Question:
Two ideal polyatomic gases at temperatures T1 and T2 are mixed so that there is no loss of energy. If F1 and F2, m1 and m2, n1 and n2 be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is :
Two ideal polyatomic gases at temperatures T1 and T2 are mixed so that there is no loss of energy. If F1 and F2, m1 and m2, n1 and n2 be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is :
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Total internal energy is conserved on mixing; each molecule has average energy (F/2)kT by equipartition.
Updated On: Nov 4, 2026
- \(\frac{n_1F_1T_1+n_2F_2T_2}{F_1+F_2}\)
- \(\frac{n_1F_1T_1+n_2F_2T_2}{n_1F_1+n_2F_2}\)
- \(\frac{n_1T_1+n_2T_2}{n_1+n_2}\)
- \(\frac{n_1F_1T_1+n_2F_2T_2}{n_1+n_2}\)
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The Correct Option is B
Solution and Explanation
Internal energy of a gas of n molecules with F degrees of freedom at temperature T is \(U=\frac{F}{2}nkT\).
No energy is lost, so the total energy before mixing equals that after mixing (final temperature T, total energy of both gases).
\(\frac{F_1}{2}n_1kT_1+\frac{F_2}{2}n_2kT_2=\frac{F_1}{2}n_1kT+\frac{F_2}{2}n_2kT\)
\(T=\frac{n_1F_1T_1+n_2F_2T_2}{n_1F_1+n_2F_2}\)
Answer: (2) \(\frac{n_1F_1T_1+n_2F_2T_2}{n_1F_1+n_2F_2}\)
No energy is lost, so the total energy before mixing equals that after mixing (final temperature T, total energy of both gases).
\(\frac{F_1}{2}n_1kT_1+\frac{F_2}{2}n_2kT_2=\frac{F_1}{2}n_1kT+\frac{F_2}{2}n_2kT\)
\(T=\frac{n_1F_1T_1+n_2F_2T_2}{n_1F_1+n_2F_2}\)
Answer: (2) \(\frac{n_1F_1T_1+n_2F_2T_2}{n_1F_1+n_2F_2}\)
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