Three ideal gases are combined, with initial absolute temperatures T1, T2, and T3. Their molecular masses are m1, m2, and m3, and the corresponding numbers of molecules are n1, n2, and n3. If no energy is lost during mixing, the equilibrium temperature of the resulting mixture is

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Accepted Answer

(n1T1 + n2T2 + n3T3)/(n1 + n2 + n3). Internal energy of an ideal gas is (f/2) n k T, depending on molecule count n, not mass. Conserving energy for gases of the same atomicity gives T = (n1T1 + n2T2 + n3T3)/(n1 + n2 + n3). The masses are distractors.

Three ideal gases are combined, with initial absolute temperatures T1, T2, and T3. Their molecular masses are m1, m2, and m3, and the corresponding numbers of molecules are n1, n2, and n3. If no energy is lost during mixing, the equilibrium temperature of the resulting mixture is

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