Question

$1$ mole of ice at $0$ and $4.6mm$ $Hg$ pressure is converted to water vapor at a constant temperature and pressure. Find $△H$ and $△E$ if the latent heat of fusion of ice is $80Cal/gm$ and latent heat of vaporisation of liquid water at $0$ is $536$ Cal per gram and the volume of ice in comparison to that of water(vapour) is neglected.

Answer 1

Hence ice get converted to vapour

$\Delta H=\Delta H_f+\Delta H_v=80\times 18+596\times 18$

$\Delta H=12168cal/mol$

$\Delta H=\Delta U+P\Delta V$ $-$ According to first law of thermodynamics

Now $PV=nRT$

$P\Delta V=nRT=1\times 8-314\times \frac{278}{0.18}=543cal$

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

$\Delta U=12168-543cal$

$\Delta U=11625cal$

Hence, $\Delta H$ is $12168cal$ and $\Delta U$ is $11625cal$.