- #1

Frigus

- 337

- 157

If its answer is that it is not only for a special case in which pressure is constant then how can we even use it because we cannot measure the internal energy.

Thanks

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- Thread starter Frigus
- Start date

- #1

Frigus

- 337

- 157

If its answer is that it is not only for a special case in which pressure is constant then how can we even use it because we cannot measure the internal energy.

Thanks

- #2

Chestermiller

Mentor

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- #3

DrStupid

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If we know that at constant pressure heat absorbed or gained is path independent then what was the need for introducing this new enthalpy term.

The change of internal energy is dU = dq + dw. It is easy to eliminate non-volumetric work. But than you still have dU = dq - p·dV. Eliminating the volumetric work would require keeping the volume constant which is difficult in practice. That means that an accurate calorimetric measurement would always need to be accompanied by a measurement of the change in volume.

This problem is solved by the definition of enthalpy H = U + p·V. That results in dH = dU + p·dV + V·dp, without non-volumentric work in dH = dq + V·dp and under constant pressure in dH = dq. If you manage to keep the pressure constant (which is quite easy in practice) the calorimetric measurement directly gives you the change of enthalpy.

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