The Laplace's Law in thermodynamics states that an adabatic reversible transformation of a perfect gas verifies the following identity : $$ PV^{\gamma} = cte \qquad \left( \gamma = \frac{C_p}{C_v} \right) $$ It seems to me that I can derive it without using the "reversible" requirement : adiabatic means $\delta Q = 0$, meaning I can write the energy and enthalpy variations as : $$ dU = 0 - P dV = n C_v dT \quad and \quad dH = 0 + V dP = n C_p dT. $$ Dividing one equation by the other leads to : $$ \gamma \frac{dV}{V} = \frac{dP}{P} \quad \Rightarrow \quad \gamma \ ln \left( \frac{V}{V_0} \right) = ln \left( \frac{P}{P_0} \right) + cte \quad i.e. \quad PV^{\gamma} = cte. $$

Nowhere in this did it feel like I was using reversibility. What bothers me is that the Joule expansion is an adiabatic irrevesible process where $T = cte$, which is not compatible with the Laplace's Law. Then I should not be able to derive it without using the requirement of reversibility.

I seems I am making a basic mistake in the definition of enthalpy, or heat transfer, or reversibility. Can anyone explain what is wrong in the derivation above ?

Edit 1: One of the answers states that using the first principle in the form $dU = T dS - P dV$ solves the problem (i.e. now adiabaticity and reversibility are required to cancel the first term). But why is there a difference between the first principle in those two forms (with $\delta Q$ and $T dS$) ?

When dealing with the Joule expansion, why is it now legal to use the first principle in the form $dU = \delta Q - P dV$ ? The Joule expansion is also associated with creation of entropy, meaning the term $\delta S_{creation}$ is not zero, and you cannot say $dU = 0$ if you start from the first principle with $T dS$.

Edit 2: I think all my problems boil down to what is explained in this post, provided by Chemomechanics (it is in a comment, I cannot accept it as best answer...)

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    $\begingroup$ You assumed reversibility when you equated the work to $P\,dV$. See here. $\endgroup$ Commented Aug 28, 2022 at 15:34
  • $\begingroup$ The very fundamental statement of the first principle is $d E^{tot} = \delta Q^{ext} + \delta W^{ext}$ i.e. the total energy of the system varies because of heat transfer $\delta Q^{ext}$ from the external environment or work done by external forces $\delta W^{ext}$. Kinetic energy theorem (from classical Mechanics) reads $d K = \delta W^{tot} = \delta W^{in} + \delta W^{ext}$. Subtracting Kinetic Energy Theorem from First Principle, you get $d E = \delta Q^{tot} - \delta W^{in}$, being $E = E^{tot} - K$ the internal energy (see next comment) $\endgroup$
    – basics
    Commented Aug 28, 2022 at 17:38
  • $\begingroup$ Now, you can write the internal work as the sum of the reversible and the irreversible contributions (from friction and viscosity), that must be $\delta W^{in, irr} := -\delta D \le 0$, s.t.$d E = \delta Q^{tot} - \delta W^{in, irr} - \delta W^{in, rev} = \delta Q^{tot} + \delta D - \delta W^{in, rev}$. You can collect $\delta Q^{tot} + \delta D$ in a term $T dS$ (see the documents I attached before for more details), that can be used as the definition of entropy. For a fluid, you can write the reversible elementary internal work as $\delta W^{in, rev} = p dV$. (next) $\endgroup$
    – basics
    Commented Aug 28, 2022 at 17:51
  • $\begingroup$ Putting everything together, we can write the first principle of Thermodynamics for fluids as $d E = T dS - p dV$. Following this procedure, you should be able to go from the general statement of the First Principle with heat transfer and external work to the statement for fluid with the exact differential of entropy and volume. $\endgroup$
    – basics
    Commented Aug 28, 2022 at 17:53
  • $\begingroup$ Are you aware of the fact that a reversible and irreversible adiabatic process can't connect the same two equilibrium states? If you are, then you must also know that the adiabatic equation cannot apply to both a reversible and irreversible adiabatic process. It has to be one or the other. It can't be both. $\endgroup$
    – Bob D
    Commented Aug 28, 2022 at 19:21

2 Answers 2


Adiabatic isn't enough, but full reversibility isn't necessary (although, in practice, it's usually the assumption people go for and it's fine).

You need $dU=-P\,dV$. The starting point is, for an adiabatic process: $$dU=\delta W+\delta Q=\delta W$$ Assuming that work comes only from pressure forces this becomes: $$\delta W=-P_\text{ext}\,dV$$ Now let's assume mechanical reversibility. First this implies that the process is quasistatic, so that pressure $P$ is defined at all time. Second, by definition of mechanical equilibrium, $P=P_\text{ext}$, so: $$\delta W=-P\,dV$$ As a result, $dU=-P\,dV$ and you can finish the proof.

Conclusion: although full reversibility is enough to get the result, mechanical reversibility is the real necessary hypothesis (along with no source of work other than pressure).


The assumptions is $dS =0$, since you manipulate the First Principle $dU = T dS - P dV$ to get $dU = - P dV$, then further manipulated using perfect gas law.

So, the problem is how to get $dS = 0$, and the answer comes from the Second Principle Thermodynamics, i.e.

$d S \ge \dfrac{\delta Q}{T}$,

that can be written as an equality if we consider the effects of dissipation of energy (always positive or at least equal to zero, $\delta D \ge 0$, since a system can't produce power from nothing) by internal forces, as an example friction and the effect of viscosity in viscous fluids that are not performing roto-translations (it can be considered a kind of definition of entropy, derived from everyday life experience)

$d S = \dfrac{\delta Q}{T} + \dfrac{\delta D}{T}$.

To get $d S = 0$ you need to have those two independent contributions equal to zero, i.e. no heat transfer and no energy dissipation. Only if those two conditions are both satisfied, the process is isentropic (adiabatic and reversible).

This second contribution is usually never explicitly treated in Thermodynamics, since it deals with equilibrium states, of uniform medium, where no influence of friction or viscosity of the fluids are usually considered.

  • $\begingroup$ Thank you for your answer. What you say is indeed true if you start with $T dS$ instead of $\delta Q$ in the first principle. But my understanding is that the first principle written with $\delta Q$ is somewhat more fundamental than with $T dS$. This might be where I'm wrong. I'll think about what you said and come back if I'm still confused. $\endgroup$
    – bernihl
    Commented Aug 28, 2022 at 15:43
  • $\begingroup$ Try to check if this can help, basics.altervista.org/test/Physics/TD/td_principles.html . I wrote it down a couple of months ago, trying to connect everyday life experience, with mechanics and the principles of Thermodynamics. I have never been satisfied with the explanation given by professore at university and many books about the presentation of the principles of Thermodynamics. I suggest to have a look at it, because I still remember the pain to get some formalization, that now make me satisfied. Let me know $\endgroup$
    – basics
    Commented Aug 28, 2022 at 15:57

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