When a system is in equilibrium for all possible processes, the differential or finite change of entropy is:

negative

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positive

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zero

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none of these

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Solution

The correct option is A negative
For an equilibrium process, $△ S_{t o t a l}$ is zero i.e.
$△ S_{t o t a l} = 0$
We know that, $△ S_{t o t a l} = △ S_{s y s t e m} + △ S_{s u r r o u n d i n g}$
From 1st law of thermodynamics, we know that heat lost by the system must be absorbed by the surrounding
$△ S_{s u r r o u n d i n g} = \frac{q_{s u r r o u n d i n g}}{T_{s u r r o u n d i n g}}$
Since, heat is absorbed, $q_{s u r r o u n d i n g}$ is positive and hence $△ S_{s u r r o u n d i n g}$ will be positive
Hence $△ S_{s y s t e m} = - △ S_{s u r r o u n d i n g}$
i.e. $△ S_{s y s t e m}$ will be negative

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Similar questions

For a reversible process at equilibrium, the change in entropy may be expressed as:

Δ S_{t o t a l}

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