Question

What is $\Delta E$ when $2.0$ mole of liquid water vaporises at ${100}^{o}C$? The heat of vaporisation, ($\Delta H$) of water at ${100}^{o}C$ is $40.66kJ{mol}^{-1}$

Answer 1

$\Delta vap{H}^o=\Delta U^o+\Delta ngRT$

(internal energy)

$\Delta vap{H}^o=40.66$ kJ/mol

$=2\times 40.66$ ($2$ moles mentioned in the question)

$=81.32kJ$

$2H_{2}O\left(l\right)\to 2H_{2}O\left(g\right)$

Change in no. of molecules moles $\Delta ng=2-0=2$

$\Delta U^o=\Delta vap{H}^o-RT=81.32-(2\times 8.314\times {10}^{-3}kJ/km$)

$\times 373K$

$\therefore \Delta U^o=75.12kJ/mol$.