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application-of-law-of-equipartitio-to-heat-capacities

🚀 The behavior of perfect gases can be understood through the kinetic theory of gases, which explains how gas particles move and interact. The law of equipartition of energy states that energy is distributed equally among all degrees of freedom in a system at thermal equilibrium. This principle can be applied to understand heat capacities of gases, which are crucial in thermodynamics. The specific heat capacity of a substance is the amount of heat required to change its temperature by one degree Celsius. For ideal gases, the specific heat capacities at constant volume (C_v) and constant pressure (C_p) can be derived using the equipartition theorem.

Theory Explanation

Understanding Degrees of Freedom

In kinetic theory, each degree of freedom corresponds to a way in which a particle can store energy. For a monatomic gas, there are 3 translational degrees of freedom (movement in x, y, and z directions). For diatomic gases, there are additional rotational degrees of freedom, leading to a total of 5 degrees of freedom at room temperature. The equipartition theorem states that each degree of freedom contributes (1/2)kT to the total energy, where k is the Boltzmann constant and T is the temperature in Kelvin.

\[ E = \frac{f}{2} k T \]
Applying Equipartition to Heat Capacities

The specific heat capacity at constant volume (C_v) can be derived from the equipartition theorem. For a monatomic ideal gas, C_v = (3/2)R, where R is the universal gas constant. For diatomic gases, C_v = (5/2)R, accounting for both translational and rotational motion. The specific heat capacity at constant pressure (C_p) is related to C_v by the equation C_p = C_v + R, leading to C_p = (5/2)R for monatomic gases and C_p = (7/2)R for diatomic gases.

\[ C_p = C_v + R \]

Key Points

  • 🎯 The law of equipartition states that energy is equally distributed among all degrees of freedom.
  • 🎯 Monatomic gases have 3 translational degrees of freedom, while diatomic gases have 5 at room temperature.
  • 🎯 The specific heat capacity at constant volume (C_v) and constant pressure (C_p) can be derived from the equipartition theorem.

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Examples:💡

Calculate the specific heat capacity at constant volume (C_v) for a monatomic ideal gas.

Solution:

Step 1: Identify that for a monatomic ideal gas, C_v = (3/2)R.

\[ C_v = \frac{3}{2} R \]

Step 2: Substitute R = 8.314 J/(mol·K) to find C_v.

\[ C_v = \frac{3}{2} \times 8.314 = 12.471 J/(mol·K) \]

Determine the specific heat capacity at constant pressure (C_p) for a diatomic ideal gas.

Solution:

Step 1: For a diatomic gas, C_v = (5/2)R.

\[ C_v = \frac{5}{2} R \]

Step 2: Calculate C_p using the relation C_p = C_v + R.

\[ C_p = \frac{5}{2} R + R = \frac{7}{2} R \]

Step 3: Substitute R = 8.314 J/(mol·K) to find C_p.

\[ C_p = \frac{7}{2} \times 8.314 = 24.79 J/(mol·K) \]

Common Mistakes

  • Mistake: Confusing C_v and C_p for different types of gases.

    Correction: Always remember that C_v is for constant volume and C_p is for constant pressure. Use the correct formulas for each type of gas.

  • Mistake: Not accounting for all degrees of freedom in diatomic gases.

    Correction: Ensure to include both translational and rotational degrees of freedom when calculating C_v and C_p for diatomic gases.

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