1. Given mixture

• Monoatomic gas: $1$ mole
• Diatomic gas: $3$ moles
• Total moles in mixture: $n=1+3=4$

2. Molar specific heat at constant volume of each gas

• For a monoatomic ideal gas: $C_{V,\,mono}=\frac{3}{2}R$

• For a diatomic ideal gas (no vibrational mode): $C_{V,\,dia}=\frac{5}{2}R$

3. Total heat capacity of the mixture at constant volume

Since heat capacities add: $C_{V,\,total}=1\left(\frac{3}{2}R\right)+3\left(\frac{5}{2}R\right)$

$=\frac{3}{2}R+\frac{15}{2}R=\frac{18}{2}R=9R$

4. Molecular (molar) specific heat of the mixture

The molar specific heat of the mixture is total heat capacity divided by total moles:

$C_{V,\,mix}=\frac{9R}{4}$

Given that $C_{V,\,mix}=\frac{\alpha^2}{4}R$

So, $\frac{\alpha^2}{4}R=\frac{9}{4}R$

Cancel $\frac{R}{4}$ from both sides: $\alpha^2=9$

Thus, $\alpha=3$

(We take the positive value since the question asks for the value of $\alpha$.)

5. Comparison with stored answer

Derived answer: $3$

Stored correct answer: $3$

They match.