If the volume of a gas is doubled either isothermally or adiabatically, in which case change in mean free path is more.

Options

A. (a)Adiabatic
B. (b)Isothermal
C. (c)Same in both
D. (d)No change

Explanation

To solve the problem of determining in which case the change in mean free path is more when the volume of a gas is doubled either isothermally or adiabatically, we can follow these steps:

### Step 1: Understand Mean Free Path
The mean free path (λ) of a gas is given by the formula:
\[ \lambda = \frac{V}{2 \pi d^2 n} \]
where:
- \(V\) is the volume of the gas,
- \(d\) is the diameter of the gas molecules,
- \(n\) is the number density of the gas molecules (number of molecules per unit volume).

### Step 2: Analyze the Conditions
1. **Isothermal Process**: In an isothermal process, the temperature of the gas remains constant. When the volume is doubled, the number density \(n\) (which is the number of molecules per unit volume) will decrease because the number of molecules remains constant while the volume increases.
- If the volume doubles, \(V' = 2V\).
- The new number density \(n' = \frac{n}{2}\) (since \(n = \frac{N}{V}\) where \(N\) is the number of molecules).

2. **Adiabatic Process**: In an adiabatic process, the gas expands without heat exchange. When the volume is doubled, the temperature of the gas decreases.
- The number of molecules remains constant, so the number density will also decrease similarly to the isothermal case.

### Step 3: Calculate Mean Free Path for Both Processes
1. **Isothermal Case**:
\[ \lambda' = \frac{V'}{2 \pi d^2 n'} = \frac{2V}{2 \pi d^2 \frac{n}{2}} = \frac{2V}{\pi d^2 n} = 2 \lambda \]

2. **Adiabatic Case**:
\[ \lambda' = \frac{V'}{2 \pi d^2 n'} = \frac{2V}{2 \pi d^2 \frac{n}{2}} = \frac{2V}{\pi d^2 n} = 2 \lambda \]

### Step 4: Conclusion
In both the isothermal and adiabatic processes, the mean free path doubles when the volume is doubled. Therefore, the change in mean free path is the same in both cases.

### Final Answer
Both processes result in the same change in mean free path. Thus, the correct option is **D**.

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