Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Thermodynamic Entropy Variation is defined as $$\Delta S = \int_i^f \frac{dQ}{T},$$

where $i$ and $f$ are the initial and final states of the process.

My question is: does this equation apply to quasi-static irreversible processes, or only to reversible processes?

Obviously, it does not apply to processes that go through non-equilibrium states, since Temperature (or any state variable) is not even well defined in these states. But I'm unsure of whether it applies to irreversible processes that are quasi-static (and therefore don't go through non-equilibrium states).

share|cite|improve this question

2 Answers 2

up vote 3 down vote accepted

1) Irreversible processes are the ones, which by definition increase the entropy. And the increase, similarly to non-equilibrium cases, is added to the reversible $\mathrm{d} S$. Hence, for all irreversible processes: $\delta S > \dfrac{\delta Q}{T}$

2) For at least some non-equilibrium cases thermodynamic variables like $T$ actually may well be defined, but for example only locally, or separately for different chemical species etc. Consider a chemical reaction, which goes quasi-statically, is not in equilibrium, hence produces entropy additional entropy and hence is irreversible.

share|cite|improve this answer

The equation is only valid for quasistatic processes, though entropy change is defined for irreversible processes (it's a state function)

I personally have been taught physics from a pure reveraible point of view--where all processes are implied to be reversible. I'm not sure, but I think that this view is carried on throughout physics. In chemistry, on the other hand, I was taught both reversible and irreversible processes, so the chem equation I know is: $$\Delta S = \int_i^f \frac{\rm dq_{rev}}{T}$$ and sometimes $$\Delta S = \int_i^f \frac{q_{rev}}{T}$$ The $q$ is lowercase because it's a path function(convention in chem), and it's not really a differential in one of the equations because $q$ and $\rm dq$ are used interchangeably.

Edit: Realized that I missed the 'ir' in quasistatic irreversible :/

No, the equation will not hold--as entropy is a state function, while q is a path function. You must carry out the process $i \to f$ in a reversible manner. Otherwise you could potentially get two values for entropy.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.