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

Answer: (C)

4

Answer 2

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