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 change in internal energy ($\Delta U$) for water vaporization is determined using the equation $\Delta H = \Delta U + P\Delta V$. Given $\Delta_{vap}H^\circ = +40.49 \, \text{kJ mol}^{-1}$ at 1 bar (approximately 100 kPa). The work done ($P\Delta V$) for vaporization at constant temperature and pressure ($\Delta T = 0$) from liquid to gas is $P\Delta V = nRT$. For 1 mole of water at $100^\circ\text{C}$ (373.15 K), using $R = 8.3 \, \text{J K}^{-1} \text{mol}^{-1}$, the work done is $P\Delta V = 8.3 \times 373.15 \, \text{J mol}^{-1}$. Converting to kJ, $P\Delta V = \frac{8.3 \times 373.15}{1000} \, \text{kJ mol}^{-1} \approx 3.096 \, \text{kJ mol}^{-1}$. Solving for $\Delta U$: $\Delta U = \Delta H - P\Delta V = 40.49 - 3.096 \approx 37.39 \, \text{kJ mol}^{-1}$. Rounded to the nearest integer, $\Delta U = 37 \, \text{kJ mol}^{-1}$. This value of $37 \, \text{kJ mol}^{-1}$ is not strictly within the provided range of 38,38. The calculation for $\Delta U$ under these conditions is complete.