For irreversible expansion of an ideal gas under isothermal condition, the correct option is :

For an ideal gas the internal energy $U$ depends only on temperature. In an isothermal process the temperature is constant, therefore the change in internal energy is zero:

$\Delta U\; =\; 0$

During the expansion the gas goes from an initial volume $V\_1$ to a final volume $V\_2$ (with $V\_2\; >\; V\_1$). If the process were reversible, the entropy change of the gas would be

$\Delta S\_\{\backslash text\{gas\}\}\; =\; \backslash frac\{Q\_\{\backslash text\{rev\}\}\}\{T\}\; =\; \backslash frac\{nR\backslash ,\backslash ln\backslash !\backslash left(\backslash dfrac\{V\_2\}\{V\_1\}\backslash right)\}\{T\}$

Because the temperature is constant, the heat absorbed from the surroundings equals the work done by the gas, and the above expression is positive (since $\backslash ln(V\_2/V\_1)\; >\; 0$).

In an irreversible expansion, additional entropy is generated inside the system due to the lack of equilibrium during the change. This entropy production, often denoted $\sigma$, is always non‑negative and is zero only for a reversible path. Hence the total entropy change (system plus surroundings) becomes

$\Delta S\_\{\backslash text\{total\}\}\; =\; \Delta S\_\{\backslash text\{gas\}\}\; +\; \Delta S\_\{\backslash text\{surroundings\}\}\; +\; \sigma$

For an isothermal process, the heat exchanged with the surroundings is the same magnitude as in the reversible case, so

$\Delta S\_\{\backslash text\{surroundings\}\}\; =\; -\backslash frac\{Q\}\{T\}\; =\; -\Delta S\_\{\backslash text\{gas\}\}$

Substituting, we obtain

$\Delta S\_\{\backslash text\{total\}\}\; =\; \sigma \; \backslash ;\; >\; \backslash ;\; 0$

Thus, while the internal energy remains unchanged ($\Delta U\; =\; 0$), the total entropy of the universe increases because of irreversibility ($\Delta S\_\{\backslash text\{total\}\}\; \ne \; 0$).

Therefore the correct statement for irreversible isothermal expansion of an ideal gas is:

$\Delta U\; =\; 0,\backslash ;\; \Delta S\_\{\backslash text\{total\}\}\; \backslash neq\; 0$