Question

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

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

Answer 1

Answer: (B)

$188.494kJ$

Answer 2

: $\Delta H$ for 18 g water = $40.8kJ$

For 90 g water $= \frac{40.8}{18} \times 90 = 204kJ$

n for 90 g water = 90 /18 = 5

$\Delta n_{g} = 5 - 0 = 5$

$\Delta H = \Delta U + \Delta n_{g}RT$

$\Delta U = 204000 - (5 \times 8.314 \times 373)$

= 188494 J or $188.494kJ$