Question

When $1$ mole of ice melt at $0^{\circ }C$ and at constant pressure of $1atm$. $1440$ calories of heat are absorbed by the system. The molar volumes of ice and water are $0.0196$ and $0.0180$ litre respectively. Calculate $△H$ and $△E$ for the reaction.

Answer 1

We know that, $\Delta H=\Delta E+P\Delta V$

now, $P\Delta V=1atm(0.0180-0.0196)=-1.6\times {10}^{-3}lit-atm$

$=-0.039cals$ (since 1 lit-atm = 24.4cal)

now, $\Delta H=1440Cal$

$\Delta E=\Delta H-P\Delta V$

$=1440-(-0.039)$

$\approx 1440Cals$