# How does enthalpy change differ from heat flow?

May 23, 2015

Well, Enthalpy is defined as follows:

$H = U + P V$

$\Delta H = \Delta U + \Delta \left(P V\right)$

$= \Delta U + \left({P}_{1} + \Delta P\right) \left({V}_{1} + \Delta V\right) - {P}_{1} {V}_{1}$

$= \Delta U + {P}_{1} \Delta V + {V}_{1} \Delta P + \Delta P \Delta V$

$= q + w + {P}_{1} \Delta V + {V}_{1} \Delta P + \Delta P \Delta V$

$= q \cancel{- {P}_{1} \Delta V + {P}_{1} \Delta V} + {V}_{1} \Delta P + \Delta P \Delta V$
(with $w = - P \Delta V$ for volume expansions/contractions)

As a result,

$\Delta H = q + {V}_{1} \Delta P + \Delta P \Delta V$

where:

• $H =$ enthalpy
• $U =$ internal energy
• $q =$ heat flow
• $w =$ work
• $P =$ pressure
• $V =$ volume

So you can see that enthalpy is heat flow along with a constant-volume pressure change, and enthalpy and heat flow are only the same if the pressure is constant (${q}_{p} = \Delta H$) (like in a coffee-cup calorimeter).

We can then say that enthalpy is equivalent to heat flow in a constant-pressure system open to the air, whereas internal energy is equivalent to heat flow in a constant-volume system closed to the air.