In only pressure-volume work is done at constant pressure, then according to first law of thermodynamics,
∆E = qp – P∆V
qp = ∆E + P∆V
qp = (E2 – E1) – (PV2 + PV1)
For constant pressure, q is represented by qp.
Here, E, P and V, all are state functions, E + PV is also a state function. This is known as enthalpy or heat content.
Enthalpy of a system is equal to sum of its internal energy and pressure – volume energy. The enthalpy change for initial and final state are:
H2 – H1 = (E2 -E1 + PV2 – PV1)
∆H = ∆E + P∆V = qp
Thus, heat absorbed by the system at constant pressure is equal to its enthalpy change. H is a state function. Hence, qp is also a state function.
If V2 = V1 then, ∆H = ∆E
In gaseous state, there is an appreciable distance between H and E, then according to ideal gas equation,
PVA = nART
PVB = nBRT
PVB – PVA – nBRT – nART = (nB – nA )RT
P(VB – VA) = (nB – nA)RT
P∆V = ngRT
Here, ∆ng refers to the number of moles of gaseous products minus the number of moles of gaseous reactants. Substituting the value of P∆V from equation, we get
∆H = ∆E + ∆ngRT