Question

$1$ mole of ice at $0^{o}C$ and $4.6mm$ $Hg$ pressure is converted to water vapour at a constant temperature and pressure. Find $\Delta H$ and $\Delta E$ if the latent heat of fusion of ice is $80$ Cal/gm and latent heat of vaporisation of liquid water at $0^{o}C$ is $596$ Cal per gram and the volume of ice in comparison to that of water vapour is neglected.

Answer 1

$1$ mole of ice$=18g.H_{2}O\left(s\right)$

$△H=80\times 18+596\times 18=12.168Kcal$

$H_{2}O\left(s\right)⟶H_{2}O\left(g\right)$

$△H=△E+△n_{g}RT$

here ${△}_{ng}=$(gaseous moles of product)-(gaseous moles of reactant)

$=1$

$△E=△H-Rt=12.168-2\times 273\times {10}^{-3}$

$△E=11.62Kcal$