Avogadro’s law is a specific version of the ideal gas law. It says equal volumes at equal temperatures of an ideal gas all have the same number of molecules. This Avogadro’s law example problem will show how to use Avogadro’s law to find the number of moles in a given volume or the volume of a given number of moles.

Example Problem

Three balloons filled with different amounts of an ideal gas.

Question: Three balloons are filled with different amounts of an ideal gas. One balloon is filled with 3 moles of the ideal gas, filling the balloon to 30 L. a) One balloon contains 2 moles of gas. What is the volume of the balloon? b) One balloon encloses a volume of 45 L. How many moles of gas are in the balloon?

Solution:

Avogadro’s law says the volume (V) is directly proportional to the number of molecules of gas (n) at the same temperature.

n ∝ V

This means the ratio of n to V is equal to a constant value.

Since this constant never changes, the ratio will always be true for different amounts of gas and volumes.

where n_{i} = initial number of molecules V_{i} = initial volume n_{f} = final number of molecules V_{f} = final volume.

Part a) One balloon has 3 moles of gas in 30 L. The other has 2 moles in an unknown volume. Plug these values into the above ratio:

The larger volume means there is more gas in the balloon. In this case, there are 4.5 moles of the ideal gas in the larger balloon.

An alternative method would be to use the ratio of the known values. In part a, the known values were the number of moles. There was second balloon had ^{2}⁄_{3} the number of moles so it should have ^{2}⁄_{3} of the volume and our final answer is ^{2}⁄_{3} the known volume. The same is true of part b. The final volume is 1.5 times larger so it should have 1.5 times as many molecules. 1.5 x 3 = 4.5 which matches our answer. This is a great way to check your work.

Here’s a balanced equation for the combustion of butane, interpreted in terms of MOLES, MASS, and GAS VOLUMES at rtp. C4H10(g) + 6½ O2(g) 4CO2(g) + 5H2O(l) MOLES: 1 6½ 4 5 MOLES/grams 58g 208g 176g 90g GAS VOLS/dm3 1×24 6½x24 4×24 (negligible – liquid at rtp)

Dividing throughout by 24, we find that ratios of MOLES & GAS VOLUMES are the same, ie, ratio of C4H10 : O2 : CO2 = 1 : 6½ : 4

PS: (58g + 208g) must equal (176g + 90g) (Law of Conservation of MASS) but VOLUMES of GASES don’t have to be equal because they vary with MOLES, which also don’t have to be equal.

Here’s a balanced equation for the combustion of butane,

interpreted in terms of MOLES, MASS, and GAS VOLUMES at rtp.

C4H10(g) + 6½ O2(g) 4CO2(g) + 5H2O(l)

MOLES: 1 6½ 4 5

MOLES/grams 58g 208g 176g 90g

GAS VOLS/dm3 1×24 6½x24 4×24 (negligible –

liquid at rtp)

Dividing throughout by 24, we find that ratios of MOLES & GAS VOLUMES are the same, ie, ratio of C4H10 : O2 : CO2 = 1 : 6½ : 4

PS: (58g + 208g) must equal (176g + 90g) (Law of Conservation of MASS)

but VOLUMES of GASES don’t have to be equal because they vary with MOLES, which also don’t have to be equal.