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 \(3170 \mathrm{~Pa} ; \mathrm{R}=8.314 \mathrm{JK}^{-1}\) \(\mathrm{mol}^{-1}\) ) (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 number of moles in the vapor phase at equilibrium is \(1.27 \times 10^{-3} \text{ mol}\).

Step 1: Understand the Problem

We need to find how many moles of water vapor are at equilibrium in a flask that initially contains a small volume of liquid water. We'll use the ideal gas law to find the moles.

Step 2: Apply the Ideal Gas Law

The ideal gas law is given by \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is number of moles, \( R \) is the gas constant, and \( T \) is temperature in Kelvins.

Step 3: Set Up the Equation

Use the given values: \( P = 3170 \text{ Pa} \), \( V = 1.0 \, \mathrm{dm}^{3} \) (converted to \( 1.0 \times 10^{-3} \text{ m}^3 \)), \( R = 8.314 \text{ JK}^{-1}\text{mol}^{-1} \), and \( T = 300 \text{ K} \). Substitute these into the equation \( n = \frac{PV}{RT} \).

Step 4: Calculate the Moles

Substitute the values into the formula:\[n = \frac{3170 \times 1.0 \times 10^{-3}}{8.314 \times 300}\]Calculate:\[n = \frac{3.170}{2494.2} \approx 0.00127 \text{ moles}\]

Step 5: Choose the Correct Answer

The closest answer choice to our calculation is (d) \( 1.27 \times 10^{-3} \text{ mol} \).

Key Concepts

Vapor Pressure

Vapor pressure refers to the pressure exerted by a vapor in equilibrium with its liquid phase at a given temperature. When you add water to a closed system, like a flask, some of it will evaporate, and the vapor molecules will exert a pressure against the walls. This pressure is known as the vapor pressure.

At 300 K, the vapor pressure of water is given as 3170 Pa. This value helps us determine how much of the water has transitioned into vapor, since it reflects the point where the rate of evaporation equals the rate of condensation, meaning equilibrium is achieved.

• Vapor pressure is unique to each substance at a specified temperature.
• This value affects how much of the substance is present as vapor.
• It is crucial for calculating the number of moles in the vapor phase, as used in the ideal gas law.

Equilibrium

Equilibrium in a physical context, especially in closed systems like flasks, refers to the state where the phase changes between liquids and gases (or solids) balance out. In simpler terms, it's when the rate at which water evaporates equals the rate at which vapor condenses back into liquid water.

This state is essential for solving problems involving vapor pressure and moles calculation.

• Equilibrium is reached when there's no net change in the amount of liquid and vapor in the flask.
• It's crucial for determining the system's conditions to apply the ideal gas law correctly.
• Understanding equilibrium helps in predicting how substances behave under specific conditions.

Knowing equilibrium points is vital for correctly applying formulas like the ideal gas equation to get accurate answers.

Moles Calculation

In chemistry, moles are a way to quantify the amount of a substance. For gases, the ideal gas law, given by the formula \( PV = nRT \), is used to find the moles present at specific conditions. This formula can be rearranged to solve for the number of moles (\( n \)) in terms of pressure (\( P \)), volume (\( V \)), the universal gas constant (\( R \)), and temperature (\( T \)):

\[ n = \frac{PV}{RT} \]

Here's how it applies to the problem:

• Pressure \( P \) is the vapor pressure of the substance, which is 3170 Pa in this case.
• Volume \( V \) is the space the vapor occupies, here it's given as \( 1.0 \times 10^{-3} \text{ m}^3 \).
• Temperature \( T \) must be in Kelvin, here it's 300 K.
• The gas constant \( R \) is 8.314 JK^-1mol^-1, a universal value for calculations.

Plug these into the equation to determine the moles of water vapor in equilibrium inside the flask.

Temperature Conversion

Temperature conversion is the process of converting between temperature units, which is often necessary in science because different formulas require temperatures to be in specific units. For gas law calculations, temperatures need to be in Kelvin. This is because Kelvin is an absolute scale where 0 K is absolute zero, the lowest possible temperature where particles have minimum thermal motion.

To convert Celsius to Kelvin, which is the most common conversion, use the formula:

\[ K = °C + 273.15 \]

In gas law applications, like the one in this problem, ensuring the temperature is in Kelvin is crucial for the accuracy of calculations. This prevents errors and allows use of the universal gas constant \( R \) in its standard form.

• Always check the problem's given temperature and convert if needed.
• Kelvin removes negative temperatures, ensuring consistent calculations.
• It's used globally in scientific calculations for its precision and universality.