JEE Main 2015 — Thermodynamics (C) Question with Solution

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{H}}_2\mathrm{O}\left(\mathcal{l}\right)$

At melting point of ${\mathrm{H}}_2\mathrm{O}$ at (273 K = T)

dG = VdP - SdT

At constant temperature

$\mathrm{d}\left(∆\mathrm{G}\right)=∆\mathrm{V}\left(\mathrm{d}\mathrm{P}\right)=\left(\frac{\mathrm{M}}{{\mathrm{\rho }}_{\mathrm{L}}}-\frac{\mathrm{M}}{{\mathrm{\rho }}_{\mathrm{S}}}\right)\mathrm{d}\mathrm{P}$

${\mathrm{\rho }}_{\mathrm{L}}=Density of Liq. H_{2}O$

${\mathrm{\rho }}_{\mathrm{S}}=Density of ice$

$\mathrm{d}\left(∆\mathrm{G}\right)=\mathrm{M}\left(\frac{{\mathrm{\rho }}_{\mathrm{S}}\mathrm{ }-\mathrm{ }{\mathrm{\rho }}_{\mathrm{L}}}{{\mathrm{\rho }}_{\mathrm{L}}\mathrm{ }\cdot \mathrm{ }{\mathrm{\rho }}_{\mathrm{S}}}\right)\mathrm{d}\mathrm{P}$

Integrating

$∆\mathrm{G}=\mathrm{M}\left(\frac{{\mathrm{\rho }}_{\mathrm{S}}\mathrm{ }-\mathrm{ }{\mathrm{\rho }}_{\mathrm{L}}}{{\mathrm{\rho }}_{\mathrm{L}}\mathrm{ }\cdot \mathrm{ }{\mathrm{\rho }}_{\mathrm{S}}}\right)(\mathrm{P}-1)$

Pressure increased from 1 atm to P atm

${\mathrm{\rho }}_{\mathrm{S}}<{\mathrm{\rho }}_{\mathrm{L}}\mathrm{ },\mathrm{ }\mathrm{P}>1$

$∆\mathrm{G}$ is -ve, so melting in forward direction spontaneous.

  Method 2 :
                              $Ice\stackrel{}{\rightleftharpoons }Water$

On Increasing P : Molecule  Come closer & density increasing; water is formed;

                              Ice → Water;

                     Since  $S_{ice}<S_{water}$

 $\because Process\; From\; Ice\; to\; water\; is\; spontaneous;$
    $\Delta G<0$

In physical equilibrium, on increasing pressure, equilibrium shift toward more density side.