Question

Show that $dS > = 0$ , from the second law of thermodynamics and any other relation that is required.

Answer 1

Using the second law of thermodynamics, find the total entropy of the system and its surrounding. Then, find the total entropy when the process is spontaneous, non – spontaneous and when it is in equilibrium.

Complete step-by-step answer: We have to show that $dS > = 0$ from the second law of thermodynamics, so, you should know what is the second law of thermodynamics.
The statement of the second law of thermodynamics is, “the total entropy of an isolated system can never decrease over time and is constant if and only if all processes are reversible.” Isolated systems spontaneously evolve towards thermodynamic equilibrium, the state with maximum entropy. The total entropy of the surrounding system remains constant in ideal cases where the system is in thermodynamic equilibrium or it undergoes the reversible process. The total entropy of the system and surrounding increases and processes are irreversible in thermodynamic sense in all the processes which occur which also includes spontaneous processes.
The second law of thermodynamics can be expressed as –
$\Rightarrow \Delta {S_{universe}} = \Delta {S_{total}} = \Delta {S_{system}} + \Delta {S_{surrounding}} > 0$
There are many other cases for the total entropy which defines the type of process –
If, $\Delta {S_{total}} > 0$ , process is spontaneous
$\Delta {S_{total}} < 0$ , process is non – spontaneous
$\Delta {S_{total}} = 0$ , process is at equilibrium

Note: The thermodynamic property of all substances that is proportional to their degree of disorder is known as entropy. The more the number of possible microstates for the system the greater will be disorder and high will be entropy.