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I know that enthalpy is defined as $H=E+PV$. So for a system the enthalpy is the sum of internal energy and the amount of work needed at constant pressure $P$ to give it's volume $V$. But the differential of enthalpy is given by legendre transformation, $$dH=TdS+VdP$$ For an adiabatic system ($dQ=0$) if work is done on the system at a constant pressure the mechanical equilibrium is achived when enthalpy is minimized. So whenever there is some work done on the system at constant pressure with no heat exchange, $dH$ is negative.

So my question is, if $dQ=0$, then $TdS$ is also zero and at constant pressure $dP=0$. So by the definion of enthalpy, $dH$ is not zero it's negative, but from the differential $dH=0$ always. Can anyone explain how do I interpret this?

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    $\begingroup$ If the process is adiabatic, you can’t change the volume unless the pressure changes. And if the volume doesn’t change and no heat is transferred, the process is null. $\endgroup$ Commented Sep 2, 2021 at 18:40

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For a reversible adiabatic (constant entropy) process for an ideal gas, pressure varies with volume according to

$$PV^{\gamma}=C$$

Where $C$ is a constant. So pressure cannot be constant.

A constant pressure adiabatic process is an irreversible process. See:

https://physics.stackexchange.com/questions/577884/is-a-constant-pressure-adiabatic-irreversible-expansion-possible#:~:text=That%2C%20of%20course%2C%20can',external%20pressure%20expansion%20or%20contraction.

So a constant pressure adiabatic process generates entropy, i.e., $dS>0$, in spite of the fact that $Q=0$.

Hope this helps.

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    $\begingroup$ The Laplace equation holds only for an adiabatic and reversible transformation. At constant pressure, the transformation cannot be reversible, so the equation doesn't hold $\endgroup$
    – Pen
    Commented Sep 2, 2021 at 17:55
  • $\begingroup$ @Pen Isn't that what this answer is saying as well? $\endgroup$ Commented Sep 2, 2021 at 17:59
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I believe the issue comes from mixing different ways of calculating the quantities at stake.

On the one hand, the real transformation where $\delta Q = 0$ and some work is performed so as to change the system from an initial to a final state.

On the other hand, you can consider a fictive reversible transformation, where heat and work are provided to the system until it reaches the same final state.

Because energy, enthalpy and entropy are state variables, their increment is the same regardless of the transformation path, and this fictive transformation an easier way to estimate dS as $\delta Q_{rev}/T$.

It appears to me that you considered $dS = \delta Q/T$ with the real heat exhanged (=0). This estimation is incorrect, because some entropy is created here. This can be intuited from the transformation: As work is provided to the system, the system's pressure changes so a constant applied pressure cannot provide a reversible transformation'

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