**Avogadro’s law formula** states that the **volume **(V) of a given quantity of gas directly depends upon its **number of moles** (n), which means that the volume of the gas increases as its number of moles increases. Here’s the formula of avogadro’s law: **V**_{1}**/n**_{1}** = V**_{2}**/n**_{2}

Let’s solve some problems based on this formula, so you’ll get a clear idea.

## Avogadro’s Law Practice Problems

**Problem 1:** A 5 L of gas containing 0.4 moles of helium gas is filled in a rubber ball. Calculate the final volume of the gas, when the additional 2 moles of helium gas are added in a rubber ball.

Solution:

Given data:

Initial volume of the gas, V_{1} = 5 L

Initial number of moles of the gas, n_{1} = 0.4 mol

Final volume of the gas, V_{2} = ?

Final number of moles of the gas, n_{2} = 0.4 mol + 2 mol = 2.4 mol

Using the formula of avogadro’s law,

V_{1}/n_{1} = V_{2}/n_{2}

V_{2} = (V_{1} × n_{2}) ÷ n_{1}

V_{2} = (5 × 2.4) ÷ 0.4

V_{2} = 12 ÷ 0.4

V_{2} = 30 L

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

**Problem 2:** A 6 L of gas containing 0.8 moles of hydrogen gas is released in an empty tube. Calculate the final number of moles of the gas, when the volume of the gas increases to 12 L.

Solution:

Given data:

Initial volume of the gas, V_{1} = 6 L

Initial number of moles of the gas, n_{1} = 0.8 mol

Final number of moles of the gas, n_{2} = ?

Final volume of the gas, V_{2} = 12 L

Using the formula of avogadro’s law,

V_{1}/n_{1} = V_{2}/n_{2}

n_{2} = (n_{1} × V_{2}) ÷ V_{1}

n_{2} = (0.8 × 12) ÷ 6

n_{2} = 0.8 × 2

n_{2} = 1.6 mol

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

**Problem 3:** One container filled with a few liters of gas contains 0.7 moles of lithium gas. If the volume of the gas is increased to 9 L and the extra 1.2 moles of lithium gas are added in the container, then what is the initial volume of the gas?

Solution:

Given data:

Initial number of moles of the gas, n_{1} = 0.7 mol

Final volume of the gas, V_{2} = 9 L

Final number of moles of the gas, n_{2} = 0.7 mol + 1.2 mol = 1.9 mol

Initial volume of the gas, V_{1} = ?

Using the formula of avogadro’s law,

V_{1}/n_{1} = V_{2}/n_{2}

V_{1} = (V_{2} × n_{1}) ÷ n_{2}

V_{1} = (9 × 0.7) ÷ 1.9

V_{1} = 6.3 ÷ 1.9

V_{1} = 3.31 L

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

**Problem 4:** A helium balloon filled with 3 L of gas contains some moles of argon gas. The volume of the gas is increased to 9 L and the additional 3 moles of argon gas are added in a helium balloon. Calculate the initial number of moles of the gas.

Solution:

Given data:

Initial volume of the gas, V_{1} = 3 L

Final volume of the gas, V_{2} = 9 L

Final number of moles of the gas, n_{2} = 3 mol

Initial number of moles of the gas, n_{1} = ?

Using the formula of avogadro’s law,

V_{1}/n_{1} = V_{2}/n_{2}

n_{1} = (V_{1} × n_{2}) ÷ V_{2}

n_{1} = (3 × 3) ÷ 9

n_{1} = 9 ÷ 9

n_{1} = 1 mol

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

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