Question

If $\alpha$ moles of a monoatomic gas are mixed with $\beta$ moles of a polyatomic gas and mixture behaves like diatomic gas, then [ neglect the vibrational mode of freedom ]

• A. $2\alpha =\beta$
• B. $\alpha =2\beta$
• C. $\alpha =-3\beta$
• D. $3\alpha =-\beta$

Answer 1

The correct option is A $2\alpha =\beta$

Given: Moles of monoatomic gas = $\alpha$, moles of diatomic gas is $\beta$.

Solution:

The mixture is behaving as diatomic gas. If we neglect the vibrational mode of freedom then degree of freedom ($f_{mix}$) of mixture is 3 (for translational)+2(for rotational)=5.

We know that,

$f_{mix}=\frac{f_1{n}_1+f_2{n}_2}{n_1+n_2}$

$\Rightarrow 5=\frac{3\alpha +6\beta }{\alpha +\beta }$

$\Rightarrow 5\alpha +5\beta =3\alpha +6\beta$

$\therefore 2\alpha =\beta$

Hence the correct option is A.