**Constant Pressure Processes and Enthalpy**

In chemistry constant pressure processes are more common than constant volume processes since most reactions are usually carried out in open vessels under atmospheric pressure. Under a certain constraint pressure, gases can expand or contract; depending on the type of constraint, the final state of the gas may change.

A convenient thermodynamic function *enthalpy, or heat content* is used to describe a constant pressure process. This function is represented by the letter H and is defined by;

H = U + PV … … … (1)

Like any other thermodynamic state function it is impossible to determine the absolute value of enthalpy of a system. The enthalpy of a system depends only on the state of the system and is in no way affected by the way in which the state is reached. When a system changes from state 1 to state 2 the value of ∆H, (i.e., H_{2} – H_{1}), is constant, no matter how the system undergoes change in going from the initial state to the final state. In thermodynamic treatments we are concerned with ∆H. As will be seen, for solids and liquids ∆H values are usually almost equal to the change in internal energy, ∆U. This is because solids and liquids are only slightly compressible; i.e., their volume changes only to a small extent when the pressure is changed. Under constant pressure conditions equation may be written as,

dU = q_{P} – P.dV

q_{P} = dU + P.dV (as w = P.∆V)

= U_{2} – U_{1} + P (V_{2} – V_{1})

= (U_{2} + PV_{2}) – (U_{1} +PV_{1}) … … … (2)

q_{P}, being the heat absorbed at constant pressure.

As H = U + PV, equation (2) takes the form,

q_{P} = H_{2} – H_{1} = ∆H

In other words, the heat absorbed under constant pressure conditions is equal to the increase in enthalpy or heat content, if the only work done is the pressure volume work.

**Example:** What will be the change in enthalpy when 5.0 mol of iron is heated from 90°C to 140°C at atmospheric pressure?

(specific heat of Fe is 0.4476 J g^{-1} deg^{-1})

Ans: We know, ∆H = q_{P} = (sp.heat) m ∆T

∆H = (5.0 x 55.8) (0. 0.4476) (140 – 90) = 6.25 kJ