Problem 108 - A Complete Resource Book in Chemistry for JEE Main

Question

Equal weights of methane and oxygen are mixed in an empty container at \(25^{\circ} \mathrm{C}\). the fraction of the total pressure exerted by oxygen is (a) \(1 / 2\) (b) \(2 / 3\) (c) \(1 / 3 \times 273 / 298\) (d) \(1 / 3\)

Step-by-Step Solution

Answer

(d) \(1/3\)

Step 1: Understanding the problem

We are given equal masses of methane (CH₄) and oxygen (O₂) mixed in a container, and we need to find the fraction of the total pressure that is exerted by oxygen at 25°C.

Step 2: Determine moles from the masses

Let's assume the mass of methane is 'm' grams. Since the masses are equal, the oxygen also has a mass of 'm' grams. The molecular weight of methane (CH₄) is 16 g/mol and that of oxygen (O₂) is 32 g/mol.

Step 3: Calculate moles of each gas

The moles of methane, n(CH₄), are given by \( \frac{m}{16} \) and moles of oxygen, n(O₂), are \( \frac{m}{32} \).

Step 4: Calculate total moles

The total moles within the container is the sum of moles of CH₄ and O₂: \( n_{\text{total}} = \frac{m}{16} + \frac{m}{32} \). To simplify, we find a common denominator. This gives \( n_{\text{total}} = \frac{2m}{32} + \frac{m}{32} = \frac{3m}{32} \).

Step 5: Calculate mole fraction of oxygen

The mole fraction of oxygen is the moles of oxygen divided by the total moles: \( X_{\text{O}_2} = \frac{\left(\frac{m}{32}\right)}{\left(\frac{3m}{32}\right)} \). Simplifying, we get \( X_{\text{O}_2} = \frac{1}{3} \).

Step 6: Use the ideal gas law to justify

Since gas pressure is proportional to mole fraction (assuming ideal gas laws in a closed system), the fraction of the total pressure exerted by oxygen is equal to its mole fraction. Therefore, the fraction of pressure exerted by oxygen is also \( \frac{1}{3} \).

Key Concepts

Mole Fraction

The mole fraction is a crucial concept in understanding gas mixtures. It tells us how much of a particular gas is present compared to the total amount of gas. In mathematical terms, it's a ratio. You divide the number of moles of a particular gas by the total number of moles in the mixture. This value is always less than 1, as it represents a fraction of the total. For example, if you have a mixture of two gases, A and B, and you want to find the mole fraction of gas A, the formula is given by:

• \( X_{A} = \frac{n_{A}}{n_{A} + n_{B}} \)

Here, \( n_{A} \) is the number of moles of gas A, and \( n_{B} \) is the number of moles of gas B. In the exercise, we learned that to find the mole fraction of oxygen when mixed with methane, we used the formula \( X_{ ext{O}_2} = \frac{1}{3} \). This calculated mole fraction further helps in finding the proportion of pressure exerted by each gas in the mixture.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation that relates the pressure, volume, temperature, and number of moles of a gas. The formula is expressed as \( PV = nRT \), where:

• \( P \) is the pressure of the gas
• \( V \) is the volume
• \( n \) is the number of moles
• \( R \) is the ideal gas constant
• \( T \) is the temperature in Kelvin

This equation helps us understand how gases behave under various conditions. In a mixture of gases, each gas exerts pressure independently. This pressure is directly proportional to the number of moles of the gas, assuming the volume and temperature remain constant. In the original exercise, the pressure exerted by oxygen was determined equivalent to its mole fraction of the total mix, due to the properties defined by the Ideal Gas Law.

Gas Mixtures

Gas mixtures occur when two or more gases are combined. Each gas in a mixture behaves as if it is the only gas present, a concept explained by Dalton's Law of Partial Pressures. This law states that the total pressure exerted by a mixture is the sum of the partial pressures of each constituent gas. Mathematically, it is expressed as:

• \( P_{total} = P_{1} + P_{2} + P_{3} + ext{...} \)

Each partial pressure is calculated based on the mole fraction and total pressure:

• \( P_{x} = X_{x} \times P_{total} \)

In the solution to the exercise, understanding how gas mixtures work allowed us to derive that the fraction of the pressure exerted by oxygen was the same as its mole fraction, \( \frac{1}{3} \). This is because in a perfect system where gases behave ideally, the contribution of each gas to the total pressure is proportional to its mole fraction.