Question

$\Delta_{vap}H^\circ$ for water is $+40.49 \, \text{kJ mol}^{-1}$ at 1 bar and $100^\circ\text{C}$. Change in internal energy for this vaporisation under same condition is .............. kJ mol$^{-1}$. (Integer answer)
(Given R = 8.3 J K$^{-1}$ mol$^{-1}$)

Answer 1

Correct Answer: 38

The enthalpy of vaporization $\Delta_{\text{vap}}H^\circ$ is related to the internal energy change $\Delta_{\text{vap}}U^\circ$ by the equation:
\[ \Delta_{\text{vap}}H^\circ = \Delta_{\text{vap}}U^\circ + \Delta n_gRT. \]
For the vaporization of water:
\[ \text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(g), \]
\[ \Delta n_g = 1 \quad (\text{1 mol of gas is formed}). \]
Substitute the given values:
\[ \Delta_{\text{vap}}H^\circ = 40.49 \, \text{kJ/mol}, \]
\[ R = 8.3 \, \text{J K}^{-1} \text{mol}^{-1}, \quad T = 100^\circ \text{C} = 373.15 \, \text{K}. \]
Calculate $\Delta_{\text{vap}}U^\circ$:
\[ \Delta_{\text{vap}}H^\circ = \Delta_{\text{vap}}U^\circ + \Delta n_gRT, \]
\[ 40.49 = \Delta_{\text{vap}}U^\circ + \frac{1 \times 8.3 \times 373.15}{1000}. \]
Simplify: \[ 40.49 = \Delta_{\text{vap}}U^\circ + 3.0971, \]
\[ \Delta_{\text{vap}}U^\circ = 40.49 - 3.0971 = 37.6929 \, \text{kJ/mol}. \]
Thus: \[ \Delta_{\text{vap}}U^\circ \approx 38 \, \text{kJ/mol}. \]