Choose the correct statement(s):

Options

A. Temperature , enthalpy and entropy are state functions
B. For reversible and irreversible both isothermal expansion of an ideal gas, change in internal energy and enthalpy is zero
C. for a reaction in which `Deltan_(2) =0` , entropy change is not always zero
D. The entropy change associated with reversible isothermal expansion of an ideal gas is equal to `2.303` R `"log"_(10)(P_(1))/(P_(2))`

Answer

To solve the question, we need to evaluate each of the four statements provided and determine if they are correct or not.

### Step-by-Step Solution:

1. **Evaluate Statement A**: "Temperature, enthalpy, and entropy are state functions."
- **Explanation**: State functions are properties that depend only on the current state of the system, not on how it got there. Temperature, enthalpy, and entropy are all state functions.
- **Conclusion**: This statement is **correct**.

2. **Evaluate Statement B**: "For a reversible and irreversible both isothermal expansion of an ideal gas, the change in internal energy and enthalpy is zero."
- **Explanation**: In an isothermal process for an ideal gas, the temperature remains constant. Therefore, the change in internal energy (which depends only on temperature for an ideal gas) is zero. Since enthalpy is a function of internal energy and pressure, it is also zero in this case.
- **Conclusion**: This statement is **correct**.

3. **Evaluate Statement C**: "For a reaction in which ΔN2 is zero, entropy change is not always zero."
- **Explanation**: ΔN2 refers to the change in the number of moles of gas. Even if ΔN2 is zero (meaning the number of moles of gaseous reactants and products is the same), the entropy change can still be non-zero due to factors like changes in temperature or phase.
- **Conclusion**: This statement is **correct**.

4. **Evaluate Statement D**: "The entropy change associated with reversible isothermal expansion of an ideal gas is equal to 2.303 R log10(P1/P2)."
- **Explanation**: The formula for the change in entropy (ΔS) for a reversible isothermal expansion of an ideal gas is indeed given by ΔS = nR ln(V2/V1) or ΔS = nR ln(P1/P2). When using log base 10, it can be expressed as ΔS = 2.303 R log10(P1/P2).
- **Conclusion**: This statement is **correct**.

### Final Conclusion:
All four statements (A, B, C, and D) are correct.