N moles of a polyatomic gas (f=6) must be mixed with two moles of a monoatomic gas so that the mixture behaves as a diatomic gas. The value of N is :

Solution

Physics sir · 4 steps

Treat the mixture's degrees of freedom as a mole-weighted average and set that average equal to 5.

1. Average degrees of freedom. Each mole carries its own f. So the mixture behaves as if every molecule had the mole-weighted average fₘᵢₓ. A diatomic gas has f = 5, a monoatomic gas has f = 3, and the polyatomic gas here has f = 6.

   fₘᵢₓ = (n₁ f₁ + n₂ f₂)/(n₁ + n₂) = (6N + 2 × 3)/(N + 2)

2. Set it to 5. (6N + 6)/(N + 2) = 5 gives 6N + 6 = 5N + 10, so N = 4.
3. Check. (6 × 4 + 6)/6 = 30/6 = 5. The same result comes from C_V: (4 × 3R + 2 × 3/2R)/6 = 5/2R, which is C_V for a diatomic gas.
4. So. 4 moles of the polyatomic gas are needed, which is option (c).

The idea to keep. For any gas mixture, average f or C_V by weighting with moles, not by a plain average. Then match the result to the gas type you want.

Answer from the NTA final answer key. Solution by PrepWiser.