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

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