How do I calculate the entropy change for an irreversible isothermal expansion against $p_\text{ext}=p_0$?

For a reversible isothermal expansion we know $\delta S_\text{uni}=0$ so $\delta S_\text{sys}=-\delta S_\text{surr}=nR\ln\frac{V_\mathrm f}{V_\mathrm i}$

However I can't figure out why $\delta S_\text{sys}=nR\ln\frac{V_\mathrm f}{V_\mathrm i}$ remains for irreversible expansion.

I also found out an equation for irreversible expansion $\delta S_\text{uni}=q_\text{rev}/T-q_\text{irrev}/T$ Is it correct? If yes? Can you explain how do we get it?


Is this correct ?

$\delta S_\text{uni}=\delta S_\text{sys}+\delta_\text{surr}$

$\delta S_\text{sys}=\frac{q_\text{rev}}{T}$

$\delta S_\text{surr}=-\frac{q_\text{irrev}}{T}=p_\text{ext}\delta V/T$


Here are the steps to determining the change in entropy for an irreversible process on a closed system:

  1. Use the first law of thermodynamics to determine the final thermodynamic equilibrium state of the system for the irreversible path.

  2. Totally forget about the irreversible path. It is of no further use. Focus only on the initial equilibrium state of the system and the final equilibrium state.

  3. Devise a reversible path for the system that takes it from the initial equilibrium state to the final equilibrium state. This reversible path does not have to bear any resemblance whatsoever to the real irreversible path, other than it must pass through the same initial and final end points. (Entropy is a function of state).

  4. Calculate the integral of dq/T for the reversible path that you have devised. This will give you the change in entropy of the system for the irreversible path as well as for the reversible path.

For your problem, this procedure will give you the equation that you have written. Note that, for the irreversible path, the system temperature matches the surroundings temperature (isothermal) only at the interface with the surroundings, but not throughout the interior of the system. For an irreversible expansion, this is what we define as "isothermal."

  • $\begingroup$ So is then change in entropy of universe 0 for isothermal irreversible process? Since by 4th point qrev and qirrev are same. $\endgroup$ Oct 22 '17 at 14:46
  • $\begingroup$ Btw can you also include how to deal when expansion is against a constant pressure? $\endgroup$ Oct 22 '17 at 14:47
  • $\begingroup$ Are you saying that you don't know how to find the final equilibrium state for this case using the first law of thermodynamics? $\endgroup$ Oct 22 '17 at 15:07
  • $\begingroup$ I have never used the concept of initial and final states. I was taught to work with changes ( which I know is basically the same thing). Can you please show how to solve this question (I am not asking it for homework. I am asking because it would help me understand better). Determine the entropy change in sys, surr, uni, when a sample of helium has of mass M grams at 298K and 1 bar doubles its volume in isothermal irreversible expansion against external pressure =p $\endgroup$ Oct 22 '17 at 15:14
  • $\begingroup$ Well, if you know the mass M, then you know the number of moles. And you know the initial temperature and pressure, so, from the ideal gas law, you know the initial volume. What is that equal to? At time zero, you suddenly drop the pressure to p and hold it at this value until the system reequilibrates. at 298 and p. What is the final volume? What is the ratio of the volumes? $\endgroup$ Oct 22 '17 at 15:29

I think by thermodynamical definition of entropy you can calculate entropy change exactly only in reversible changes only. As Clausius inequality refers we can not calculate exactly the entropy change in an irreversible process. We can only estimate its limiting value. Statistically we can determin it by using Boltzmann theorem in terms of most probable thermodynamic probability of the state.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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