1. **State the problem:** We are given two hydrocarbon fuel families, Paraffins with formula $C_nH_{2n+2}$ and Cycloalkanes with formula $C_nH_{2n}$. 2. The relation between their equivalence ratios $\Phi$ is given by: $$\Phi_{C_nH_{2n+2}} = \Phi_{C_nH_{2n}} + \frac{a}{n}$$ where $a$ is a constant and $n$ the number of carbon atoms. 3. Both fuels have the same number of carbon atoms $n$ and use the same actual moles of air for combustion. 4. The equivalence ratio $\Phi$ relates the actual fuel-to-air ratio to the stoichiometric fuel-to-air ratio: $$\Phi = \frac{(\text{Fuel/Air})_\text{actual}}{(\text{Fuel/Air})_\text{stoich}}$$ 5. Since the actual moles of air are the same, and the number of carbons $n$ is fixed, the difference in equivalence ratio arises from difference in stoichiometric air requirements. 6. Stoichiometric combustion for Paraffin $C_nH_{2n+2}$ requires: $$C_nH_{2n+2} + \left( n + \frac{n+1}{2} \right)O_2 \to nCO_2 + \left(n+1\right) H_2O$$ So, moles of $O_2$ required = $$\frac{3n+1}{2}$$ 7. Stoichiometric combustion for Cycloalkane $C_nH_{2n}$ requires: $$C_nH_{2n} + \frac{3n}{2} O_2 \to n CO_2 + n H_2O$$ So moles of $O_2$ required = $$\frac{3n}{2}$$ 8. Since equivalence ratio is inversely proportional to stoichiometric air, the Paraffin has a higher $\Phi$ by amount proportional to $$\frac{1}{n}$$, matching the given formula. **Final conclusion:** The difference in equivalence ratio between Paraffin and Cycloalkane fuels with the same carbon number $n$ and same actual air amount is: $$\boxed{\Phi_{C_nH_{2n+2}} = \Phi_{C_nH_{2n}} + \frac{a}{n}}$$ where $a$ represents the difference in stoichiometric air coefficients per carbon atom, confirming the given relation.