If `gamma` be the ratio of specific heats `(C_(p) & C_(v))` for a perfect gas. Find the number of degrees of freedom of a molecules of the gas?

Explanation

To solve the problem of finding the number of degrees of freedom of a molecule of a perfect gas given the ratio of specific heats (γ = Cp/Cv), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definitions**:
- \( C_p \): Specific heat at constant pressure.
- \( C_v \): Specific heat at constant volume.
- \( \gamma \): Ratio of specific heats, defined as \( \gamma = \frac{C_p}{C_v} \).

2. **Relate Specific Heats to Degrees of Freedom**:
- The specific heat at constant volume \( C_v \) for a perfect gas can be expressed in terms of the number of degrees of freedom \( f \):
\[ C_v = \frac{f}{2} R \]
- The specific heat at constant pressure \( C_p \) can be expressed as:
\[ C_p = C_v + R = \frac{f}{2} R + R = \left(\frac{f}{2} + 1\right) R = \frac{f + 2}{2} R \]

3. **Substitute into the Ratio**:
- Now, substituting \( C_p \) and \( C_v \) into the expression for \( \gamma \):
\[ \gamma = \frac{C_p}{C_v} = \frac{\frac{f + 2}{2} R}{\frac{f}{2} R} \]
- The \( R \) cancels out:
\[ \gamma = \frac{f + 2}{f} \]

4. **Rearranging the Equation**:
- Rearranging the equation gives:
\[ \gamma = 1 + \frac{2}{f} \]
- From this, we can isolate \( f \):
\[ \frac{2}{f} = \gamma - 1 \]
- Therefore, solving for \( f \):
\[ f = \frac{2}{\gamma - 1} \]

5. **Conclusion**:
- The number of degrees of freedom \( f \) of a molecule of the gas is given by:
\[ f = \frac{2}{\gamma - 1} \]