A gas expands adiabatically and its volume doubles while its absolute temperature drops 1.32 times. What number of degrees of freedom do the gas molecules have ?

Answer

To find the number of degrees of freedom of the gas molecules given that a gas expands adiabatically, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Adiabatic Process**: In an adiabatic process, the relationship between temperature (T) and volume (V) is given by the formula:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
where \( \gamma \) is the ratio of specific heats (C_p/C_v).

2. **Identify Initial and Final Conditions**:
- Initial volume \( V_1 = V \)
- Final volume \( V_2 = 2V \)
- Initial temperature \( T_1 = T \)
- Final temperature \( T_2 = \frac{T}{1.32} \)

3. **Substitute Values into the Adiabatic Equation**:
Substitute the known values into the adiabatic equation:
\[ T \cdot V^{\gamma - 1} = \left(\frac{T}{1.32}\right) \cdot (2V)^{\gamma - 1} \]

4. **Simplify the Equation**:
Cancel \( T \) from both sides:
\[ V^{\gamma - 1} = \frac{(2V)^{\gamma - 1}}{1.32} \]
This simplifies to:
\[ 1 = \frac{2^{\gamma - 1}}{1.32} \]

5. **Rearranging to Find \( \gamma \)**:
Rearranging gives:
\[ 1.32 = 2^{\gamma - 1} \]
Taking the logarithm of both sides:
\[ \ln(1.32) = (\gamma - 1) \ln(2) \]

6. **Solve for \( \gamma \)**:
Rearranging gives:
\[ \gamma - 1 = \frac{\ln(1.32)}{\ln(2)} \]
Thus,
\[ \gamma = 1 + \frac{\ln(1.32)}{\ln(2)} \]

7. **Calculate \( \gamma \)**:
Using a calculator:
- \( \ln(1.32) \approx 0.276 \)
- \( \ln(2) \approx 0.693 \)
Therefore:
\[ \gamma = 1 + \frac{0.276}{0.693} \approx 1 + 0.398 \approx 1.398 \approx 1.4 \]

8. **Relate \( \gamma \) to Degrees of Freedom**:
The relationship between \( \gamma \) and degrees of freedom \( F \) is given by:
\[ \gamma = 1 + \frac{2}{F} \]
Setting \( \gamma = 1.4 \):
\[ 1.4 = 1 + \frac{2}{F} \]
Rearranging gives:
\[ 0.4 = \frac{2}{F} \implies F = \frac{2}{0.4} = 5 \]

9. **Conclusion**:
The number of degrees of freedom of the gas molecules is \( F = 5 \).