In an adiabatic process, volume is doubled then find the ratio of final average relaxation time & initial relaxation time. Given `C_p/C_v = gamma`

Options

A. `1/2`
B. `2`
C. `(1/2)^gamma`
D. `(2)^(gamma+1)`/2

Explanation

To solve the problem, we need to find the ratio of the final average relaxation time (\( \tau_f \)) to the initial relaxation time (\( \tau_i \)) in an adiabatic process where the volume is doubled. We are given that \( \frac{C_p}{C_v} = \gamma \).

### Step-by-Step Solution:

1. **Understanding Relaxation Time**:
The relaxation time (\( \tau \)) is directly proportional to the volume (\( V \)) divided by the square root of the temperature (\( T \)):
\[ \tau \propto \frac{V}{\sqrt{T}} \]

2. **Adiabatic Process Relation**:
In an adiabatic process, the relationship between pressure (\( P \)), volume (\( V \)), and temperature (\( T \)) can be expressed as:
\[ P V^\gamma = \text{constant} \]
From the ideal gas law, we also know that:
\[ PV = nRT \implies T = \frac{PV}{nR} \]

3. **Expressing Temperature in Terms of Volume**:
For an adiabatic process, we can derive that:
\[ T \propto \frac{1}{V^{\gamma - 1}} \]
This means that temperature is inversely proportional to the volume raised to the power of \( \gamma - 1 \).

4. **Substituting Temperature into Relaxation Time**:
We can substitute this expression for temperature into the equation for relaxation time:
\[ \tau \propto \frac{V}{\sqrt{T}} \propto \frac{V}{\sqrt{\frac{1}{V^{\gamma - 1}}}} = V^{\frac{\gamma - 1}{2}} V = V^{\frac{\gamma + 1}{2}} \]

5. **Finding Initial and Final Relaxation Times**:
Let \( V_i \) be the initial volume and \( V_f = 2V_i \) be the final volume (since the volume is doubled). Thus:
\[ \tau_i \propto V_i^{\frac{\gamma + 1}{2}} \]
\[ \tau_f \propto (2V_i)^{\frac{\gamma + 1}{2}} = 2^{\frac{\gamma + 1}{2}} V_i^{\frac{\gamma + 1}{2}} \]

6. **Calculating the Ratio**:
Now, we can find the ratio of the final relaxation time to the initial relaxation time:
\[ \frac{\tau_f}{\tau_i} = \frac{2^{\frac{\gamma + 1}{2}} V_i^{\frac{\gamma + 1}{2}}}{V_i^{\frac{\gamma + 1}{2}}} = 2^{\frac{\gamma + 1}{2}} \]

### Final Result:
The ratio of the final average relaxation time to the initial relaxation time is:
\[ \frac{\tau_f}{\tau_i} = 2^{\frac{\gamma + 1}{2}} \]