In thermodynamics work can be done by moving alone a reversible or irreversible path. Physical definitions of reversible and irreversible processes is uncommon in thermodynamics textbooks. The main idea is that I have to imagine something I can't. It is better to work with formal things than try to imagine. Work is line integral of differential form, $$W=\int_{L} P(x, y, z) dx+Q(x, y, z) dy+R(x, y, z) dz\tag{1}.$$ where $L$ is the equation of the curve. Does this mean that some curves $L$ have the property of being reversible paths and some don't? What does taking a line integral alone a reversible path mean mathematically? How can I formalize it?

  • $\begingroup$ If I understand you mean the "curve" is the path over which the work is done, then an irreversible process will give you a different path (shape curve) between two equilibrium states than a reversible process. The irreversible path produces less work. $\endgroup$ – Bob D Mar 29 at 18:14
  • $\begingroup$ I have read many physical chemistry textbooks and I don't understand chapters about second law. I don't understand intuitive descriptions with reversible and irreversible process, because differentials are not intuitive for me. I am chemist.I cant't imagine it. I spent few years trying to read it over and over again and to understand it. I need formula. $\endgroup$ – Alex Alex Mar 29 at 18:31
  • $\begingroup$ I will post an answer to help you visualize this without differentials. If it doesn't work for you, i'll withdraw it. $\endgroup$ – Bob D Mar 29 at 18:32
  • $\begingroup$ I can visualize differentials. I suppose I should solve many problems connected to differential forms and integration in connection with physical problems to gain intuition. I am chemist I didn't have such practice, but I can learn how to solve equations. $\endgroup$ – Alex Alex Mar 29 at 18:39
  • $\begingroup$ Let me know if the answer below helps. $\endgroup$ – Bob D Mar 29 at 18:47

Does this mean that some curves 𝐿 have the property of being reversible paths and some don't?

Yes. The work done by a reversible process between the same two equilibrium states will be greater than for the same process carried out irreversibly between the same two equilibrium states. Each has a different shape "curve" between the start and end points.

For a closed system the work of interest is called "pdV" work and the work between two states is $\int pdV$. The diagram below shows two different paths (or "curves" as you call them) between equilibrium states 1 and 2. Path 1-2 is for a reversible isothermal process. That's a process carried out very slowly gradually reducing the external pressure while the gas expands so the temperature of the ideal gas is constant (pV=constant) while it gradually takes in heat from the surroundings and performs an equal amount of work.

Path 1-1a-2 is an irreversible isothermal process. The external pressure is abruptly reduced from P1 to P1a and then the gas allowed to expand to the same final state 2. Although the temperature at the interface between the gas and surroundings is constant, there are temperature and pressure gradients within the gas that makes the process irreversible.

Note that the work done ($\int pdv$), or area under the curve, for path 1-2 is greater than that for path 1-1a-2. The difference is the light gray shaded area.

Hope his helps.

enter image description here

| cite | improve this answer | |

The way I see it, thermodynamical variables like $p$ and $T$ are only really defined in equilibrium. So when you draw curves in the $pV$ plane, you are implicitly saying that the process is quasistatic. Whether a quasistatic process is reversible or not is a confusing issue (to me); in the purest of situations it will be, but it depends on what happens in the environment (see here). So every curve $L$ has at least the potential to describe a reversible process. There is no such thing as an inherently irreversible curve.

A general irreversible process will start and end in equilibrium states (points on the $pV$ diagram), but will go through nonequilibrium states to get there. That will look like a "jump" between two points on the diagram, with no curve connecting them. To calculate properties like work and entropy change of irreversible processes, you need some additional information about what's happening during the process. For example, in the example by Bob D, it's not that the system actually goes along the full vertical and horizontal lines of constant $V$ and $p$, but rather that it somehow goes from 1 to 1a under constant volume, and then somehow on to 2 under constant external pressure (even though the "internal pressure" $p$ is not defined). This allows you to compute the work as $W = p_\mathrm{ext} \Delta V$ even though you can't write $\Delta V = \int dV$ along a nice curve of equilibrium states.

Apologies if that sounds a bit rambling.

| cite | improve this answer | |
  • 1
    $\begingroup$ The pressure in the $pV$ diagram is always the external pressure (pressure of the surroundings) and at the interface between the system and surroundings for both a quasistatic and non quasistatic process. The gas pressure is only in equilibrium and the same as the surroundings for the reversible (quasistatic) process. $\endgroup$ – Bob D Mar 31 at 10:26

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.