Q121 / 200 · MHT CET - Medical 2026 (21 April Shift 1) · Chemistry
Standard enthalpy of vaporization $\Delta_{vap}H^0$ for water at $100^\circ\text{C}$ is 40.66 kJ $\text{mol}^{-1}$. What is the internal energy of vaporization of water at $100^\circ\text{C}$ (in kJ $\text{mol}^{-1}$)?
- Correct answerA$+37.56$
- B$-43.76$
- C$+43.76$
- D$+40.66$
Solution
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: $+37.56$
$\text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(g)$
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: $+37.56$
Hence, the correct answer is option: 'A'.