Q.The density of water is $1000\ \text{kg m}^{-3}$. The density of water vapour at $100\ ^\circ\text{C}$ and $1\ \text{atm}$ pressure is $0.6\ \text{kg m}^{-3}$. The volume of a molecule multiplied by the total number gives, what is called, molecular volume. Estimate the ratio (or fraction) of the molecular volume to the total volume occupied by the water vapour under the above conditions of temperature and pressure.

Concept: Molecular Volume Fraction

The molecular volume is the actual space occupied by the molecules themselves, while the total volume includes all the empty space between them. For a liquid, molecules are closely packed so molecular volume ≈ total volume. For a gas, molecules are far apart so molecular volume ≪ total volume.

Reasoning:

Consider a fixed mass $m$ of water.

As liquid water at density $\rho_\text{liquid} = 1000\ \text{kg m}^{-3}$, it occupies volume:

Since liquid molecules are tightly packed, this volume is essentially the molecular volume $V_\text{mol}$.

When the same mass vaporizes at $100\ ^\circ\text{C}$ and $1\ \text{atm}$, the vapour has density $\rho_\text{vapour} = 0.6\ \text{kg m}^{-3}$ and occupies: …

Why this approach works

When water evaporates, the same molecules that were tightly packed in liquid form spread out into a much larger volume as vapour. The molecular volume — the actual physical space the molecules themselves occupy — doesn't change; what changes is how much empty space surrounds them.

In the liquid phase, molecules are essentially touching, so the molecular volume is nearly equal to the total volume. In the vapour phase, molecules are far apart, so the molecular volume is a tiny fraction of the total volume. The key insight: the ratio of densities tells us how much the same mass has expanded, which directly gives us the molecular volume fraction.

Step-by-step reasoning

1. Understand what stays constant

Consider a fixed mass $m$ of water. Whether it's liquid or vapour, the number of molecules $N$ is the same, and each molecule has the same intrinsic volume $v_0$. Therefore, the molecular volume $V_{\text{mol}} = N v_0$ is identical in both phases.

2. Relate molecular volume to liquid water

In liquid water at $1000\ \text{kg m}^{-3}$, molecules are closely packed with negligible gaps. The volume occupied by mass $m$ is:

Since molecules are touching, $V_{\text{mol}} \approx V_{\text{liquid}}$, so:

3. Find the total volume of vapour

The same mass $m$ as water vapour at $0.6\ \text{kg m}^{-3}$ occupies:

4. Calculate the molecular volume fraction

The fraction we seek is: …

A quicker route skips computing $V_{mol}$ separately. Since the same mass $m$ of water is used in both states, and liquid molecules are packed with virtually no gaps, the molecular volume is essentially $V_{liquid}$. So the required fraction is simply the ratio of the two densities taken in reverse: $\frac{V_{mol}}{V_{vapour}} = \frac{m/\rho_{liquid}}{m/\rho_{vapour}} = \frac{\rho_{vapour}}{\rho_{liquid}} = \frac{0.6}{1000} = 6\times10^{-4}$. This one-line proportional argument — fraction $= \rho_{vapo …