The experimentally determined latent enthalpy of vaporization of water at exactly () is . Calculate — Thermodynamics and Thermochemistry Chemistry Question
Question
The experimentally determined latent enthalpy of vaporization of water at exactly $100^\circ\text{C}$ ($373 \text{ K}$) is $40.66 \text{ kJ mol}^{-1}$. Calculate the corresponding internal energy of vaporization ($\Delta U_{vap}$) for exactly 1 mole of water at this temperature in $\text{kJ}$. (Assume perfect ideal gas behavior for water vapor and use $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$. Format the final answer rounded to two decimal places).
💡 Solution & Explanation
The phase transition is represented as $H_2O(l) \rightarrow H_2O(g)$. The net change in the number of gaseous moles is $\Delta n_g = 1 - 0 = 1$. The fundamental thermodynamic relationship correlating enthalpy and internal energy is $\Delta H = \Delta U + \Delta n_g RT$. Rearranging this equation to solve for internal energy gives: $\Delta U = \Delta H - \Delta n_g RT$. Substituting the knowns: $\Delta U = 40.66 \text{ kJ} - (1 \times 8.314 \times 10^{-3} \text{ kJ K}^{-1}\text{mol}^{-1} \times 373 \text{ K}) = 40.66 - 3.101 = 37.559 \text{ kJ}$. Rounding to two decimal places provides $37.56$.