Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increase as `V^q`, where V is the volume of the gas. The value of q is : `(gamma=(C_p)/(C_v))`

Options

A. `(gamma+1)/2`
B. `(gamma-1)/2`
C. `(3gamma+5)/6`
D. `(3gamma-5)/6`

Explanation

To find the value of \( q \) in the expression for the average time of collision between molecules during an adiabatic expansion of an ideal gas, we will follow these steps:

### Step 1: Understand the relationship between volume and time of collision
The average time of collision between gas molecules is related to the mean free path (\( \lambda \)) and the root mean square velocity (\( v_{rms} \)). The formula for the average time of collision (\( t \)) can be expressed as:
\[ t = \frac{\lambda}{v_{rms}} \]

### Step 2: Define the mean free path
The mean free path (\( \lambda \)) is given by:
\[ \lambda = \frac{1}{\sqrt{2} \pi D^2 n} \]
where \( D \) is the diameter of the gas molecules and \( n \) is the number density of the gas, defined as:
\[ n = \frac{N}{V} \]
where \( N \) is the number of molecules and \( V \) is the volume of the gas.

### Step 3: Relate number density to volume
Substituting the expression for number density into the mean free path formula gives:
\[ \lambda = \frac{V}{\sqrt{2} \pi D^2 N} \]

### Step 4: Calculate the root mean square velocity
The root mean square velocity (\( v_{rms} \)) for an ideal gas is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the gas constant, \( T \) is the temperature, and \( M \) is the molar mass of the gas.

### Step 5: Substitute and simplify the average time of collision
Now substituting \( \lambda \) and \( v_{rms} \) into the expression for \( t \):
\[ t = \frac{\lambda}{v_{rms}} = \frac{V}{\sqrt{2} \pi D^2 N} \cdot \frac{M}{\sqrt{3RT}} \]
This can be simplified to:
\[ t \propto \frac{V}{\sqrt{T}} \]

### Step 6: Relate temperature to volume during adiabatic expansion
For an adiabatic process, we know that:
\[ PV^\gamma = \text{constant} \]
From the ideal gas law, we can express \( P \) as:
\[ P = \frac{nRT}{V} \]
Substituting this into the adiabatic condition gives:
\[ \frac{nRT}{V} V^\gamma = \text{constant} \]
This leads to:
\[ T \propto \frac{1}{V^{\gamma - 1}} \]

### Step 7: Substitute back to find the relationship with volume
Substituting this relationship into the expression for \( t \):
\[ t \propto \frac{V}{\sqrt{\frac{1}{V^{\gamma - 1}}}} = V^{\frac{1}{2}} V^{\frac{\gamma - 1}{2}} = V^{\frac{\gamma + 1}{2}} \]

### Conclusion
Thus, we find that the average time of collision increases as \( V^q \), where:
\[ q = \frac{\gamma + 1}{2} \]

### Final Answer
The value of \( q \) is:
\[ q = \frac{\gamma + 1}{2} \]