1. Basic facts for an ideal gas

For an ideal gas:

• Internal energy depends only on temperature.
• Hence, $\Delta U = nC_V(T_2-T_1)$
• Work done by the gas in expansion is $w_{\text{by}}=\int p_{\text{ext}}\,dV$ and work done on the gas is the negative of this.

We now check each option.

2. Option A

Statement: The work done on the gas is maximum when it is compressed irreversibly from $({p_2},{V_2})$ to $({p_1},{V_1})$ against constant pressure $p_1$.

For compression from $V_2$ to $V_1$ ($V_2>V_1$), if the external pressure is constant and equal to $p_1$, then work done on the gas is $w_{\text{on}}=p_1(V_2-V_1)$ which is the maximum possible for irreversible compression between these end states, because to complete the compression up to the final state, the largest constant opposing pressure possible is $p_1$.

Also, reversible compression gives the minimum work on the gas among compression paths, while irreversible compression against a suitably larger constant pressure requires more work.

So A is correct.

3. Option B

Statement: If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabatic.

In free expansion:

• External pressure is zero, so $w=0$
• If the system is insulated, then $q=0$ so the process is adiabatic.

For an ideal gas, $\Delta U=q+w=0$ Hence, $nC_V(T_2-T_1)=0 \Rightarrow T_2=T_1$ So temperature remains constant, i.e. the process is isothermal as well.

Thus for an ideal gas, free expansion is both adiabatic and isothermal.

So B is correct.

4. Option C

Statement: The work done by the gas is less when it is expanded reversibly from $V_1$ to $V_2$ under adiabatic conditions as compared to that when expanded reversibly from $V_1$ to $V_2$ under isothermal conditions.

For reversible expansion, work done by gas is area under the $p$-$V$ curve.

• Reversible isothermal curve: $p=\frac{nRT}{V}$
• Reversible adiabatic curve: $pV^\gamma=\text{constant}$

During expansion from the same initial state, adiabatic pressure falls more rapidly than isothermal pressure, so at each intermediate volume, $p_{\text{adiabatic}}<p_{\text{isothermal}}$ Therefore area under adiabatic curve is smaller: $w_{\text{by, adiabatic}}<w_{\text{by, isothermal}}$

So C is correct.

5. Option D

Statement: The change in internal energy of the gas is (i) zero, if it is expanded reversibly with $T_1=T_2$, and (ii) positive, if it is expanded reversibly under adiabatic conditions with $T_1\, e \, T_2$.

Part (i): Since for an ideal gas internal energy depends only on temperature, $\Delta U=nC_V(T_2-T_1)=0$ if $T_1=T_2$. So part (i) is true.

Part (ii): In reversible adiabatic expansion of an ideal gas, temperature decreases: $T_2<T_1$ Hence, $\Delta U=nC_V(T_2-T_1)<0$ So the change in internal energy is negative, not positive.

Therefore the combined statement in D is false.

6. Final conclusion

The correct options are: $\boxed{A,\ B,\ C}$

7. Comparison with stored correct answer

Stored correct answer: A, B, C

This matches the derived answer exactly.