A cubical vessel of side 15cm contains oxygen gas at `27^(@)C`. How long does it take a typical molecule to cross the container?

Explanation

To solve the problem of how long it takes a typical molecule to cross a cubical vessel containing oxygen gas, we can follow these steps:

### Step 1: Convert the temperature to Kelvin
The temperature given is 27°C. To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
\[ T = 27 + 273 = 300 \text{ K} \]

### Step 2: Convert the side length of the cubical vessel to meters
The side length of the cubical vessel is given as 15 cm. To convert this to meters:
\[ A = 15 \text{ cm} = 0.15 \text{ m} \]

### Step 3: Calculate the RMS speed of the oxygen molecules
The RMS speed (\(V_{RMS}\)) can be calculated using the formula:
\[ V_{RMS} = \sqrt{\frac{3RT}{M}} \]
Where:
- \(R = 8.31 \, \text{J/(mol K)}\) (universal gas constant)
- \(T = 300 \, \text{K}\)
- \(M = 0.032 \, \text{kg/mol}\) (molar mass of oxygen)

Substituting the values:
\[ V_{RMS} = \sqrt{\frac{3 \times 8.31 \times 300}{0.032}} \]
Calculating the numerator:
\[ 3 \times 8.31 \times 300 = 7479 \]
Now, substituting this back into the equation:
\[ V_{RMS} = \sqrt{\frac{7479}{0.032}} = \sqrt{233718.75} \approx 483.44 \, \text{m/s} \]

### Step 4: Calculate the time taken for a molecule to cross the container
The time taken (\(t\)) for a molecule to cross the container can be calculated using the formula:
\[ t = \frac{\text{Distance}}{\text{Speed}} \]
Here, the distance is the side length of the cube, and the speed is the RMS speed we just calculated:
\[ t = \frac{0.15 \, \text{m}}{483.44 \, \text{m/s}} \approx 0.00031 \, \text{s} \]
This can be expressed in scientific notation:
\[ t \approx 3.1 \times 10^{-4} \, \text{s} \]

### Final Answer
The time taken for a typical molecule to cross the container is approximately \(3.1 \times 10^{-4} \, \text{s}\).

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