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Step-by-Step Solution
Step 1: Write the Balanced Combustion Reaction
The combustion of heptane (C7H16) can be represented as:
$ C_{7}H_{16}(l) + 11\,O_{2}(g) \rightarrow 7\,CO_{2}(g) + 8\,H_{2}O(l) $
Ensure the reaction is balanced carefully, accounting for all atoms of carbon, hydrogen, and oxygen.
Step 2: Identify the Gas-Phase Species
Here, the gaseous species before the reaction are 11 moles of $O_{2}(g)$, and after the reaction are 7 moles of $CO_{2}(g)$; water is in liquid form ($H_{2}O(l)$), so it does not contribute to the moles of gas.
Step 3: Determine the Change in Moles of Gas
The change in moles of gas, $ \Delta n_{g} $, is given by (total moles of gaseous products) – (total moles of gaseous reactants):
$ \Delta n_{g} = (7) - (11) = -4
$
Step 4: Use the Thermodynamic Relation
For reactions occurring in the gaseous phase, the relation between enthalpy change ($\Delta H$) and internal energy change ($\Delta U$) is:
$ \Delta H - \Delta U = \Delta n_{g}\, R\, T
$
Substituting $\Delta n_{g} = -4$:
$ \Delta H - \Delta U = -4\, R\, T
$
Conclusion
The difference between $ \Delta H $ and $ \Delta U $ for the combustion of one mole of heptane at temperature T is $ -4RT $, matching the correct answer.
