Question

Two perfect monoatomic gases at temperatures \(300\) K and \(410\) K are mixed without any loss of heat. If \(10^{24}\) and \(10^{23}\) are the number of molecules in the respective gases, then the temperature of the mixture is

• A. $340$ K
• B. $310$ K
• C. $360$ K
• D. $350$ K
• E. $370$ K

Hint:

For mixing gases: - Use weighted average of temperatures - Weight depends on number of moles or molecules

Answer 1

The Correct Option is B

Concept: For mixing of gases without heat loss: \[ T = \frac{N_1 T_1 + N_2 T_2}{N_1 + N_2} \] Since both are monoatomic gases, specific heats are same.

Step 1: Substitute given values.
\[ T = \frac{(10^{24} \times 300) + (10^{23} \times 410)}{10^{24} + 10^{23}} \]

Step 2: Factor out $10^{23}$.
\[ T = \frac{10^{23}(10 \times 300 + 410)}{10^{23}(10 + 1)} \]

Step 3: Simplify.
\[ T = \frac{3000 + 410}{11} = \frac{3410}{11} = 310\ \text{K} \]