$$N$$ 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:

N moles of a polyatomic gas (f = 6) mixed with 2 moles of a monoatomic gas (f = 3) results in a mixture with effective degrees of freedom given by the weighted average: $$f_{mix} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}$$. Since the mixture behaves as a diatomic gas (fₘᵢₓ = 5), substituting the values yields $$5 = \frac{N \times 6 + 2 \times 3}{N + 2}$$. Multiplying both sides by (N + 2) gives $$5(N + 2) = 6N + 6$$, which simplifies to $$5N + 10 = 6N + 6$$. Solving for N leads to $$N = 4$$.

The value of $$N$$ is 4, which corresponds to Option 3.