Question

What is $△U$ when $3.0\text{mol}$ of liquid water evaporates at $100{}^{\circ }C?$
The heat of evaporation, $△H_{vap}$ of water at $100{}^{\circ }C$ is $40.66$ ${\text{kJ mol}}^{-1}$.

• A. $75.11\text{kJ}$
• B. $85.46\text{kJ}$
• C. $105.78\text{kJ}$
• D. $112.68\text{kJ}$

Answer 1

The correct option is D $112.68\text{kJ}$

$H_{2}O\left(l\right)\to H_{2}O\left(g\right):△n_g=1$
for 3 moles
$3H_{2}O\left(l\right)\to 3H_{2}O\left(g\right):△n_g=3$
$△U=△H-△n_g\times RT=3\times 40.66-3\times 8.314\times 373\times {10}^{-3}=112.68\text{kJ}$