We know that change in internal energy is equal to heat supplied - work done in a cyclic process. The change in internal energy is zero and work done is non-zero. This implies that the heat given is non zero then how can the enthalpy change be 0?
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1$\begingroup$ Enthalpy is a state variable, so it’s net change during a cycle must be zero. I suspect you are mixing this up with the fact that during an isobaric process, change in enthalpy is equal to heat, but that is only for an isobaric process. $\endgroup$– marchSep 18, 2020 at 7:17
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$\begingroup$ I have updated my answer. Maybe it will help. $\endgroup$– Bob DSep 19, 2020 at 11:54
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Internal energy, pressure and volume are all state properties. A cycle returns all properties to their original values. So the change in each of these is zero. Enthalpy is defined as
$$H=U+PV$$
Therefore it too is a state property so its change is also zero.
Work and heat are energy transfers and are not properties. They depend on the processes carried out during the cycle. since $\Delta U=0$ then $W=Q$. Meaning the net work done in a cycle equals the net heat added.
This implies that the heat given is non zero then how can the enthalpy change be 0?
Enthalpy is defined as above. It only equals the heat transferred for a constant pressure process, as follows.
$$\Delta H=\Delta U+\Delta (PV)$$ $$\Delta H=Q-W+P\Delta V+V\Delta P$$
For a constant pressure process, $W=P\Delta V$ and $V\Delta P=0$, and therefore
$$\Delta H=Q=mC_{p}\Delta T$$
And $\Delta H$ is positive if heat is added to the system resulting in a temperature increase. But a process does not constitute a cycle. For a cycle the temperature must return to its original value. This require us to reverse the constant pressure process to transfer the same amount of heat out of the system making $\Delta H=0$ for the cycle.
Hope this helps
