Avogadro’s law formula, expressed as V1/n1 = V2/n2, relates the volume of a given quantity of gas to its number of moles. In this formula, V1 and n1 represent the initial volume and number of moles of the gas, while V2 and n2 represent the final volume and number of moles, respectively. This formula can be used to calculate the volume or number of moles of a gas under different conditions, given the initial volume and number of moles of the gas.

Contents

## Practice problems

### Problem #1

A rubber ball contains 5 L of gas with 0.4 moles of helium at a given temperature and pressure. Determine the final volume of the gas when an additional 2 moles of helium are added to the rubber ball.

Solution

Given data:

• Initial volume of the gas, V1 = 5 L
• Initial number of moles of the gas, n1 = 0.4 mol
• Final volume of the gas, V2 = ?
• Final number of moles of the gas, n2 = 0.4 mol + 2 mol = 2.4 mol

Applying the formula:

• V1/n1 = V2/n2
• V2 = (V1 × n2) ÷ n1
• V2 = (5 × 2.4) ÷ 0.4
• V2 = 12 ÷ 0.4
• V2 = 30 L

Therefore, the final volume of the gas is 30 L.

### Problem #2

A tube contains 6 L of gas with 0.8 moles of hydrogen gas at a certain temperature and pressure. Find the final number of moles of gas when the volume of the gas increases to 12 L.

Solution

Given data:

• Initial volume of the gas, V1 = 6 L
• Initial number of moles of the gas, n1 = 0.8 mol
• Final number of moles of the gas, n2 = ?
• Final volume of the gas, V2 = 12 L

Applying the formula:

• V1/n1 = V2/n2
• n2 = (n1 × V2) ÷ V1
• n2 = (0.8 × 12) ÷ 6
• n2 = 0.8 × 2
• n2 = 1.6 mol

Therefore, the final number of moles of the gas is 1.6 mol.

### Problem #3

A container is filled with a few liters of gas containing 0.7 moles of helium gas. If the volume of the gas is increased to 9 L and an additional 1.2 moles of argon gas are added to the container, then what was the initial volume of the gas?

Solution

Given data:

• Initial number of moles of the gas, n1 = 0.7 mol
• Final volume of the gas, V2 = 9 L
• Final number of moles of the gas, n2 = 0.7 mol + 1.2 mol = 1.9 mol
• Initial volume of the gas, V1 = ?

Applying the formula:

• V1/n1 = V2/n2
• V1 = (V2 × n1) ÷ n2
• V1 = (9 × 0.7) ÷ 1.9
• V1 = 6.3 ÷ 1.9
• V1 = 3.31 L

Therefore, the initial volume of the gas is 3.31 L.

### Problem #4

A balloon is initially filled with 3 L of gas containing some moles of argon gas. If the volume of the gas is increased to 9 L and an additional 3 moles of argon gas are added to the balloon, what was the initial number of moles of the gas?

Solution

Given data:

• Initial volume of the gas, V1 = 3 L
• Final volume of the gas, V2 = 9 L
• Final number of moles of the gas, n2 = 3 mol
• Initial number of moles of the gas, n1 = ?

Applying the formula:

• V1/n1 = V2/n2
• n1 = (V1 × n2) ÷ V2
• n1 = (3 × 3) ÷ 9
• n1 = 9 ÷ 9
• n1 = 1 mol

Therefore, the initial number of moles of the gas is 1 mol.