# The heat content of a system is equal to the enthalpy only for a system that is at constant what?

Jun 14, 2017

Only for a system at constant pressure... By definition, $q = n \Delta \overline{H} = \Delta H$ at constant pressure, where $\Delta \overline{H}$ is the molar enthalpy in $\text{J/mol}$.

We can begin from the first law of thermodynamics

$\Delta U = q + w$,

where $q$ and $w$ are the heat flow and work, respectively,

and the relation of the internal energy $U$ to the enthalpy $H$,

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

where $P$ and $V$ are pressure and volume.

When one plugs in the expression for $\Delta U$:

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

The common convention is that the work is defined by $w = - P \Delta V$ when work is with respect to the system (i.e. negatively-signed when the system does work on the surroundings by expanding, with $\Delta V > 0$).

Furthermore, we apply the product rule from calculus (plus a bit extra) to see that $\Delta \left(P V\right) = P \Delta V + V \Delta P + \Delta P \Delta V$. Therefore:

$\Delta H = q - \cancel{P \Delta V + P \Delta V} + V \Delta P + \Delta P \Delta V$

As a result, the change in enthalpy is related to the heat flow as:

$\textcolor{b l u e}{\Delta H = q + V \Delta P + \Delta P \Delta V}$

Clearly, when the pressure is constant, we can see that $V \Delta P = 0$. We designate this as

$\Delta H = {q}_{P}$,

if both quantities are in $\text{J}$. Multiplying the left by $\frac{n}{n}$, where $n$ is the mols, gives the first equation presented,

$\textcolor{b l u e}{n \Delta \overline{H} = {q}_{P}}$