Premise 1: Definition of Enthalpy

$H = U + PV$ This is the definition. Enthalpy is defined to be convenient for constant-pressure processes.

Premise 2: Change in Enthalpy

For a process at any conditions: $\Delta H = \Delta U + \Delta(PV)$ This follows directly from differentiating the definition.

Premise 3: Ideal Gas Law

For ideal gases: $PV = nRT$

Therefore: $\Delta(PV) = \Delta(nRT) = RT\,\Delta n_g$ (at constant temperature $T$)

Here $\Delta n_g$ represents the CHANGE in the number of moles of gas.

Premise 4: Substitution

$\Delta H = \Delta U + \Delta(PV) = \Delta U + \Delta n_g RT$

Conclusion

$\boxed{\Delta H = \Delta U + \Delta n_g RT}$

Why Only Gaseous Moles?

Solids and liquids have negligible PV compared to gases:

• 1 mol gas at 298 K, 1 atm: $PV = RT = 2479\ \text{J}$
• 1 mol liquid water: $PV = 18\ \text{cm}^3 \times 10^5\ \text{Pa} = 1.8\ \text{J}$ (1000× smaller)

Therefore, only gaseous moles contribute significantly to $\Delta(PV)$.

Practical Check

For $CaCO_{3}$(s) → CaO(s) + $CO_{2}$(g): $\Delta n_g = 1$ $\Delta H = \Delta U + 1 \times 8.314 \times 10^{-3}\ \text{kJ/(mol·K)} \times T$

At 298 K: $\Delta H - \Delta U = 2.48$ kJ/mol (small but measurable). At 500 K: $\Delta H - \Delta U = 4.16$ kJ/mol.