Question

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

Answer 1

Only for a system at constant pressure... By definition, `q = nDeltabarH = DeltaH` at constant pressure, where `DeltabarH` is the molar enthalpy in `"J/mol"`.

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We can begin from the first law of thermodynamics

`DeltaU = 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`,

`DeltaH = DeltaU + Delta(PV)`

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

When one plugs in the expression for `DeltaU`:

`DeltaH = q + w + Delta(PV)`

The common convention is that the work is defined by `w = -PDeltaV` when work is with respect to the system (i.e. negatively-signed when the system does work on the surroundings by expanding, with `DeltaV > 0`).

Furthermore, we apply the product rule from calculus (plus a bit extra) to see that `Delta(PV) = PDeltaV + VDeltaP + DeltaPDeltaV`. Therefore:

`DeltaH = q - cancel(PDeltaV + PDeltaV) + VDeltaP + DeltaPDeltaV`

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

`color(blue)(DeltaH = q + VDeltaP + DeltaPDeltaV)`

Clearly, when the pressure is constant, we can see that `VDeltaP = 0`. We designate this as

`DeltaH = q_P`,

if both quantities are in `"J"`. Multiplying the left by `n/n`, where `n` is the mols, gives the first equation presented,

`color(blue)(nDeltabarH = q_P)`