An ideal gas is enclosed in a cylinder at pressure of 2atm and temperature, 300K. The mean time between two successive collisions is `6xx10^(-8)` s. If the pressure is doubled and temperature is increased to 500K, the mean time between two successive collisions will be close to:

Options

A. `2xx10^(-7)s`
B. `4xx10^(-8)s`
C. `0.5xx10^(-8)s`
D. `3xx10^(-6)s`

Explanation

To solve the problem, we need to determine how the mean time between two successive collisions of gas molecules changes when the pressure and temperature of the gas are altered.

### Step-by-Step Solution:

1. **Understanding Mean Time Between Collisions**:
The mean time between two successive collisions (τ) is directly proportional to the volume (V) of the gas and inversely proportional to the velocity (v) of the gas molecules. Mathematically, we can express this as:
\[ \tau \propto \frac{V}{v} \]

2. **Using the Ideal Gas Law**:
From the ideal gas law, we know:
\[ PV = nRT \]
Rearranging gives:
\[ V = \frac{nRT}{P} \]
Thus, we can see that:
\[ V \propto \frac{T}{P} \]

3. **Velocity of Gas Molecules**:
The average velocity (v) of gas molecules is proportional to the square root of the temperature (T):
\[ v \propto \sqrt{T} \]

4. **Combining the Relationships**:
Substituting the expressions for volume and velocity into the mean time equation, we get:
\[ \tau \propto \frac{V}{v} \propto \frac{T/P}{\sqrt{T}} = \frac{\sqrt{T}}{P} \]

5. **Calculating the New Mean Time**:
Let the initial conditions be \( P_1 = 2 \, \text{atm} \) and \( T_1 = 300 \, \text{K} \) with an initial mean time \( \tau_1 = 6 \times 10^{-8} \, \text{s} \). The new conditions are \( P_2 = 4 \, \text{atm} \) (pressure doubled) and \( T_2 = 500 \, \text{K} \).

The ratio of the new mean time to the old mean time can be expressed as:
\[ \frac{\tau_2}{\tau_1} = \frac{\sqrt{T_2}}{P_2} \cdot \frac{P_1}{\sqrt{T_1}} \]

Substituting the values:
\[ \frac{\tau_2}{6 \times 10^{-8}} = \frac{\sqrt{500}}{4} \cdot \frac{2}{\sqrt{300}} \]

6. **Calculating the Square Roots and Final Value**:
- Calculate \( \sqrt{500} = 22.36 \)
- Calculate \( \sqrt{300} = 17.32 \)
- Now substituting these values:
\[ \frac{\tau_2}{6 \times 10^{-8}} = \frac{22.36}{4} \cdot \frac{2}{17.32} \]
- Simplifying gives:
\[ \frac{\tau_2}{6 \times 10^{-8}} = \frac{22.36 \times 2}{4 \times 17.32} \approx \frac{44.72}{69.28} \approx 0.645 \]
- Therefore:
\[ \tau_2 \approx 0.645 \times 6 \times 10^{-8} \approx 3.87 \times 10^{-8} \, \text{s} \]

7. **Final Result**:
Rounding this value gives us:
\[ \tau_2 \approx 4 \times 10^{-8} \, \text{s} \]

### Conclusion:
The mean time between two successive collisions when the pressure is doubled and the temperature is increased to 500K will be approximately \( 4 \times 10^{-8} \, \text{s} \).