An isolated system does not exchange neither matter nor energy with its surroundings.

To determine the entropy variation of these systems, let’s consider a system that undergoes a cycle that consists of two processes: the irreversible process AB and then the reversible process BA to close the cycle (see the figure below).

We can split the cycle into N elementary Carnot cycles and, since it is irreversible, we will have the following inequality (if it were reversible we would have an equality):

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The integral over the whole cycle can be split into two integrals (one for each process) as given below:

But the second integral is the definition of the entropy change, so we have:

This result means that, when a system undergoes an irreversible process AB, the entropy change it undergoes is greater than the second term integral.

For an **isolated system**, the second term integral is always zero, because it does not exchange heat with its surroundings. Therefore, we have the following inequality (the equal sign is valid when the process is reversible):

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##### Entropy of the universe

The universe is an isolated system (if we consider the universe to be everything, then there is nothing for it to exchange matter and/or energy with). Moreover, all real life processes are irreversible, so for any process AB occurring in the universe, the entropy of the latter will increase. Always.

For practical purposes, and going back to a more earthly scale, **we will consider the universe to consist of the thermodynamic system and its surroundings**.

For instance, when we consider a heat engine, the universe consists of the heat engine (system) and the thermal reservoirs (surroundings) it exchanges heat with. If the heat engine works reversibly, then the entropy of the universe will not change; if the heat engine works irreversibly, the entropy of the universe will increase.

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