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:
Option 1 -
(n₁T₁ + n₂T₂) / (n₁ + n₂)
Option 2 -
(n₁F₁T₁ + n₂F₂T₂) / (n₁F₁ + n₂F₂)
Option 3 -
(n₁F₁T₁ + n₂F₂T₂) / (F₁ + F₂)
Option 4 -
(n₁F₁T₁ + n₂F₂T₂) / (n₁ + n₂)
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1 Answer
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Correct Option - 2
Detailed Solution:U = U? + U? = (n? /N_A) (F? R/2)T? + (n? /N_A) (F? R/2)T?
For the mixture: U = (n? +n? )/N_A * (FR/2)T
F = (n? F? + n? F? ) / (n? + n? )
Equating the expressions for U and solving for T gives:
T = (n? F? T? + n? F? T? ) / (n? F? + n? F? )
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