Question

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 :

• A. 6
• B. 3
• C. 4
• D. 2

Answer 1

The Correct Option is C

To determine the required number of moles \(N\) of a polyatomic gas with \(f=6\) degrees of freedom that must be combined with 2 moles of a monoatomic gas (with \(f=3\)) to achieve a mixture behaving as a diatomic gas (with \(f=5\)), we utilize the formula for the degrees of freedom of a gas mixture:

\(f_{\text{mix}} = \frac{N_1 \cdot f_1 + N_2 \cdot f_2}{N_1 + N_2}\)

Where:

• \(N_1\) and \(f_1\) represent the moles and degrees of freedom of the first gas (polyatomic).
• \(N_2\) and \(f_2\) represent the moles and degrees of freedom of the second gas (monoatomic).

Given:

• Polyatomic gas: \(f_1 = 6\), \(N_1 = N\)
• Monoatomic gas: \(f_2 = 3\), \(N_2 = 2\)
• Target mixture degrees of freedom: \(f_{\text{mix}} = 5\)

Substituting these values into the formula yields:

\(5 = \frac{N \cdot 6 + 2 \cdot 3}{N + 2}\)

Solving for \(N\):

\(5(N + 2) = 6N + 6\)

Expanding the equation:

\(5N + 10 = 6N + 6\)

Rearranging terms to isolate \(N\):

\(10 - 6 = 6N - 5N\)

\(4 = N\)

Therefore, \(N = 4\). The required number of moles of the polyatomic gas is 4.