An irreversible process occurring isothermally in an isolated system leads to
A. Zero entropy
B. An increase in the total entropy of the system
C. A decrease in the total entropy of the system
D. None of these

Hint: The second law of thermodynamics is that "entropy of the universe increases for an irreversible process." It remains constant for a reversible process.

Complete Step by Step Solution:
Entropy denoted by "S" is the degree of disorderness in a state of a thermodynamic system.
It simply indicates the abnormality in the structure or lack of uniformness in the structure of a thermodynamic system.

It is an extensive property as the measuring of entropy or entropy change relies on the amount of matter in a thermodynamic system.

It is a state function as its value does not rely on the path by which the system acquired a particular state but only on the initial and final state of the system.

A process is said to be isothermal if the temperature of the system remains constant during any step of the process.

An isolated system is a system that cannot exchange energy or matter with its surroundings.
We know that for n moles of an ideal gas, the entropy change is given by:-
\[{\rm{\Delta S = n}}{{\rm{C}}_{\rm{v}}}{\rm{ln}}\left( {\dfrac{{{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}}}} \right){\rm{ + nRln}}\left( {\dfrac{{{{\rm{V}}_{\rm{2}}}}}{{{{\rm{V}}_1}}}} \right)\]
For an isolated system, isothermal compression is not possible as if the surrounding performs work on the system, it will not be isolated anymore.

For an irreversible isothermal process occurring in an isolated system the entropy is given by the expression: -
\[{\rm{\Delta S = nRln}}\left( {\dfrac{{{{\rm{V}}_{\rm{2}}}}}{{{{\rm{V}}_1}}}} \right)\]
In this case, \[{{\rm{V}}_{\rm{2}}}{\rm{ > }}{{\rm{V}}_{\rm{1}}}\] which means there is an isothermal expansion of the ideal gas.
Considering the above equation, the entropy change is positive.
If the entropy change is positive then the entropy of the system increases.

So, option B is correct.

Note: Entropy of an irreversible reaction increases. For an isothermal irreversible reaction occurring in an isolated system, the entropy increases. This happens because in this case, only expansion can happen and expansion leads to an increase in entropy.