Standard enthalpy of vaporization Δ_ vap H^0 for water at 100^ ^ C...

The vaporization of water is represented by the equation:

$\text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(g)$

The change in the number of moles of gas is $\Delta n_g = 1 - 0 = 1$.

Using the relation between enthalpy change and internal energy change:

\Delta_{vap}H^{}^{\circ} = \Delta_{vap}U^{}^{\circ} + \Delta n_g RT

Rearranging for \Delta_{vap}U^{}^{\circ}:

\Delta_{vap}U^{}^{\circ} = \Delta_{vap}H^{}^{\circ} - \Delta n_g RT

Substituting the given values (T = 100^{}^{\circ}\text{C} = 373 \text{ K}, $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$):

\Delta_{vap}U^{}^{\circ} = 40.66 - (1 \times 8.314 \times 10^{-3} \times 373)

\Delta_{vap}U^{}^{\circ} = 40.66 - 3.101

\Delta_{vap}U^{}^{\circ} = +37.559 \text{ kJ mol}^{-1}

Rounding to two decimal places, we get $+37.56 \text{ kJ mol}^{-1}$.

Answer: $+37.56$