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

Step 1: Understanding the Question:
We are mixing two quantities of the same ideal gas contained in identical boxes of volume \(V\).
One box is emptied into the other, keeping the volume of the mixture constant at \(V\).
We must find the new mean free path and the final temperature of the gas.

Step 2: Key Formula or Approach:

• Mean free path formula:
\[ \lambda = \frac{1}{\sqrt{2} \pi d^2 n} \implies \lambda \propto \frac{1}{n} \]

• Conservation of internal energy for ideal gases:
\[ U_{\text{final}} = U_1 + U_2 \implies N_{\text{final}} T_{\text{final}} = N_1 T_1 + N_2 T_2 \]

Step 3: Detailed Explanation:

• Let the volume of each box be \(V\).

• The total number of particles in the final mixture is:
\[ N = N_1 + N_2 \implies n V = n_1 V + n_2 V \implies n = n_1 + n_2 \]

• Since \(\lambda \propto \frac{1}{n}\), we can write \(n = \frac{C}{\lambda}\) for some constant \(C\).:
\[ \frac{C}{\lambda} = \frac{C}{\lambda_1} + \frac{C}{\lambda_2} \implies \frac{1}{\lambda} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2} \] \[ \lambda = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2} \]
• Now, we apply conservation of internal energy to find the final temperature:
\[ U_{\text{final}} = U_1 + U_2 \] \[ \frac{f}{2} N k_B T = \frac{f}{2} N_1 k_B T_1 + \frac{f}{2} N_2 k_B T_2 \] \[ N T = N_1 T_1 + N_2 T_2 \]
• Substitute \(N_1 = n_1 V\), \(N_2 = n_2 V\), and \(N = (n_1 + n_2) V\).:
\[ (n_1 + n_2) V T = (n_1 T_1 + n_2 T_2) V \] \[ T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2} \]

Step 4: Final Answer:
The mean free path is \(\lambda = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}\) and the temperature is \(T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}\).