Strictly speaking, the ideal-gas law applies only to gases in thermal equilibrium, meaning that the state variables are constant and not changing. But, by definition, an ideal-gas process causes some of the state variables to change. The gas is not in thermal equilibrium while the process of changing from state 1 to state 2 is under way.

To use the ideal-gas law throughout, we will assume that the process occurs so slowly that the system is never far from equilibrium. In other words, the values of $p$, $V$, and $T$ at any point in the process are essentially the same as the equilibrium values they would assume if we stopped the process at that point. A process that is essentially in thermal equilibrium at all times is called a quasi-static process. It is an idealization, like a frictionless surface, but one that is a very good approximation in many real situations.

An important characteristic of a quasi-static process is that the trajectory through the $pV$ diagram can be reversed. If you quasi-statically expand a gas by slowly pulling a piston out, as shown in FIGURE 18.10a, you can reverse the process by slowly pushing the piston in. The gas retraces its $pV$ trajectory until it has returned to its initial state. Contrast this with what happens when the membrane bursts in FIGURE 18.10b. That is a sudden process, not at all quasi-static. The expanding gas is not in thermal equilibrium until some later time when it has completely filled the larger container, so the irreversible process of Figure 18.10b cannot be represented on a $pV$ diagram.

FIGURE 18.10 The slow motion of the piston is a quasi-static process. The bursting of the membrane is not.

The critical question is: How slow must a process be to qualify as quasi-static? That is a difficult question to answer. This textbook will always assume that processes are quasi-static. It turns out to be a reasonable assumption for the types of examples and homework problems we will look at. Irreversible processes will be left to more advanced courses.