Gas laws
Avogadro's law of volume

Avogadro’s Law states that equal volumes of all gases, when measured at the same temperature and pressure, contain an equal number of molecules. This means that the volume of a gas is directly proportional to the number of moles (or molecules) of the gas present, provided temperature and pressure remain constant.

Mathematically, Avogadro’s Law is expressed as:
$$V \propto n \quad \text{(at constant T and P)}$$ or
$$\frac{V_1}{n_1} = \frac{V_2}{n_2}$$ where:
\(V_1\) and \(V_2\) are the initial and final volumes of the gas, and
\(n_1\) and \(n_2\) are the corresponding number of moles (or molecules).

Illustration: If 1 mole of any gas occupies 22.4 dm3 (or 22,400 cm3) at standard temperature and pressure (S.T.P.), then 2 moles of the same gas will occupy twice that volume, i.e. 44.8 dm3, under the same conditions.

Avogadro’s Number (Constant)

Avogadro’s number (also called Avogadro’s constant) is the number of particles — atoms, molecules, or ions — present in one mole of a substance. It is denoted by the symbol \(N_A\).

The value of Avogadro’s number is:
$$N_A = 6.022 \times 10^{23} \, \text{particles per mole}$$
Illustration: One mole of hydrogen molecules contains \(6.022 \times 10^{23}\) H2 molecules, and one mole of oxygen molecules contains the same number of O2 molecules.

Calculations

Example 1: Calculate the number of molecules present in 5 moles of carbon dioxide (CO2).

Solution

$$ N_A = 6.022 \times 10^{23} $$ $$ \text{No. of molecules} = \text{moles} \times N_A $$ $$ = 5 \times 6.022 \times 10^{23} $$ $$ = 3.011 \times 10^{24} \, \text{molecules of CO}_2 $$

Example 2: How many oxygen atoms are present in 3 moles of oxygen gas (O2)?

Solution

$$ N_A = 6.022 \times 10^{23} $$

Each O2 molecule contains 2 atoms of oxygen.

$$ \text{No. of atoms} = 3 \times N_A \times 2 $$ $$ = 3 \times 6.022 \times 10^{23} \times 2 $$ $$ = 3.613 \times 10^{24} \, \text{oxygen atoms} $$

Example 3: Determine the number of sodium ions (Na⁺) in 0.5 mole of sodium chloride (NaCl).

Solution

Each formula unit of NaCl contains one Na⁺ ion. $$ \text{No. of Na}^+ ions = 0.5 \times N_A $$ $$ = 0.5 \times 6.022 \times 10^{23} $$ $$ = 3.011 \times 10^{23} \, \text{ions of Na}^+ $$

Example 4: How many molecules are contained in 11.2 dm³ of oxygen gas (O2) at S.T.P.?

Solution

At S.T.P., 1 mole of gas occupies 22.4 dm³. $$ \text{Moles of O}_2 = \frac{11.2}{22.4} = 0.5 \, \text{mol} $$ $$ \text{No. of molecules} = 0.5 \times N_A $$ $$ = 0.5 \times 6.022 \times 10^{23} $$ $$ = 3.011 \times 10^{23} \, \text{molecules of O}_2 $$

Example 5: Find the number of chloride ions (Cl⁻) present in 2.5 moles of calcium chloride (CaCl2).

Solution

Each formula unit of CaCl2 contains 2 Cl⁻ ions. $$ \text{No. of Cl ions} $$ $$= 2.5 \times N_A \times 2 $$ $$ = 2.5 \times 6.022 \times 10^{23} \times 2 $$ $$ = 3.011 \times 10^{24} \, \text{Cl ions} $$

Example 6: Calculate the number of atoms in 4.5 g of aluminium (Al). [Molar mass of Al = 27 g/mol]

Solution

$$ \text{Moles of Al} = \frac{\text{mass}}{\text{molar mass}} $$ $$= \frac{4.5}{27} = 0.1667 \, \text{mol} $$ $$ \text{No. of atoms} = 0.1667 \times N_A $$ $$ = 0.1667 \times 6.022 \times 10^{23} $$ $$ = 1.00 \times 10^{23} \, \text{atoms of aluminium} $$

Example 7: What volume (in dm³ and cm³) at S.T.P. is occupied by 5.00 g of carbon dioxide (CO2)? [Molar mass of CO2 = 44.01 g mol⁻¹; at S.T.P. 1 mol = 22.4 dm³]

Solution

$$\text{Moles of CO}_2 = \frac{\text{mass}}{\text{molar mass}} $$ $$ = \frac{5.00}{44.01}$$ $$ = 0.1136\ \text{mol}$$ $$\text{Volume at S.T.P. }(V) = n \times 22.4\ \text{dm}^3 $$ $$ = 0.1136 \times 22.4 = 2.545\ \text{dm}^3$$ $$V = 2.545 \times 10^3\ \text{cm}^3 = 2545\ \text{cm}^3$$

Final Answer: \(V = 2.545\ \text{dm}^3\) (≈ 2545 cm³) at S.T.P.

Example 8: Calculate the mass of nitrogen (N2) that contains \(2.00 \times 10^{24}\) molecules. [Molar mass of N2 = 28.02 g mol⁻¹; \(N_A = 6.022\times10^{23}\)]

Solution

$$\text{Moles }(n) = \frac{\text{number of molecules}}{N_A} $$ $$= \frac{2.00\times10^{24}}{6.022\times10^{23}} $$ $$ = 3.32\ \text{mol}$$ $$\text{Mass} = n \times \text{molar mass} $$ $$ = 3.32 \times 28.02 = 93.06\ \text{g}$$

Final Answer: Mass of N2 ≈ 93.06 g.

Example 9: A sample of ammonia gas (NH3) has a volume of 250 cm³ at S.T.P. Calculate (a) the number of molecules present and (b) the mass of the sample.
[Molar mass NH3 = 17.03 g mol⁻¹; \(N_A = 6.022\times10^{23}\); 1 mol gas at S.T.P. = 22.4 dm³]

Solution

Convert volume to dm³: $$V = 250\ \text{cm}^3 = 0.250\ \text{dm}^3$$ Moles of NH3: $$n = \frac{V}{22.4} = \frac{0.250}{22.4} $$ $$ = 0.01116\ \text{mol}$$ Number of molecules: $$\text{No. of molecules} = n \times N_A $$ $$ = 0.01116 \times 6.022\times10^{23} $$ $$ = 6.72\times10^{21}$$ Mass of NH3: $$\text{Mass} = n \times \text{molar mass} $$ $$ = 0.01116 \times 17.03 $$ $$ = 0.190\ \text{g}$$ Final Answers: (a) Number of molecules = \(6.72\times10^{21}\)

(b) Mass = \(0.190\ \text{g}\)

Avogadro's Law & Molar Calculator


Formulas Used:
  1. \( N = n \times N_A \) → To find number of particles

  2. \( n = \frac{N}{N_A} \) → To find number of moles

  3. \( n = \frac{m}{M} \) → Moles from mass

  4. \( m = n \times M \) → Mass from moles
Use scientific notation (e.g., 6.022e23). All results shown to 3 significant figures.

Summary