To find the change in internal energy for the vaporization of water under the given conditions, we'll use the following relationship between enthalpy change ($$\Delta_{\text{vap}} H$$) and internal energy change ($$\Delta_{\text{vap}} U$$):

$$\Delta_{\text{vap}} H = \Delta_{\text{vap}} U + P \Delta V$$

For vaporization, the change in volume ($$\Delta V$$) can be approximated by considering the volume of the vapor because the volume of liquid water is relatively small compared to the volume of the vapor.

The ideal gas law gives us:

$$P V = n R T$$

Since we are dealing with 1 mole of water:

$$V = \frac{R T}{P}$$

Plugging this into the enthalpy change equation, we get:

$$\Delta_{\text{vap}} H = \Delta_{\text{vap}} U + P \left( \frac{R T}{P} \right)$$

This simplifies to:

$$\Delta_{\text{vap}} H = \Delta_{\text{vap}} U + R T$$

Rearranging for $$\Delta_{\text{vap}} U$$:

$$\Delta_{\text{vap}} U = \Delta_{\text{vap}} H - R T$$

Given:

$$\Delta_{\text{vap}} H = 40.79 \, \text{kJ mol}^{-1}$$ (or 40790 J/mol)

$$R = 8.3 \, \text{JK}^{-1} \text{mol}^{-1}$$

$$T = 100^\circ \text{C} + 273.15 = 373.15 \, \text{K}$$

Now, substitute the values into the equation:

$$\Delta_{\text{vap}} U = 40790 \, \text{J mol}^{-1} - (8.3 \, \text{JK}^{-1} \text{mol}^{-1} \times 373.15 \, \text{K})$$

$$\Delta_{\text{vap}} U = 40790 \, \text{J mol}^{-1} - 3097.145 \, \text{J mol}^{-1}$$

$$\Delta_{\text{vap}} U = 37692.855 \, \text{J mol}^{-1}$$

Converting back to kJ:

$$\Delta_{\text{vap}} U = 37.69 \, \text{kJ mol}^{-1}$$

Rounding to the nearest integer, the change in internal energy for the vaporization of water under the given conditions is:

38 kJ mol^-1