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From my previous question What is enthalpy? I learned that enthalpy is the amount of energy required to set up a thermodynamic system plus the energy required to make space for it.

It is true that $\Delta H = Q$ when pressure is constant while the change happens. It seems indeed to be the most delicious use of the concept.

Which now I interpret as: Enthalpy change in a process that doesn't require a change in the space the system needs will be equal to the heat flow.

Which means: there is no work flow when there is no change in the space the system needs.

That confuses me a bit. And that confusion made me find this question (Does work always imply a change in volume? - Answer: yes for mechanical work but no for other cases.)

My question is parallel, but it regards the pressure: Why is there no work flow for constant pressure processes? Can I not expand a gas while increasing its temperature to keep the pressure constant?

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Álvaro N. Franz
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  • Who says there is no work done in a constant pressure process? – Chet Miller Aug 18 '16 at 17:16
  • @ChesterMiller I interpreted it from the fact that all the enthalpy change is equal to heat flow in a constant pressure process. If enthalpy considers all the internal energy and internal energy considers heat and work, and there is only heat, then there is no work. Of course i might be wrong, but that is the core of my question. Thank you Chester, with all my heart. – Álvaro N. Franz Aug 18 '16 at 17:26
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    In an isobaric process: $W=\int_{V_1}^{V_2}pdV=p(V_2-V_1)$. – Gert Aug 18 '16 at 17:33
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    To expand on Gert's comment, for a constant pressure process, $\Delta U=Q-W=Q-P\Delta V$ and $$\Delta H=\Delta U+\Delta (PV)=\Delta U+P\Delta V=Q-P\Delta V+P\Delta V=Q$$ – Chet Miller Aug 18 '16 at 18:53
  • @ChesterMiller When you write $\Delta U=Q-W=Q-P\Delta V$ we are considering W as work that is done by the system. But if we consider it as work that enters the system, then the rest will not be the same. My thought is that it doesn't matter because a positive increment in volume is done by the system and a decrement in volume is done by an external factor, and it will flip the signs anyways. But I don't know if that is always true. – Álvaro N. Franz Aug 19 '16 at 08:51
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    Sure it is. The work done by the system on the surroundings is equal to minus the work done by the surroundings on the system. – Chet Miller Aug 19 '16 at 11:33

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Objects and systems usually expand as they are heated up, or receive an input of heat. So they will do work on the environment around them. This means $W$ is negative, using the standard convention.

So internal energy is lost as work, therefore you need to add in heat to restore that energy.

Now assume the pressure surrounding your system is constant, then the heat required per degree is called the heat capacity at constant pressure.

Heat capacity is $C = Q/\Delta T = (\Delta U -W)/\Delta T $

So $C_p = [(\Delta U - (- P\Delta V)]/\Delta T $ holding pressure constant.

= $\left(\frac{\partial U}{\partial T} \right) $ + $P\left(\frac{\partial V}{\partial T} \right) $ (pressure of partial derivatives assumed held constant in both expressions).

The final term $P\left(\frac{\partial V}{\partial T} \right) $ is the additional heat we need to supply to make up for the energy lost as work on the environment

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    Álvaro Could I make two suggestions to you please. 1. Make a list of all the thermodynamic related equations on a separate page, away from your notes, as it is very easy to get lost with these formulae 2. Read over my previous answer about enthalpy, and then see can you relate it to this above answer. It does require a bit of study to follow the argument, but doing the exercises in your textbook is always worthwhile. Best of luck with it. –  Aug 18 '16 at 18:44