If a process is taking place at constant temperature and pressure, delta E=0 but what about delta H?

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Answer 1 · 2017-06-03T14:52:15

$H = E + pV$

$implies dH = dE + pdV + Vdp$

Now from combined mathematical statement of the first and second laws of thermodynamics (for reversible transformations),

$TdS = dE + pdV$

Using the above relationships,

$dH = TdS + Vdp$

Integrating, one can obtain the change in $H$ .