Question

What is \( C_v \) for a monoatomic gas?

Hint:

For a monoatomic ideal gas, \( C_v = \frac{3}{2} R \), which is derived from the translational degrees of freedom.

Answer 1

Step 1: Understanding the heat capacities.
For a monoatomic ideal gas, the heat capacity at constant volume \( C_v \) is related to the ideal gas constant \( R \). In the case of a monoatomic gas, the degrees of freedom are three translational degrees of freedom, and the heat capacity is derived using the equipartition theorem.
Step 2: Formula for \( C_v \).
For a monoatomic ideal gas, the formula for \( C_v \) is: \[ C_v = \frac{3}{2} R \] where \( R \) is the universal gas constant.
Step 3: Conclusion.
Thus, the heat capacity at constant volume for a monoatomic gas is \( \frac{3}{2} R \).

Answer 2

Step 1: A monoatomic gas molecule can only move in three directions (translational motion) — it has 3 degrees of freedom.

Step 2: By the equipartition theorem, each degree of freedom contributes (1/2)R to the molar heat capacity.

Step 3: So Cᵥ = 3 × (1/2)R = (3/2)R.

Answer 3

Via Mayer's relation and the adiabatic index.
For an ideal gas, \( C_p - C_v = R \), and the adiabatic index is defined as \( \gamma = C_p/C_v \).
For a monoatomic gas, \( \gamma = 5/3 \) is a known standard value. Substituting \( C_p = \gamma C_v \) into Mayer's relation, \[ \gamma C_v - C_v = R \implies C_v(\gamma - 1) = R \]
\[ C_v = \frac{R}{\gamma - 1} = \frac{R}{\tfrac{5}{3} - 1} = \frac{R}{\tfrac{2}{3}} = \frac{3}{2}R \] which matches the equipartition-theorem result.