In the most simple case of a closed system with no chemical reaction and interacting with the environment through pressure only, we have: $$dU=\delta Q+\delta W=TdS-PdV$$ (first law + fundamental equation of thermodynamics)

Reading some elementary texts in thermodynamics, I see the following idea often suggested but I have not read it explicitly:

$$TdS\geq\delta Q$$ $$-PdV\leq\delta W$$

A possible argumentation to justify it would be the following:

Consider doing the same mechanical action on the system while isolating the system thermally. To highlight this is a different process than the original one, we use denotation $dV'$, $dS'$.... The work $\delta W'=\delta W$ and volume change $dV'=dV$ would be the same. For this process, we would have $\delta W=dU'=-PdV+TdS'$ and since the system is thermally isolated, $dS'\geq 0$ (second law) so that $-PdV\leq\delta W$.

It follows that $TdS\geq\delta Q$.

Is the reasoning and conclusion correct?

  • $\begingroup$ Is the process reversible or irreversible? $\endgroup$ Apr 9, 2023 at 11:37
  • $\begingroup$ Either. The only assumption is that the change is continuous, the change in all state variables is small, and the equilibrium is sufficient to consider intensive state variables are defined and almost constant across the system (quasistatic). Initial and final states are equilibrium ones. Also, all variables mentioned here are the inner variables of the system. For example T is the internal temperature, not the temperature of an external bath. Same for pressure. Neither the mechanical action nor the heat exchange are assumed reversible (even though they can be). $\endgroup$
    – Benoit
    Apr 9, 2023 at 11:58
  • $\begingroup$ Are you aware that, if you have an adiabatic irreversible path between two thermodynamic equilibrium ends states, there is no adiabatic reversible path between the same two end states? $\endgroup$ Apr 9, 2023 at 12:22
  • $\begingroup$ I am. The question does not mention anything about adiabaticity, even less reversibility. My reasoning to try to prove the inequalities with full generality is saying "considering any process (quasistatic) X, then an adiabatic process with the same mechanical action as X would result in the same work and the same volume change as X". Not the same final state. Anyway they both can be reversible or irreversible. This argument is debatable, even suspicious to me, that is why I asked the question. $\endgroup$
    – Benoit
    Apr 9, 2023 at 12:39
  • $\begingroup$ What do think isolating the system thermally means if not adiabatic? $\endgroup$ Apr 9, 2023 at 12:49

1 Answer 1


Yes, your argument is valid, but your proof is for adiabatic process only. The inequality is more general and applies whether heat is exchanged or not.

Start with entropy generation, which is defined as the entropy change of the system ($dS$) plus that of the surroundings. Assuming the surroundings to be at the temperature of the system the entropy of the surroundings is $dS_\text{surr}=-dQ/T$. Then, $$ dS_\text{gen} = dS -\frac{dQ}{T}\geq 0 \Rightarrow dQ \leq T dS $$ which is the inequality we seek. If the surroundings are at a temperature different from that of the system there is additional entropy generation due to heat transfer between the system and the surroundings, and the inequality is even stronger.


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