I have trouble wrapping my head around the thermodynamic identity.

From Schroeder's book "An Introduction to Thermal Physics," I understand that entropy is defined as $$\mathrm dS=\frac{δQ}T\tag1$$ Later in the book, however, the author warns that $Q=T\,\mathrm dS$ only holds when the change in volume of the process takes place quasi-statically because we then know that $W=-p\,\mathrm dV$. So, from the thermodynamic entity $$\mathrm dU=T\,\mathrm dS−p\,\mathrm dV\tag2$$ we can associate the heat and work with the first and second terms, respectively.

These two affirmations seem contradictory to me, and I can't seem to figure it out.

My Questions:

  • Under which assumptions is equation $(1)$ true?

  • When this equation isn't true, how can we find the change in entropy of a system?

  • 1
    $\begingroup$ More specifically, $\mathrm{d}S=\dfrac{\delta Q_\mathrm{rev}}{T}$ or $\mathrm{d}S \gt \dfrac{\delta Q_\mathrm{irrev}}{T}$ $\endgroup$
    – Poutnik
    Mar 8 at 17:18

2 Answers 2


The two equations you presented are valid only for reversible processes (i.e., processes in which the system passes through a continuous sequence of thermodynamic equilibrium states).

For a system experiencing an irreversible process, your two equations are not valid. For an irreversible process, the change in entropy of the system is determined by doing the following:

  1. Use the first law of thermodynamics to solve for the final thermodynamic equilibrium state of your system (starting from the initial state).

  2. Completely forget about the irreversible process in step 1. because it is no longer of interest, except for the initial and final states.

  3. Devise (dream up or conceive of) a reversible thermodynamic process between the same two end states.

  4. Calculate the integral of dQ/T for the reversible process you devised in step 2. This is the change in entropy of the system.

  • $\begingroup$ Thank you for the answer! So you are saying that once I know of the initial and final state of my system, I can then approximate it by a reversible process, right? Just to make sure, wouldn't that cause some issues with non-state variables that depend on the process? Or am I missing something? $\endgroup$ Mar 8 at 19:10
  • $\begingroup$ You are not approximating the entropy change of the irreversible process. You are determining it exactly. The entropy change is a physical property of the substances comprising the system, and the change is independent of any process. I might also mention that there are an infinite number of reversible processes going between the two end states, and the all give the same value for the integral of dQ/T. $\endgroup$ Mar 8 at 20:15

The entropy of a system is only defined (strictly for equilibrium thermodynamics) only for reversible processes as:

$$\text{d} S=\frac{\delta Q}{T}$$

Due to Clausius inequality, for any other process entropy is actually defined by the statement: $$\text{d}S\geq \frac{\delta Q}{T}$$ Where the equality is true for perfectly reversible processes (those that you call Quasi-Static).

For any other process that is not reversible, you might however still obtain a value for Entropy by considering an analogous process that is reversible and has the same starting and ending point in the space of thermodynamics variables.

  • $\begingroup$ Thank you, I'm going to edit that. $\endgroup$ Mar 9 at 9:24
  • $\begingroup$ Thanks for the answer, I didn't understand the full meaning of Clausius' inequality, now it's clearer! $\endgroup$ Mar 9 at 12:49

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