Relation Between Enthalpy and Internal Energy

Relation Between Enthalpy (H) and Internal Energy (U)

Let H₁, U₁, V₁ be the enthalpy, internal energy and volume of the system in the initial state and H₂, U₂, V₂ be the corresponding values in the final state.
We know that, Enthalpy
H = U + PV
in the initial state
H₁ = U₁ + PV₁ --- Equation:1
in the final stete
H₂ = U₂ + PV₂ --- Equation:2

Subtracting equation:1 from equation:2 we get
H₂ − H₁ = (U₂ + PV₂) − (U₁ + PV₁)
ΔH = (U₂ − U₁) + P(V₂ − V₁)
ΔH = ΔU + PΔV
where, ΔH is change in enthalpy, ∆U is change in internal energy, and P∆V is change in work energy.

Thus, the change in enthalpy at constant pressure is equal to the increase in internal energy plus pressure-volume work done (work energy).

For, ideal gas, we know that PV = nRT
Putting the value of PV in the above equation we get
ΔH = ΔU + Δn.RT
This equation shows that the change in enthalpy is equal to the change in internal energy plus changes in moles at constant temperature and pressure.