The difference between ΔH and ΔU is ΔH - ΔU, when the combustion of one mole of heptane l is carried out at a temperature T, is equal to:

We start with the thermodynamic relation that connects the enthalpy change $$\Delta H$$ with the internal energy change $$\Delta U$$ for a chemical reaction carried out at temperature $$T$$ and pressure $$P$$.

The formula is stated as

$$\Delta H = \Delta U + \Delta n_g\,R\,T$$

where $$\Delta n_g$$ is the algebraic change in the number of moles of gaseous species between products and reactants, $$R$$ is the universal gas constant and $$T$$ is the absolute temperature. Rearranging, we obtain

$$\Delta H - \Delta U = \Delta n_g\,R\,T$$

So, to evaluate $$\Delta H - \Delta U$$ for the combustion of one mole of liquid heptane, $$\mathrm{C_7H_{16}(l)}$$, we must first write and balance the chemical equation.

The complete combustion equation is

$$\mathrm{C_7H_{16}(l)} \;+\; 11\;\mathrm{O_2(g)} \;\longrightarrow\; 7\;\mathrm{CO_2(g)} \;+\; 8\;\mathrm{H_2O(l)}$$

Now we calculate $$\Delta n_g$$, remembering that only gaseous species are counted:

Number of moles of gaseous products  =  $$7$$ (from $$7\;\mathrm{CO_2(g)}$$)

Number of moles of gaseous reactants  =  $$11$$ (from $$11\;\mathrm{O_2(g)}$$)

Hence,

$$\Delta n_g = n_{\text{products (g)}} - n_{\text{reactants (g)}} = 7 - 11 = -4$$

Substituting this value of $$\Delta n_g$$ into the rearranged relation, we get

$$\Delta H - \Delta U = (-4)\,R\,T = -4RT$$

This result exactly matches the expression given in Option A.

Hence, the correct answer is Option A.