Problem 108
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
Verified Answer
The fraction of total pressure exerted by oxygen is \(\frac{1}{3}\) (option d).
1Step 1: Understanding the Problem
We are given equal weights of methane (5) and oxygen (9) mixed in a container, and we need to find the fraction of the total pressure exerted by oxygen.
2Step 2: Calculate Moles of Methane and Oxygen
The molar mass of methane is 16 g/mol, and the molar mass of oxygen is 32 g/mol. Given the same mass of each, the number of moles of methane is double the number of moles of oxygen since methane is half as heavy.
3Step 3: Determine Mole Fractions
If we have 1 mole of methane, we must have 0.5 moles of oxygen by weight equivalency. So, the mole fraction of oxygen, \(X_{O_2} = \frac{\text{moles of } O_2}{\text{moles of } CH_4 + \text{moles of } O_2} = \frac{0.5}{1 + 0.5} = \frac{1}{3}.\)
4Step 4: Conclusion on Partial Pressure Fraction
The partial pressure is directly proportional to the mole fraction, thus the fraction of the total pressure exerted by oxygen is also \(\frac{1}{3}\).
Key Concepts
Mole FractionPartial PressureIdeal Gas Law
Mole Fraction
The concept of mole fraction is essential when dealing with mixtures of gases. It tells us how much of a particular component is present in the gas mixture compared to the total amount.
The mole fraction, represented by the symbol \( X \), is calculated using this equation:
\[ X_i = \frac{n_i}{n_{total}} \]where \( n_i \) is the number of moles of the component of interest, and \( n_{total} \) is the total number of moles of all components in the mixture.
In our case with methane and oxygen, you calculate the mole fraction of oxygen as \( X_{O_2} = \frac{0.5}{1 + 0.5} \), resulting in \( \frac{1}{3} \).
This means one-third of the moles in the container are oxygen. The mole fraction is useful for understanding other properties, such as partial pressure.
The mole fraction, represented by the symbol \( X \), is calculated using this equation:
\[ X_i = \frac{n_i}{n_{total}} \]where \( n_i \) is the number of moles of the component of interest, and \( n_{total} \) is the total number of moles of all components in the mixture.
In our case with methane and oxygen, you calculate the mole fraction of oxygen as \( X_{O_2} = \frac{0.5}{1 + 0.5} \), resulting in \( \frac{1}{3} \).
This means one-third of the moles in the container are oxygen. The mole fraction is useful for understanding other properties, such as partial pressure.
Partial Pressure
Partial pressure is the pressure that a gas in a mixture would exert if it were alone in the container. This concept helps us understand how individual gases contribute to the total pressure of a gas mixture.
According to Dalton's Law, the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases. The partial pressure of a gas \( P_i \) is given by:
\[ P_i = X_i \times P_{total} \]where \( X_i \) is the mole fraction, and \( P_{total} \) is the total pressure.
In this exercise, since the mole fraction of oxygen is \( \frac{1}{3} \), the partial pressure it exerts is also \( \frac{1}{3} \) of the total pressure. This demonstrates that the partial pressure is directly proportional to the gas's mole fraction.
According to Dalton's Law, the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases. The partial pressure of a gas \( P_i \) is given by:
\[ P_i = X_i \times P_{total} \]where \( X_i \) is the mole fraction, and \( P_{total} \) is the total pressure.
In this exercise, since the mole fraction of oxygen is \( \frac{1}{3} \), the partial pressure it exerts is also \( \frac{1}{3} \) of the total pressure. This demonstrates that the partial pressure is directly proportional to the gas's mole fraction.
Ideal Gas Law
The ideal gas law is a fundamental equation in describing the behavior of gases. It relates the pressure, volume, temperature, and number of moles of a gas. The ideal gas law is expressed as:
\[ PV = nRT \]where:
Although real gases may deviate from this behavior, especially under high pressure and low temperature, the ideal gas law serves as an excellent approximation for many practical situations. In the context of our exercise, while we didn't directly use the ideal gas law, it underpins the principles that allow us to understand the behavior of the gas mixture in terms of mole fractions and partial pressures.
\[ PV = nRT \]where:
- \( P \) is the pressure of the gas.
- \( V \) is the volume of the container.
- \( n \) is the number of moles of the gas.
- \( R \) is the ideal gas constant (approximately 0.0821 L·atm/mol·K).
- \( T \) is the temperature in Kelvin.
Although real gases may deviate from this behavior, especially under high pressure and low temperature, the ideal gas law serves as an excellent approximation for many practical situations. In the context of our exercise, while we didn't directly use the ideal gas law, it underpins the principles that allow us to understand the behavior of the gas mixture in terms of mole fractions and partial pressures.
Other exercises in this chapter
Problem 106
Equal weights of ethane and hydrogen are mixed in an empty container at \(25^{\circ} \mathrm{C}\). The fraction to total pressure exerted by hydrogen is (a) \(1
View solution Problem 107
The ratio of root mean square velocity to average velocity of a gas molecule at a particular temperature is (a) \(1: 1,086\) (b) \(2: 1.086\) (c) \(1.086: 1\) (
View solution Problem 109
An L.P.G. cylinder contains \(15 \mathrm{~kg}\) of butane gas at \(27^{\circ} \mathrm{C}\) and 10 atmospheric pressure. It was leaking and its pressure fell dow
View solution Problem 112
\(10 \mathrm{~mL}\) of propane are mixed with \(70 \mathrm{~mL}\) of oxygen and exploded. What would be the volume of residual gases after explosion and after e
View solution