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$$\Delta H = \Delta U + P× \Delta V$$

Textbook says that this means a change in enthalpy could either be due to a change in internal energy or expansion/compression work.

But in a constant pressure condition, how can you change internal energy without changing volume and vice versa ? So can they be independently changed or not ?

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  • $\begingroup$ Who says volume isn't changing? Anyway, $\Delta H=\Delta U+\Delta (PV)$, not just $\Delta U+P\Delta V$ $\endgroup$ Mar 14, 2021 at 15:23
  • $\begingroup$ This is what the book says - imgur.com/a/AYneSXl that a change in enthalpy could either be due to change in energy or work due to volume change. So can they happen independently ? $\endgroup$ Mar 14, 2021 at 16:17
  • $\begingroup$ What the book is saying is incorrect. Enthalpy change between two specified thermodynamic equilibrium states of a system is independent of any particular process used to transition the system between these end states. It is strictly a physical property of the material(s) comprising what we define as the system. For example, the change in internal energy and the work done in simply increasing the pressure on an incompressible liquid are both zero, and yet the enthalpy change is $\Delta H=V\Delta P$. $\endgroup$ Mar 14, 2021 at 17:12
  • $\begingroup$ @Chet Miller Thanks for responding ! and that makes total sense, but assuming a constant pressure process, is it possible to increase inernal energy without expanding the system ? $\endgroup$ Mar 14, 2021 at 18:01
  • $\begingroup$ Sure. Just add heat to a truly incompressible (constant density) liquid. Or carry out a endothermic chemical reaction of ideal gases at constant pressure in which the number of moles of gas doesn't change, and enough heat is added to hold the temperature constant. $\endgroup$ Mar 14, 2021 at 18:09

1 Answer 1

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The Enthalpy is given by $$H=U+pV\Rightarrow dH=dU+d(pV)=TdS+Vdp$$ So the enthalpy can be changed by either change of $S$ and $p$.

The expression is given by OP

$$dH=dU+pdV \ \ \ \text{not true in general}$$

If the process is isobaric (constant pressure ) $$dH=TdS \ \ \text{(Isobaric process)}$$ The right side is just Heat (assuming reversible process). So at the constant process, the enthalpy can be changed by Heat.

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  • $\begingroup$ Enthalpy is a function of state, independent of any process. So, associating it with a process is irrelevant. $\endgroup$ Mar 14, 2021 at 15:24
  • $\begingroup$ Yeah! It's up to us what we are taking as the independent variable to change during the process. Volume, temperature, etc. are easy to control than enthalpy. $\endgroup$ Mar 14, 2021 at 15:26
  • $\begingroup$ Like I said, it is independent of any process. It is a physical property of the material comprising what we define as our system. $\endgroup$ Mar 14, 2021 at 15:30

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