1. Key idea: isolated system

   For an isolated system, $q=0, \qquad w=0$ so from the first law, $\Delta U = q+w = 0.$

2. For an ideal gas

   The internal energy of an ideal gas depends only on temperature: $U = U(T).$ Hence, $\Delta U = nC_V (T_f-T_i).$

   Since $\Delta U=0$, we get $nC_V (T_f-T_i)=0 \Rightarrow T_f=T_i.$

3. Effect of reversible vs irreversible expansion

   Whether the expansion is reversible or irreversible does affect the path, but for an ideal gas in an isolated system, the total change in internal energy remains zero in both cases.

   Therefore, in both processes, $T_f = T_i.$

4. Check options

   • A: $(T_f)_{\text{irrev}} > (T_f)_{\text{rev}}$ ❌
   • B: $(T_f)_{\text{rev}} = (T_f)_{\text{irrev}}$ ✅
   • C: $T_f>T_i$ for reversible but $T_f=T_i$ for irreversible ❌
   • D: $T_f=T_i$ for both reversible and irreversible ✅

5. Subtlety about options B and D

   Since both final temperatures equal the same initial temperature, both statements $B$ and $D$ are true mathematically: $(T_f)_{\text{rev}}=(T_f)_{\text{irrev}}=T_i.$

   If forced to choose the most complete option, D is the best statement because it explicitly gives the actual relation with $T_i$.

Conclusion: $T_f=T_i \text{ for both reversible and irreversible expansion of an ideal gas in an isolated system.}$ So the correct choice should be D (and note that B is also implied true).