# The definition of enthalpy is H=E+PV, is it only valid for (ideal) gas? As this naturally comes from the first law of thermodynamics, dE=dQ+PdV for a enclosed system of gas.

Is enthalpy also valid for liquids and solids?
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Jaylene Tyler
For solids, as well as for liquids and gases, you can define a specific heat as
${c}_{V}\equiv {\left(\frac{\mathrm{\partial }U}{\mathrm{\partial }T}\right)}_{V}$
and
${c}_{P}\equiv {\left(\frac{\mathrm{\partial }U}{\mathrm{\partial }T}\right)}_{P}$
For solids, however, V is basically a constant, so that with excellent approximation
${c}_{P}\approx {c}_{V}\equiv c$
and we can now define an internal energy E as
$dE=c\left(T\right)dT\phantom{\rule{thickmathspace}{0ex}}.$
We can proceed likewise with enthalpy:
$dH=dE+d\left(PV\right)=c\left(T\right)dT+PdV+VdP$
For a solid, dV=0 for a fixed amount of matter, and furthermore $dE\gg VdP$, thus
H=E
There is nothing wrong in these concepts (the thermodynamic potentials) for liquids and solids.