A closed box contains an ideal gas if temperature of gas increased which

Options

A. mean free path remain same
B. mean free path decreases
C. relaxation time decreases
D. relaxation time remain same

Answer

To solve the question regarding the behavior of an ideal gas in a closed box when the temperature increases, we can follow these steps:

### Step 1: Understand the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
where:
- \( P \) = pressure of the gas
- \( V \) = volume of the gas
- \( n \) = number of moles of the gas
- \( R \) = universal gas constant
- \( T \) = temperature of the gas (in Kelvin)

### Step 2: Analyze the Conditions
Since the box is closed, the volume \( V \) of the gas remains constant. Therefore, if the temperature \( T \) of the gas increases, we can analyze how this affects the pressure \( P \).

### Step 3: Relate Pressure and Temperature
From the ideal gas law, if \( V \) is constant, we can rearrange the equation to show the relationship between pressure and temperature:
\[ P = \frac{nRT}{V} \]
As \( T \) increases, \( P \) must also increase to maintain the equality, given that \( n \) and \( V \) are constant.

### Step 4: Mean Free Path
The mean free path \( \lambda \) is defined as the average distance a molecule travels between collisions. It is given by:
\[ \lambda \propto \frac{T}{P} \]
Since we established that \( P \) increases with \( T \) (and \( V \) is constant), the mean free path \( \lambda \) remains constant.

### Step 5: Relaxation Time
The relaxation time \( \tau \) is related to the mean free path and the root mean square velocity \( v_{rms} \) of the gas molecules:
\[ \tau = \frac{\lambda}{v_{rms}} \]
The root mean square velocity is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( M \) is the molar mass of the gas. As \( T \) increases, \( v_{rms} \) increases, while \( \lambda \) remains constant. Therefore, since \( \tau \) is inversely proportional to \( v_{rms} \), the relaxation time \( \tau \) will decrease as the temperature increases.

### Conclusion
1. The pressure of the gas increases as the temperature increases.
2. The mean free path of the gas remains constant.
3. The relaxation time decreases as the temperature increases.

### Final Answer
- Mean free path remains constant.
- Relaxation time decreases.