Question

The value for $\Delta U$ for the reversible isothermal evaporation of $90 \,g$ water at $100 \,{}^°C$ will be ($\Delta H_{evap}$ of water $= 40.8\, kJ\, mol^{-1}$, $R = 8.314 \,J \,K^{-1}\, mol^{-1}$)

• A. $4800 \,kJ$
• B. $188.494 \,kJ$
• C. $40.8 \,kJ$
• D. $125.03 \,kJ$

Answer 1

Answer: (B)

$\Delta H$ for $18 \,g$ water $= 40.8 \,kJ$
For $90 \,g$ water $= \frac{40.8}{18}\times90 = 204\,kJ$
$n$ for $90\, g$ water $= 90/18 = 5$
$\Delta n_{g} = 5 - 0 = 5$
$\Delta H = \Delta U+\Delta n_{g}RT$
$\Delta U = 204000-\left(5\times8.314\times373\right)$
$= 188494\, J$ or $188.494\, kJ$