Calculation of work done in different processes of thermodynamics in Physics and Chemistry I am a high school student and I am very confused in thermodynamics, My confusion is that :In physics, when we study thermodynamics, we say that we cannot find the work done in irreversible processes because in such processes the different state variables of the gas are not defined. For example, the pressure in different regions of the system is different when a gas expands quickly.
Yes, I also agree with that. We cannot directly find the work done in such a process, because we don't know what pressure of gas should we take, but I think, even if we cannot calculate this directly, we can calculate this indirectly like we do in chemistry. In chemistry for example, when we find the work done in irreversible compression, we can find how much work is done by the surroundings on the system because the pressure of the surroundings almost remains the same. Let's assume $+W$ amount of work is done on the gas, so it means that it has increased the internal energy of the system by $+W$ (assuming no heat is released), so it also means that surely the work done by the system on the surroundings will be $-W$; to absorb that much energy, it has to do $-W$ amount of work on the surroundings, and there is no other thing from which energy can be exchanged.
So, all I want to ask is: Why we don't do it like this in physics? What's wrong with it? Why do we say that we cannot calculate work done in irreversible processes like this? Please guide me towards the right path without using higher mathematics just explain it in brief, then it would be easier for me to understand.
 A: To a physicist, thermodynamics is essentially, by definition, the study of systems in thermal equilibrium.  That does not mean that it is impossible to do out-of-equilibrium calculations, but that it requires a different formulation of the dynamics in order to do those calculations.  There is a tendency among physicists when discussing thermodynamics to give a definition of pressure as
$$P=-\left.\frac{\partial E}{\partial V}\right|_{S},$$
following from the expansion of the internal energy $E=U$ in its natural variables,
$$dE=-P\,dV+T\,dS.$$
However, this definition is only applicable when a system is undergoing quasistatic evolution, because that means that the system remains on (or infinitesimally close to) the equation of state surface, making it possible to take the derivative with respect to volume.  This is a slightly more general condition than that the system needs to be evolving reversibly.  Any reversible process is necessarily quasistatic, but it is possible to have irreversible yet quasistatic evolution.  For example, a system may be allowed to expand freely in (infinitesimally) small increments; for a gas, this would mean successively removing partitions, so that at each step there is only a minuscule increase in the volume of the gas, but with many iterations, the volume does increase by a finite amount.  This is one instance in which you can calculate the work done by the gas (necessarily zero, because at each stage there is only free expansion) for a irreversible process; moreover, for an ideal gas, with internal energy dependent only on temperature, not volume, it is easy to calculate the entropy generated by the expansion, showing it is irreversible.
In a irreversible compression, you can again calculate the work done, so long as you know a complete description of the system at all times.  If you are applying a known pressure, in a nearly instantaneous compression, then there is essentially no time for the entropy of the system to change.  This means that it is possible to use $dE=-P\,dV$ to calculate the change in the energy of the system.  However, since the compression occurs very quickly, the system does not have time to thermalize as it is being compressed; during the rapid process, it is out of equilibrium (off the equation of state surface), so the definition of pressure as a partial derivative no longer applies.  That is not a problem, if you know the pressure being applied externally, which is the way chemists and engineers tend to think about systems.  However, if your natural viewpoint is the system itself—and this is the viewpoint typical of physicists—the lack of a definition of pressure that is internal to the gas becomes a problem.
Of course, that does not mean that physicists are incapable of calculating work in situations that are not quasistatic.  What it does mean is that a different approach has to be taken.  In a system that is out of equilibrium, it is still possible to calculate its pressure using kinetic theory, rather than thermodynamics.  This requires knowing the distribution of molecular velocities in a gas—not a trivial problem by any means, although one that can be solved if it is known that an ideal gas started from an equilibrium state and was rapidly compressed.  In this case, the work has to be calculated as the kinetic energy imparted to the gas molecules by the moving wall(s) of the container.  Equivalently, the pressure of a fluid out of equilibrium can be defined in terms of the momentum transferred to the external walls by particles bouncing off them.  This is, of course, unnecessarily complicated in many cases, but it is a formalism that can, in principle, be adapted to arbitrary irreversible expansion and compression processes, allowing for the calculation of the mechanical work done and the entropy generates.
