Which of the following statements/relationships is not correct in thermodynamic changes?

We are given four options and need to determine which statement is not correct in thermodynamic changes. Let's evaluate each option step by step.

Starting with option A: $$\Delta U = 0$$ for the isothermal reversible expansion of a gas. For an ideal gas, the internal energy $$U$$ depends only on temperature. Since the process is isothermal, the temperature remains constant, so the change in internal energy $$\Delta U$$ must be zero. Therefore, option A is correct.

Now, option B: $$w = -nRT\ln\frac{V_2}{V_1}$$ for the isothermal reversible expansion of an ideal gas. The work done by the gas during a reversible expansion is given by $$w = -\int_{V_1}^{V_2} P dV$$. For an ideal gas, $$P = \frac{nRT}{V}$$. Substituting this, we get:

$$w = -\int_{V_1}^{V_2} \frac{nRT}{V} dV$$

Since $$n$$, $$R$$, and $$T$$ are constant (isothermal process), we can factor them out:

$$w = -nRT \int_{V_1}^{V_2} \frac{1}{V} dV$$

The integral of $$\frac{1}{V}$$ is $$\ln V$$, so:

$$w = -nRT \left[ \ln V \right]_{V_1}^{V_2} = -nRT \left( \ln V_2 - \ln V_1 \right) = -nRT \ln \frac{V_2}{V_1}$$

This matches option B exactly, so it is correct.

Option C: $$w = nRT\ln\frac{V_2}{V_1}$$ for the same process. From our derivation in option B, we have $$w = -nRT \ln \frac{V_2}{V_1}$$. Option C has a positive sign instead of negative. Therefore, option C is incorrect because it does not match the standard expression for work done by the gas during expansion.

Option D: For a system of constant volume, heat involved directly changes to internal energy. At constant volume, no work is done (since $$w = -P \Delta V$$ and $$\Delta V = 0$$, so $$w = 0$$). The first law of thermodynamics states $$\Delta U = q + w$$. Substituting $$w = 0$$, we get $$\Delta U = q$$. This means the heat added or removed directly changes the internal energy. Therefore, option D is correct.

Hence, the incorrect statement is option C.

So, the answer is Option C.