 mol

3) Compute grams of He added:

0.675 mol - 0.500 mol = 0.175 mol0.175 mol x 4.00 g/mol = 0.7 grams of He added

There are no worksheet problems.