To solve this problem, we need to determine the number of moles \(N\) of a polyatomic gas with degrees of freedom \(f = 6\) that must be mixed with two moles of a monoatomic gas (which has degrees of freedom \(f = 3\)) so that the mixture behaves like a diatomic gas (which has degrees of freedom \(f = 5\)).
The degrees of freedom \(f\) for a mixture of gases can be determined using the formula:
\(f_{\text{mix}} = \frac{N_1 \cdot f_1 + N_2 \cdot f_2}{N_1 + N_2}\)
where:
We are given:
Substituting these values into the formula:
\(5 = \frac{N \cdot 6 + 2 \cdot 3}{N + 2}\)
We solve for \(N\):
\(5(N + 2) = 6N + 6\)
Expanding both sides:
\(5N + 10 = 6N + 6\)
Rearranging the equation:
\(5N + 10 - 6 = 6N\)
\(5N + 4 = 6N\)
Simplify to solve for \(N\):
\(4 = 6N - 5N\)
\(N = 4\)
Thus, the value of \(N\) that satisfies this condition is 4. Therefore, the correct answer is 4.
The average degrees of freedom for the mixture, \( f_{\text{eq}} \), can be expressed as:
\(f_{\text{eq}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}\)
where:
- \( n_1 = N \) (number of moles of polyatomic gas),
- \( n_2 = 2 \) (number of moles of monoatomic gas),
- \( f_1 = 6 \) (degrees of freedom for polyatomic gas),
- \( f_2 = 3 \) (degrees of freedom for monoatomic gas),
- \( f_{\text{eq}} = 5 \) (degrees of freedom for diatomic gas).
Now, we substitute these values into the equation:
\(5 = \frac{N \times 6 + 2 \times 3}{N + 2}\)
Simplify the equation:
\(5 = \frac{6N + 6}{N + 2}\)
Multiply both sides by \( (N + 2) \):
\(5(N + 2) = 6N + 6\)
\(5N + 10 = 6N + 6\)
\(10 - 6 = 6N - 5N\)
\(N = 4\)
Thus, the value of \( N \) is 4.
The Correct Answer is: 4
| List-I (Details of the processes of the cycle) | List-II (Name of the cycle) |
|---|---|
| (A) Two adiabatic, one isobaric and two isochoric | (I) Diesel |
| (B) Two adiabatic and two isochoric | (II) Carnot |
| (C) Two adiabatic, one isobaric and one isochoric | (III) Dual |
| (D) Two adiabatics and two isothermals | (IV) Otto |
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