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

The correct answer is: ∆U = 0, ∆S_total ≠ 0

Let us analyze this step by step by examining the two thermodynamic quantities — internal energy change (∆U) and total entropy change (∆S_total) — for an irreversible isothermal expansion of an ideal gas.

Step 1: Analyzing ∆U (Internal Energy Change)

For an ideal gas, internal energy (U) is a function of temperature only. This is because ideal gas molecules have no intermolecular interactions, so U depends solely on kinetic energy, which depends only on temperature.

Mathematically, for an ideal gas:

$U=f\left(T\right)$

Since the process is isothermal (constant temperature), there is no change in temperature:

$∆T=0$

Therefore:

$∆U=n{C}_v∆T=0$

This result holds true regardless of whether the process is reversible or irreversible. Internal energy is a state function — it depends only on the initial and final states, not the path. Since the temperature doesn't change, ∆U = 0 in both cases.

Step 2: Analyzing ∆S_total (Total Entropy Change)

Total entropy change is the sum of the entropy change of the system and the entropy change of the surroundings:

$∆S_{\mathrm{total}}=∆S_{\mathrm{system}}+∆S_{\mathrm{surroundings}}$

Now, here is the crucial distinction between reversible and irreversible processes:

• For a reversible process: ∆S_total = 0 (this is the definition of a thermodynamically reversible process).

• For an irreversible process: ∆S_total > 0 (this follows from the Second Law of Thermodynamics).

The Second Law states that for any spontaneous (irreversible) process, the entropy of the universe always increases:

$∆S_{\mathrm{total}}>0$

Since the question specifies an irreversible expansion, the total entropy change must be positive, meaning:

$∆S_{\mathrm{total}}\ne 0$

Step 3: Understanding why ∆S_total ≠ 0 for irreversible expansion

In an irreversible isothermal expansion, the work done by the gas against external pressure is less than the work done in a reversible expansion. Since ∆U = 0, we know from the First Law that q = W. Therefore, the heat absorbed from the surroundings in the irreversible case is less than in the reversible case.

While ∆S_system remains the same (entropy is a state function and depends only on initial and final states), ∆S_surroundings is less negative in the irreversible case (since less heat is absorbed from the surroundings). The net result is that ∆S_total > 0.

Conclusion:

For an irreversible isothermal expansion of an ideal gas:
• ∆U = 0 (because U depends only on T for an ideal gas, and T is constant)
• ∆S_total ≠ 0 (because the Second Law demands ∆S_total > 0 for any irreversible process)

Hence, the correct option is ∆U = 0, ∆S_total ≠ 0.