Three perfect gases at absolute temperatures T1, T2 and T3 are mixed. The masses of molecules are m1, m2 and m3 and the number of molecules are n1, n2 and n3 respectively. Assuming no loss of energy, the final temperature of the mixture is :

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Three perfect gases at absolute temperatures T₁, T₂ and T₃ are mixed. The masses of molecules are m₁, m₂ and m₃ and the number of molecules are n₁, n₂ and n₃ respectively. Assuming no loss of energy, the final temperature of the mixture is :

$\frac{n_{1}^{2} T_{1}^{2} + n_{2}^{2} T_{2}^{2} + n_{3}^{2} T_{3}^{2}}{n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3}}$

$\frac{(T_{1} + T_{2} + T_{3})}{3}$

$\frac{n_{1} T_{1}^{2} + n_{2} T_{2}^{2} + n_{3} T_{3}^{2}}{n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3}}$

$\frac{n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3}}{n_{1} + n_{2} + n_{3}}$

$T = \frac{n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3}}{n_{1} + n_{2} + n_{3}}$

When three perfect gases are mixed with no energy loss, the final temperature of the mixture is determined by the conservation of internal energy. For an ideal gas, the internal energy depends only on the temperature and the number of molecules, not on the mass of the molecules, because the internal energy for a monatomic gas is given by $U = \frac{3}{2} n k T$ , where $n$ is the number of molecules, $k$ is Boltzmann's constant, and $T$ is the absolute temperature. This formula assumes the gases are monatomic, but the principle of energy conservation holds generally.

Step 1: Write the total initial internal energy.
The initial internal energy of the mixture is the sum of the internal energies of each gas: $U_{initial} = \frac{3}{2} k (n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3})$ .

Step 2: Write the final internal energy after mixing.
After mixing, the total number of molecules is $n_{total} = n_{1} + n_{2} + n_{3}$ . Let the final temperature be $T_{final}$ . The final internal energy is: $U_{final} = \frac{3}{2} k (n_{1} + n_{2} + n_{3}) T_{final}$ .

Step 3: Apply conservation of energy.
Since there is no loss of energy, $U_{initial} = U_{final}$ . $\frac{3}{2} k (n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3}) = \frac{3}{2} k (n_{1} + n_{2} + n_{3}) T_{final}$ .

Step 4: Solve for the final temperature.
Cancel the common factors $\frac{3}{2} k$ from both sides: $n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3} = (n_{1} + n_{2} + n_{3}) T_{final}$ .
Therefore, the final temperature is: $T_{final} = \frac{n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3}}{n_{1} + n_{2} + n_{3}}$ .

Final Answer: The correct option is $\frac{n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3}}{n_{1} + n_{2} + n_{3}}$ .

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