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) \(\frac{1}{3}\)

Step 1: Calculate Molar Mass

First, we need to determine the molar masses of methane (CH₄) and oxygen (O₂). Methane has a molar mass of approximately 16 g/mol (12 for carbon and 4 for hydrogen), while oxygen has a molar mass of about 32 g/mol, since each oxygen atom has a mass of 16.

Step 2: Convert Weights to Moles

Since equal weights of methane and oxygen are used, let's assume each gas has a mass of 32 grams to simplify calculations (note that the actual mass does not affect the final pressure fraction as long as they are equal). Therefore, the number of moles of methane is \( \frac{32}{16} = 2 \) moles, and for oxygen is \( \frac{32}{32} = 1 \) mole.

Step 3: Find the Total Moles

Add the moles of both gases together to find the total number of moles in the container: \( 2 + 1 = 3 \) moles.

Step 4: Calculate the Mole Fraction of Oxygen

The mole fraction of oxygen is the ratio of the moles of oxygen to the total moles of gas. Thus, the mole fraction of oxygen is \( \frac{1}{3} \).

Step 5: Determine the Fraction of Total Pressure

According to Dalton's Law of Partial Pressures, the fraction of the total pressure exerted by a gas is equal to its mole fraction. Thus, the fraction of the total pressure exerted by the oxygen is also \( \frac{1}{3} \).

Key Concepts

Mole Fraction

The mole fraction is a key concept in understanding gas mixtures. It describes how much of a specific component contributes to the total number of moles. This is crucial because it allows us to determine how much pressure each gas exerts in a mixture.
In the exercise, we found that the mole fraction of oxygen was determined by dividing the moles of oxygen by the total moles of gas:

• For oxygen, the moles were 1 mole.
• The total moles of gas in the container were 3 moles.

Thus, the mole fraction of oxygen was calculated as \( \frac{1}{3} \). This means that oxygen accounts for one third of the total pressure, according to Dalton's Law of Partial Pressures. The concept itself helps in understanding how distributed gases are within a container and how they contribute to the container's total pressure.

Molar Mass Calculation

Calculating the molar mass is the first step in approaching many chemistry problems involving gases. Understanding the molar mass of a substance helps us convert grams of a substance into moles, which are a foundational component of chemical reactions and equations.
For example, to find the molar mass:

• Methane (CH₄) consists of one carbon atom (12 g/mol) and four hydrogen atoms (4 x 1 g/mol), totalling to 16 g/mol.
• Oxygen (O₂) consists of two oxygen atoms, each weighing approximately 16 g/mol, leading to a molar mass of about 32 g/mol.

In the given exercise, knowing these molar masses allowed us to convert equal masses of methane and oxygen into moles (2 moles of methane and 1 mole of oxygen) and further proceed with the calculation of mole fractions.

Ideal Gas Concepts

The ideal gas law provides a good approximation for the behavior of many gases under a variety of conditions. An important part of the ideal gas concept is Dalton's Law of Partial Pressures, which allows us to predict the contribution each gas in a mixture makes to the total pressure.
Dalton’s Law states that in a mixture of non-reacting gases, the total pressure is the sum of the partial pressures of individual gases. A gas's partial pressure is directly proportional to its mole fraction in the mixture, assuming all gases behave ideally:

• The partial pressure is found by multiplying the mole fraction by the total pressure.
• In the example provided, the mole fraction of oxygen tells us its contribution to the pressure is \( \frac{1}{3} \) of the total pressure exerted in the container.

Understanding these principles helps explain why gases exert pressure independently and why the behavior of a whole mixture can be determined from its parts.