Problem 201 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(10^{-4} \mathrm{dm}^{3}\) of water is introduced into \(1.0 \mathrm{dm}^{3}\) flask at \(300 \mathrm{~K}\), how many moles of water are in the vapour phase when equilibrium is established? (Given: Vapour pressure of \(\mathrm{H}_{2} \mathrm{O}\) at \(300 \mathrm{~K}\) is \(3170 \mathrm{~Pa} ; \mathrm{R}=8.314 \mathrm{JK}^{-1}\) \(\mathrm{mol}^{-1}\) ) \(\quad[\mathbf{2 0 0 9}]\) (a) \(5.56 \times 10^{-3} \mathrm{~mol}\) (b) \(1.53 \times 10^{-2} \mathrm{~mol}\) (c) \(4.46 \times 10^{-2} \mathrm{~mol}\) (d) \(1.27 \times 10^{-3} \mathrm{~mol}\)

Step-by-Step Solution

Answer

The correct answer is (d) 1.27 × 10⁻³ mol.

Step 1: Determine the Vapour Pressure in the Flask

Since we are provided with the vapour pressure of water at 300 K, which is \(3170 \text{ Pa}\), this is the pressure of the water vapour that will be in equilibrium with the liquid water in the flask. The volume of the flask plays a role in determining the number of moles of water in the vapour phase.

Step 2: Use Ideal Gas Law to Find Moles of Vapour Phase Water

Apply the ideal gas law formula, \(PV = nRT\), where \(P = 3170 \text{ Pa}\), \(V = 1.0 \text{ dm}^3 = 1.0 \times 10^{-3} \text{ m}^3\), \(R = 8.314 \text{ J }\text{K}^{-1}\text{mol}^{-1}\), and \(T = 300 \text{ K}\). We rearrange the formula to solve for \(n\): \[ n = \frac{PV}{RT} \].

Step 3: Calculate the Moles Using the Ideal Gas Law

Substitute the given values into the ideal gas law equation: \[ n = \frac{3170 \times 1.0 \times 10^{-3}}{8.314 \times 300} \]. By performing the calculations: \[ n \approx \frac{3.17}{2494.2} = 1.27 \times 10^{-3} \text{ mol} \].

Step 4: Select the Correct Option

The calculated number of moles of water in the vapour phase is approximately \(1.27 \times 10^{-3}\text{ mol}\). Check the options given:(a) \(5.56 \times 10^{-3} \text{ mol}\) (b) \(1.53 \times 10^{-2} \text{ mol}\) (c) \(4.46 \times 10^{-2} \text{ mol}\) (d) \(1.27 \times 10^{-3} \text{ mol}\)The correct answer is option \(d\).

Key Concepts

Vapour Pressure

Vapour pressure is a critical concept in understanding how liquids behave when vaporizing. It refers to the pressure exerted by a vapour in equilibrium with its liquid phase at a specific temperature. This means that when water in a flask reaches its vapour pressure, the rate of evaporation equals the rate of condensation. As a result, an equilibrium is established, and the amount of water in the vapour phase remains constant over time.
For the exercise in question, the given vapour pressure of water at 300 K is 3170 Pa. This indicates the pressure that the water vapor will exert once equilibrium is reached in the flask. At this point, no additional water will evaporate, since the system has already balanced itself between the liquid and vapour phases. This concept helps us understand that the vapour pressure of a substance is a unique property and depends on temperature, not the volume of the container.

Equilibrium in Chemical Processes

Equilibrium is a fundamental aspect of many chemical processes, including the transition between liquid and vapour phases. It occurs when the rates of forward and reverse processes are equal, resulting in no overall change in the concentrations of reactants and products over time. At equilibrium, the system is said to be in a stable state.
In the given scenario, when water is placed in the flask, it begins to evaporate, increasing the concentration of water molecules in the air. Eventually, the rate at which liquid water molecules evaporate equals the rate at which vapour molecules condense back into liquid form. This state of balance is what we call equilibrium, and it remains as long as temperature and pressure remain constant. This explains why, once the water reaches its vapour pressure, no net conversion between liquid and vapour occurs.
Understanding equilibrium helps predict how changes in conditions such as temperature or pressure might affect the system. For example, increasing the temperature would increase the vapour pressure, thus shifting the equilibrium towards more evaporation until a new equilibrium is reached.

Moles Calculation

The calculation of moles is a fundamental skill in chemistry, particularly when using the ideal gas law. The ideal gas law is represented by the equation \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the universal gas constant, and \(T\) is temperature.
In this exercise, students are tasked with determining how many moles of water are present in the vapour phase at equilibrium. Using the provided vapour pressure (3170 Pa), volume of the flask (1.0 dm³ or 1.0 × 10⁻³ m³), and temperature (300 K), alongside the gas constant (8.314 J K⁻¹ mol⁻¹), we rearrange the ideal gas law to solve for \(n\):
\[ n = \frac{PV}{RT} \]
Substituting the known values:

• \(P = 3170\) Pa
• \(V = 1.0 \times 10^{-3}\) m³
• \(T = 300\) K

We get:
\[ n = \frac{3170 \times 1.0 \times 10^{-3}}{8.314 \times 300} \approx 1.27 \times 10^{-3} \text{ mol} \]
This calculation confirms the number of moles of water in the vapour phase at equilibrium. By mastering these calculations, students can apply the ideal gas law to various scenarios, providing insights into gas behavior under different conditions.