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: (a) (n₁T₁+n₂T₂+n₃T₃)/(n₁+n₂+n₃) (b) (n₁T₁²+n₂T₂²+n₃T₃²)/(n₁T₁+n₂T₂+n₃T₃) (c) (n₁²T₁²+n₂²T₂²+n₃²T₃²)/(n₁T₁+n₂T₂+n₃T₃) (d) ((T₁+T₂+T₃))/(3)

Answer: (a). No energy lost; for the same type of gas, energy per molecule ∝ T: Σ nif/2kTi = (Σ ni)f/2kT. Molecular masses do not matter; only the number of molecules. T = (n₁T₁ + n₂T₂ + n₃T₃)/(n₁ + n₂ + n₃) So the answer is (a).

Three perfect gases at absolute temperatures T₁, T₂ and T₃ are mixed… | JEE Main 2011 (AIEEE)

Q38JEE Main 2011 · AIEEEMedium

Degrees of Freedom & Internal Energy

Three perfect gases at absolute temperatures $T_1, T_2$ and $T_3$ are mixed. The masses of molecules are $m_1, m_2$ and $m_3$ and the number of molecules are $n_1, n_2$ and $n_3$ respectively. Assuming no loss of energy, the final temperature of the mixture is:

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Answer: (a)

No energy lost; for the same type of gas, energy per molecule $\propto T$ : $\sum n_i\frac{f}{2}kT_i = \left(\sum n_i\right)\frac{f}{2}kT$ .
Molecular masses do not matter; only the number of molecules.
$T = \frac{n_1T_1 + n_2T_2 + n_3T_3}{n_1 + n_2 + n_3}$
So the answer is (a).

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