Physics · Thermodynamics · NEET
No, they are not identical. Every reversible process is quasi-static, but not every quasi-static process is reversible. A process is reversible only if it is quasi-static AND has no dissipative effects like friction or viscosity. So quasi-static is one of the two conditions needed for reversibility, not the whole thing.
If you change the system fast (for example, suddenly drop the external pressure), the piston accelerates, and the gas near the piston becomes different from the gas far away. Pressure and temperature are then not the same everywhere, so a single P and T cannot describe the gas. Doing it infinitely slowly keeps the system uniform and in equilibrium at every step, so P, V and T stay well defined.
Only the start point and end point can be marked. During a fast process the gas has no single pressure or temperature (it is not in equilibrium), so the in-between states are undefined. That is why the path is shown as a dashed or missing line. A quasi-static process passes through defined equilibrium states, so it gives a continuous solid curve on the P-V graph.
A very slow, gentle process is a good real-world approximation, but a truly quasi-static process is infinitely slow and is a hypothetical, ideal construct. In practice, a process that is slow enough and avoids sudden piston motion, big temperature gradients and dissipation is treated as approximately quasi-static.
Static means no change at all, the system just sits there. Quasi-static means nearly static: the state does change from initial to final, but so slowly that at every instant it is practically in equilibrium. The word quasi (meaning nearly) captures this idea of change happening almost without disturbing balance.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
A quasi-static process is an infinitely slow process in which the system stays in thermal and mechanical equilibrium with its surroundings at every stage. At each step the system's pressure and temperature differ from the surroundings only by an infinitesimal (very tiny) amount.
Because the gas stays in equilibrium throughout, every intermediate state has a definite P, V and T. This lets us plot the whole path on a P-V diagram and calculate work done as the area under that curve. Without it, work between two states could not be found from a graph.
It is an ideal, hypothetical construct because a truly infinitely slow process would take infinite time. Real processes that are slow and avoid sudden motion, large temperature gradients and friction are treated as good approximations to it.
Yes. Even though at every stage the gas has the same temperature as the surroundings, heat is still absorbed or released so the temperature stays constant while volume changes. Equal temperature does not mean zero heat flow in this ideal slow limit.
