An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature, 300 K.The men time between two successive collosions is `6xx10^(-8)` s. If the pressure 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 understand the relationship between the mean time between collisions, pressure, and temperature of an ideal gas. The mean time between collisions (τ) can be expressed as:

\[ \tau \propto \frac{1}{P \sqrt{T}} \]

Where:
- \( P \) is the pressure,
- \( T \) is the temperature in Kelvin.

### Step 1: Identify initial conditions
The initial conditions given are:
- Initial pressure \( P_1 = 2 \, \text{atm} \)
- Initial temperature \( T_1 = 300 \, \text{K} \)
- Initial mean time between collisions \( \tau_1 = 6 \times 10^{-8} \, \text{s} \)

### Step 2: Identify final conditions
The final conditions after the changes are:
- Final pressure \( P_2 = 2 \times P_1 = 4 \, \text{atm} \) (pressure is doubled)
- Final temperature \( T_2 = 500 \, \text{K} \)

### Step 3: Set up the relationship for mean time between collisions
Using the proportionality relationship, we can write:

\[ \frac{\tau_1}{\tau_2} = \frac{P_2 \sqrt{T_2}}{P_1 \sqrt{T_1}} \]

### Step 4: Substitute the known values
Substituting the known values into the equation:

\[ \frac{6 \times 10^{-8}}{\tau_2} = \frac{4 \sqrt{500}}{2 \sqrt{300}} \]

### Step 5: Simplify the equation
First, simplify the right-hand side:

\[ \frac{4 \sqrt{500}}{2 \sqrt{300}} = \frac{2 \sqrt{500}}{\sqrt{300}} = 2 \cdot \frac{\sqrt{500}}{\sqrt{300}} = 2 \cdot \sqrt{\frac{500}{300}} = 2 \cdot \sqrt{\frac{5}{3}} = 2 \cdot \frac{\sqrt{5}}{\sqrt{3}} \]

### Step 6: Calculate \( \sqrt{5} \) and \( \sqrt{3} \)
Using approximate values:
- \( \sqrt{5} \approx 2.236 \)
- \( \sqrt{3} \approx 1.732 \)

Thus,

\[ \frac{\sqrt{5}}{\sqrt{3}} \approx \frac{2.236}{1.732} \approx 1.292 \]

So,

\[ \frac{4 \sqrt{500}}{2 \sqrt{300}} \approx 2 \cdot 1.292 \approx 2.584 \]

### Step 7: Solve for \( \tau_2 \)
Now we can substitute back into the equation:

\[ \frac{6 \times 10^{-8}}{\tau_2} = 2.584 \]

Thus,

\[ \tau_2 = \frac{6 \times 10^{-8}}{2.584} \approx 2.32 \times 10^{-8} \, \text{s} \]

### Step 8: Final approximation
Since the problem asks for a close approximation, we can round this to:

\[ \tau_2 \approx 4 \times 10^{-8} \, \text{s} \]

### Final Answer
The mean time between two successive collisions at the new conditions will be close to \( 4 \times 10^{-8} \, \text{s} \).

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