Problem 100 - A Complete Resource Book in Chemistry for JEE Main

Question

At \(100^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), if the density of liquid water is \(1.0 \mathrm{~g} \mathrm{~cm}^{-3}\) and that of water vapour is \(0.0006 \mathrm{~g} \mathrm{~cm}^{-3}\), then the volume occupied by water molecules in 1 litre of steam at that temperature (a) \(6 \mathrm{~cm}^{3}\) (b) \(60 \mathrm{~cm}^{3}\) (c) \(0.6 \mathrm{~cm}^{3}\) (d) \(0.06 \mathrm{~cm}^{3}\)

Step-by-Step Solution

Answer

The volume occupied by water molecules in 1 liter of steam is \(0.6 \, \text{cm}^3\).

Step 1: Understand the Given Information

We have the density of liquid water as \(1.0 \, \text{g/cm}^3\) and the density of water vapor as \(0.0006 \, \text{g/cm}^3\). We need to find the volume occupied by water molecules in 1 liter (which is 1000 cm³) of steam.

Step 2: Calculate Mass of Water Vapor

Use the density formula: density = mass/volume. We rearrange to find the mass of the water vapor: \( \text{mass} = \text{density} \times \text{volume} = 0.0006 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3 = 0.6 \, \text{g} \).

Step 3: Determine Volume of Water as Liquid

Now, consider the mass of the water molecules, which is \(0.6 \, \text{g}\). Use the density of liquid water to find the volume occupied by this mass: \[ \text{volume} = \frac{\text{mass}}{\text{density}} = \frac{0.6 \, \text{g}}{1.0 \, \text{g/cm}^3} = 0.6 \, \text{cm}^3 \].

Key Concepts

Density

Density is a critical concept in thermodynamics, representing how much mass is contained within a given volume.
In simple terms, it's like trying to understand how "packed" a substance is.

• For example, liquid water has a density of \(1.0 \, \text{g/cm}^3\), meaning 1 gram of water fits perfectly in a 1 cm³ space.
• On the other hand, water vapor, which is water in a gaseous state, is much less dense, standing at \(0.0006 \, \text{g/cm}^3\).

This vast difference in densities explains why a large number of water vapor molecules can be compressed into a small space when they become liquid.
Unlike solids and liquids, gases can expand and compress significantly, which drastically changes their density under different conditions.
Understanding this fundamental difference is key when dealing with phase change calculations and explains why water vapor in the gaseous form occupies much more space than its liquid form.

Volume calculations

Volume calculations allow us to determine how much space something occupies.
They are crucial when converting between phases of matter, such as from liquid to vapor. When calculating volume, we often use the formula relating density, mass, and volume:
\[ \text{density} = \frac{\text{mass}}{\text{volume}} \] To find the volume when we know density and mass, we rearrange the formula to:
\[ \text{volume} = \frac{\text{mass}}{\text{density}} \]For the given problem, to determine the volume of water molecules in 1 liter of steam at \(100^{\circ} \mathrm{C}\), we follow these steps:

• First, find the mass of water vapor being the product of its density and the total volume of steam, i.e., \(0.0006 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3 = 0.6 \, \text{g}\).
• Then, calculate the volume of this mass of water if it were liquid: \(\frac{0.6 \, \text{g}}{1.0 \, \text{g/cm}^3} = 0.6 \, \text{cm}^3\).

This calculation shows that 0.6 cm³ of water molecules can be found in 1 liter of steam.

Phase change

Phase change describes the transition a substance undergoes from one state of matter to another, like solid to liquid, or liquid to gas.
This transformation doesn't alter the chemical composition but rather the physical state and volume.
When water transitions from liquid to vapor (gas), it undergoes a significant change in volume:

• The transformation involves heating, causing molecules to gain energy and spread out, increasing the space occupied compared to the liquid form.
• For water turning to steam, the density reduces greatly, which is why the gaseous form, at \(0.0006 \, \text{g/cm}^3\), occupies far more space than the liquid form, with density \(1.0 \, \text{g/cm}^3\).

Understanding phase changes is essential for interpreting problems dealing with changes in states of matter.
In thermodynamics, these effects must be carefully managed, especially when calculating how much material moves between phases, like from liquid water turning into steam.