# dS = dq/T for irreversible processes

How can ds = dqr/T (d-bar q(reversible) / T) can be used to determine the entropy increase of a system in an irreversible process. Why does this technique works?

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Suppose some system undergoes an irreversible process. It was initially in some state of thermal equilibrium at some pressure volume, temperature (thermal equilibrium means simply that quantities like temperature are well defined) and it evolves into another state of thermal equilibrium at some (other) volume, temperature, pressure. The entropy in this new state will, in general, be different. The total entropy of the system and the rest of the universe will always become larger or stay the same. If the total entropy has become larger, then the process is irreversible. This is because to restore the old state of the system plus rest of the universe would require the total entropy to go down, which is impossible.

Suppose then that we are given a certain irreversible process. The total entropy thus increases. Nevertheless, in practice we are not interested to what happens to the rest of the universe. All we care about is the system that evolves from some initial state of thermal equilibrium to some final state of thermal equilibrium. Then, we can just as well consider realizing the same change to the system in a different way so that the whole process is reversible. It turns out that this is always possible.

Simple example: Consider a cylinder filled with a gas at some temperature. Suppose you take some cloth and rub it against the cylinder, thereby heating it via friction. Clearly, this is an irreversible process. Work is being dissipated as heat. I think that it ...

#### Solution Summary

We explain how one can apply the equation dS = dq/T to compute entropy increase for irreversible processes by considering a few examples.