Question

What is $△U$ when 2.0 mole of liquid water vaporises at ${100}^{\circ }?$ The heat of vaporisation, $△H$ vap. of water at ${100}^{\circ }C$ is $40.66$ $kJmo{l}^{-1}$

• A. 60
• B. 75.11
• C. 88.2
• D. 59

Answer 1

The correct option is B 75.11

$H_{2}O\left(l\right)\to H_{2}O\left(g\right):△n_g=1$
for 2 moles
$2H_{2}O\left(l\right)\to 2H_{2}O\left(g\right):△n_g=2$
$△U=△H-△n_g\times RT=2\times 40.66-2\times 8.314\times 373\times {10}^{-3}=75.11kJ$