Question

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

Answer 1

`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`.