Problem 7
Question
Which of the following expressions correctly represents the relationship between the average molar kinetic energy, K.E. of \(\mathrm{CO}\) and \(\mathrm{N}_{2}\) molecules at the same temperature? (a) \(\overline{\mathrm{K} . \mathrm{E}}_{\cdot \mathrm{CO}}<\overline{\mathrm{K} \cdot \mathrm{E}}_{\mathrm{N}_{2}}\) (b) \(\overline{\mathrm{K} . \mathrm{E}_{\mathrm{CO}}}>\overline{\mathrm{K}} \cdot \mathrm{E}_{\mathrm{N}_{2}}\) (c) \(\mathrm{K} . \mathrm{E}_{\mathrm{CO}}=\mathrm{K} \cdot \mathrm{E}_{\mathrm{N}_{2}}\) (d) cannot be predicted unless volumes of the gases are given.
Step-by-Step Solution
Verified Answer
The correct expression is (c): \( \mathrm{K.E}_{\mathrm{CO}} = \mathrm{K.E}_{\mathrm{N}_2} \).
1Step 1: Understanding Kinetic Energy in Gases
The average molar kinetic energy of gases is related to the temperature, not directly to the kind of gas. This principle is explained by the kinetic molecular theory, where the formula for average kinetic energy per molecule in a gas is given by \(\frac{3}{2} k T\), where \(k\) is the Boltzmann constant and \(T\) is the temperature in Kelvin. Kinetic energy is independent of the type of gas.
2Step 2: Analyzing the Options
Since kinetic energy is dependent on temperature and not on the type of gas, for gases at the same temperature, their average kinetic energies are equal. This implies that the expression for kinetic energy should reflect equality.
3Step 3: Choosing the Correct Expression
Given the options, only option (c) states that the kinetic energy of \(\mathrm{CO}\) molecules is equal to that of \(\mathrm{N}_2\) molecules, which matches the known principle that gases at the same temperature have equal average kinetic energies regardless of the type of gas.
Key Concepts
Average Molar Kinetic EnergyKinetic Energy FormulaTemperature Dependence of Kinetic Energy
Average Molar Kinetic Energy
In the study of gases, understanding the concept of average molar kinetic energy is crucial. This refers to the average energy of motion found in one mole of gas particles, influenced by the kinetic molecular theory. It's important to note that the average molar kinetic energy is temperature dependent and not influenced by the type of gas. This energy can be represented by a formula that involves the Boltzmann constant. The formula for the average kinetic energy per mole of gas at a given temperature is derived from:
- Kinetic energy per molecule: \( \frac{3}{2} k T \).
- Kinetic energy per mole: multiplying by Avogadro's number \( N_A \), resulting in \( \frac{3}{2} R T \).
- Here, \( R \) is the universal gas constant.
Kinetic Energy Formula
The kinetic energy for an individual gas molecule is derived from basic principles in physics. The central formula that governs this concept is:\[K.E. = \frac{1}{2} m v^2\]Where:
While the basic kinetic energy formula involves physical parameters, in gases the macroscopic property of temperature simplifies our calculations, giving us the average kinetic energy through:\[K.E. = \frac{3}{2} k T\]
- \( m \) is the mass of the gas molecule.
- \( v \) is the velocity of the molecule.
While the basic kinetic energy formula involves physical parameters, in gases the macroscopic property of temperature simplifies our calculations, giving us the average kinetic energy through:\[K.E. = \frac{3}{2} k T\]
Temperature Dependence of Kinetic Energy
The relationship between temperature and kinetic energy is a foundational concept in thermodynamics and the kinetic molecular theory. Temperature directly affects the average kinetic energy of a gas, defining their speed and movement intensities. Here are the key points:
- As temperature increases, so does the average kinetic energy of gas molecules.
- The formula \( \frac{3}{2} k T \) reveals how intimately connected temperature and kinetic energy are, with \( T \) in Kelvin ensuring proportionality.
- This makes it independent of the type of gases under consideration. Only the temperature informs the kinetic energy under equal conditions.
Other exercises in this chapter
Problem 5
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Which of the following law leads us to arrive at the conclusion that I g-molecule of each gas at STP occupies a volume of \(22.4 \mathrm{~L}\) ? (a) Dalton's la
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Which pair of the gases diffuses with the same rate at same temperature and pressure? (a) \(\mathrm{CO}\) and \(\mathrm{NO}\) (b) \(\mathrm{NO}_{2}\) and \(\mat
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