NCERT Solutions

We need to calculate the ratio between the macroscopic volume occupied by a mole of gas and the actual volume of its constituent molecules.

Concept: Molar volume

The molar volume ($V_m$) of an ideal gas is the total space occupied by one mole of the gas at a given temperature and pressure. The atomic volume ($V_a$) is the actual volume occupied by the molecules themselves, calculated as the product of Avogadro's number ($N_A$) and the volume of a single molecule ($V_{\text{mol}}$).

• Volume of a spherical molecule:

$$V_{\text{mol}} = \frac{4}{3} \pi r^3$$

• Atomic volume:

$$V_a = N_A \cdot V_{\text{mol}}$$

• Ratio:

$$R = \frac{V_m}{V_a}$$

Given:

• Molar volume ($V_m$): $22.4\,\text{L} = 22.4 \times 10^{-3}\,\text{m}^3$
• Size (diameter $d$) of hydrogen molecule: $1\,\text{\AA} = 10^{-10}\,\text{m}$
• Radius ($r$) of hydrogen molecule: $0.5 \times 10^{-10}\,\text{m}$
• Avogadro's number ($N_A$): $6.023 \times 10^{23}\,\text{mol}^{-1}$

Solving:

1. Calculate the volume of a single hydrogen molecule:

$$\begin{aligned} V_{\text{mol}} &= \frac{4}{3} \pi r^3 \\ &= \frac{4}{3} \times 3.14159 \times (0.5 \times 10^{-10}\,\text{m})^3 \\ &\approx 5.236 \times 10^{-31}\,\text{m}^3 \end{aligned}$$

2. Calculate the total atomic volume of one mole of hydrogen:

$$\begin{aligned} V_a &= N_A \times V_{\text{mol}} \\ &= 6.023 \times 10^{23}\,\text{mol}^{-1} \times 5.236 \times 10^{-31}\,\text{m}^3 \\ &\approx 3.154 \times 10^{-7}\,\text{m}^3 \end{aligned}$$

3. Calculate the ratio of molar volume to atomic volume:

$$\begin{aligned} R &= \frac{V_m}{V_a} \\ &= \frac{22.4 \times 10^{-3}\,\text{m}^3}{3.154 \times 10^{-7}\,\text{m}^3} \\ &\approx 7.1 \times 10^4 \end{aligned}$$

4. The ratio is extremely large because, in the gaseous state, the molecules are separated by distances much greater than their own diameters. Most of the volume in a gas is empty space, which is why gases are highly compressible compared to liquids or solids.

Answer

$$7.1 \times 10^4$$