Question

Two identical boxes contain the same ideal gas. Let \((n_1, \lambda_1, T_1)\) and \((n_2, \lambda_2, T_2)\) be the number density, mean free path and temperature of the gas in the first and the second box, respectively. One of the boxes is emptied into the other one. What will be the mean free path \(\lambda\) and temperature \(T\) of the gas now?

• A. \(\lambda = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}\), \(T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}\)
• B. \(\lambda = \frac{n_1 \lambda_1 + n_2 \lambda_2}{n_1 + n_2}\), \(T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}\)
• C. \(\lambda = \frac{n_1 \lambda_1 + n_2 \lambda_2}{n_1 + n_2}\), \(T = \sqrt{T_1 T_2}\)
• D. \(\lambda = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}\), \(T = \sqrt{T_1 T_2}\)

Hint:

Mean free path \(\lambda\) is inversely proportional to the number density \(n\).
Since the final number density is the sum of the initial densities (\(n = n_1 + n_2\)), the final mean free path behaves like parallel resistors: \(\lambda^{-1} = \lambda_1^{-1} + \lambda_2^{-1}\).

Answer 1

The Correct Option is A

1. Consider two identical boxes containing an ideal gas with parameters \((n_1, \lambda_1, T_1)\) for the first box and \((n_2, \lambda_2, T_2)\) for the second box. Here, \(n\) is the number density, \(\lambda\) is the mean free path, and \(T\) is the temperature.
2. When one box is emptied into the other, the total number density of the gas, \(n\), in the combined box becomes: \(n = n_1 + n_2\).
3. The mean free path for a gas is inversely proportional to the number density. Thus, the combined mean free path \(\lambda\) is given by the harmonic mean of the individual mean free paths due to the addition of densities: \(\lambda = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}\).
4. The temperature of the ideal gas, when mixed, follows Dalton's law of partial pressures. The final temperature, \(T\), when both gases are added together in the same volume, is the weighted average based on their number densities: \(T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}\).
5. This leads to the correct choice, which states:
   • \(\lambda = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}\)
   • \(T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}\)
6. Therefore, the correct option is:

\(\lambda = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}\), \(T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}\)