concept-on-law-of-equipartition-of-energy
🚀 The law of equipartition of energy states that energy is distributed equally among all degrees of freedom in a system at thermal equilibrium. For a perfect gas, this means that each degree of freedom contributes an equal amount of energy to the total energy of the system. In classical mechanics, each translational degree of freedom contributes \(\frac{1}{2}kT\) to the energy, where \(k\) is the Boltzmann constant and \(T\) is the temperature in Kelvin. For a monatomic ideal gas, there are three translational degrees of freedom, leading to a total energy of \(\frac{3}{2}NkT\), where \(N\) is the number of particles. For diatomic gases, there are additional rotational degrees of freedom, leading to a total energy of \(\frac{5}{2}NkT\). This concept is crucial for understanding specific heats, as the specific heat at constant volume \(C_V\) and at constant pressure \(C_P\) can be derived from the equipartition theorem.
Theory Explanation
Understanding Degrees of Freedom
Degrees of freedom refer to the independent ways in which a system can possess energy. For a gas molecule, these include translational motion (movement in x, y, and z directions) and rotational motion. Monatomic gases have 3 translational degrees of freedom, while diatomic gases have 3 translational and 2 rotational degrees of freedom, totaling 5 degrees of freedom.
Energy Contribution per Degree of Freedom
According to the equipartition theorem, each degree of freedom contributes \(\frac{1}{2}kT\) to the total energy. Therefore, for a monatomic gas with 3 degrees of freedom, the total energy is \(E = \frac{3}{2}NkT\). For a diatomic gas with 5 degrees of freedom, the total energy is \(E = \frac{5}{2}NkT\).
Specific Heats from Equipartition
The specific heat at constant volume \(C_V\) is derived from the total energy. For a monatomic gas, \(C_V = \frac{3}{2}R\) and for a diatomic gas, \(C_V = \frac{5}{2}R\), where \(R\) is the universal gas constant. The specific heat at constant pressure \(C_P\) can be found using the relation \(C_P = C_V + R\).
Key Points
- 🎯 The law of equipartition states that energy is equally distributed among all degrees of freedom.
- 🎯 Each degree of freedom contributes \(\frac{1}{2}kT\) to the total energy.
- 🎯 Monatomic gases have 3 translational degrees of freedom, while diatomic gases have 5 (3 translational + 2 rotational).
- 🎯 Specific heats can be derived from the equipartition theorem: \(C_V = \frac{3}{2}R\) for monatomic and \(C_V = \frac{5}{2}R\) for diatomic gases.
- 🎯 The relationship between specific heats is given by \(C_P = C_V + R\).
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Examples:💡
Calculate the total energy of 1 mole of a monatomic ideal gas at 300 K.
Solution:
Step 1: Identify the number of particles in 1 mole: \(N = 6.022 \times 10^{23}\).
Step 2: Use the formula for total energy: \(E = \frac{3}{2}NkT\).
Step 3: Substitute values: \(E = \frac{3}{2} \times (6.022 \times 10^{23}) \times (1.38 \times 10^{-23}) \times 300\).
Step 4: Calculate \(E\): \(E \approx 6.022 \times 10^{-3} \times 300 \approx 540.6 J\).
Determine the specific heat at constant volume for a diatomic ideal gas.
Solution:
Step 1: Identify the degrees of freedom for a diatomic gas: 5.
Step 2: Use the equipartition theorem: \(C_V = \frac{5}{2}R\).
Step 3: Substitute \(R = 8.314 J/(mol K)\): \(C_V = \frac{5}{2} \times 8.314 \approx 20.79 J/(mol K)\).
Common Mistakes
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Mistake: Confusing the number of degrees of freedom for different types of gases.
Correction: Remember that monatomic gases have 3 degrees of freedom, while diatomic gases have 5.
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Mistake: Incorrectly applying the equipartition theorem to non-ideal gases.
Correction: Ensure that the gas behaves ideally and is at thermal equilibrium before applying the theorem.
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Mistake: Forgetting the relationship between \(C_P\) and \(C_V\).
Correction: Always use the relation \(C_P = C_V + R\) to find the specific heat at constant pressure.