Incorrect relation is

Options

A. for isothermal reversible change `W=-P_(ext)(V_f-V_i)`
B. for isothermal reversible change `Q=2.303nRtlogV_f/V_i`
C. for isothermal reversible change `W=-2.303nRTlogP_f/P_i`
D. for adiabatic change `DeltaU=W`

Answer

To solve the question regarding the incorrect relation in thermodynamics, we will analyze each option provided in the context of isothermal reversible changes and adiabatic processes.

### Step-by-Step Solution:

1. **Understanding Work Done in Isothermal Reversible Change:**
- For an isothermal reversible process, the work done (W) can be expressed as:
\[ W = -P_{\text{external}} (V_f - V_i) \]
- This is a correct relationship as per IUPAC definitions.

**Hint:** Recall that work done in a thermodynamic process is often related to the change in volume and the external pressure.

2. **Relating Heat (Q) and Work (W):**
- In an isothermal process, the heat exchanged (Q) is equal to the negative of the work done:
\[ Q = -W \]
- Therefore, if work done is expressed as:
\[ W = -2.303 nRT \log \left(\frac{V_f}{V_i}\right) \]
- Then, the heat exchanged can be expressed as:
\[ Q = 2.303 nRT \log \left(\frac{V_f}{V_i}\right) \]
- This relationship is also correct.

**Hint:** Remember that in isothermal processes, temperature remains constant, and thus the internal energy change is zero for ideal gases.

3. **Work Done in Terms of Pressure:**
- The work done can also be expressed in terms of initial and final pressures:
\[ W = -2.303 nRT \log \left(\frac{P_i}{P_f}\right) \]
- This means that the statement "for isothermal reversible change, work done is equal to \(-2.303 nRT \log P_f / P_i\)" is incorrect. The correct form is \(-2.303 nRT \log P_i / P_f\).

**Hint:** Pay attention to the logarithmic relationship and the order of the pressures when converting between volume and pressure forms.

4. **Enthalpy Change in Isothermal Process:**
- For an ideal gas undergoing an isothermal process, the change in internal energy (\(\Delta U\)) is zero, leading to the conclusion that:
\[ \Delta U = Q + W \]
- Therefore, since \(\Delta U = 0\), we can conclude that:
\[ Q = -W \]
- This is consistent with the previous findings.

**Hint:** Use the first law of thermodynamics to relate internal energy, heat, and work.

5. **Adiabatic Process Relationship:**
- In an adiabatic process, the relationship is given by:
\[ \Delta U = W \]
- This is correct because there is no heat exchange (\(Q = 0\)).

**Hint:** Recall that in adiabatic processes, heat transfer is absent, leading to direct work being equal to the change in internal energy.

### Conclusion:
After analyzing all the options, we find that the incorrect relation is in option 3, where the work done is incorrectly stated as:
\[ W = -2.303 nRT \log \left(\frac{P_f}{P_i}\right) \]
The correct expression should be:
\[ W = -2.303 nRT \log \left(\frac{P_i}{P_f}\right) \]

### Final Answer:
The incorrect relation is in option 3.