91Ó°ÊÓ

The initial temperature of the water $=25°\mathrm{C}$or $(273+25)\mathrm{K}$

The final temperature of the water$=30°\mathrm{C}$or $(273+30)\mathrm{K}$

Formula Used:

Gibb's free energy expression is

$\mathrm{Δ}\mathrm{G}=-\mathrm{S}\mathrm{Δ}\mathrm{T}+\mathrm{V}\mathrm{Δ}\mathrm{P}+\mathrm{μ}\mathrm{Δ}{\mathrm{P}}^{\mathrm{P}}$

The entropy of the water at atmospheric pressure is $\mathrm{S}=69.91\mathrm{J}/\mathrm{K}$.

Here, volume and pressure are constants.

Hence, the change in pressure is $\mathrm{Δ}\mathrm{P}=0$

The change in volume is $\mathrm{Δ}\mathrm{V}=0$

So, the equation $\mathrm{Δ}\mathrm{G}=-\mathrm{S}\mathrm{Δ}\mathrm{T}+\mathrm{V}\mathrm{Δ}\mathrm{P}+\mathrm{μ}\mathrm{AP}$becomes:

$\mathrm{Δ}\mathrm{G}=\mathrm{S}\mathrm{Δ}T+\mathrm{V}\left(0\right)+\mathrm{μ}\left(0\right)=\mathrm{S}\mathrm{Δ}T$

Substitute

$69.91\mathrm{J}/\mathrm{K}$for $\mathrm{S}$

$5\mathrm{K}$for $\mathrm{Δ}T$

Molar mass of a water molecule is $18\mathrm{g}/\mathrm{mol}$and one mole of water molecule is $18\mathrm{g}/\mathrm{molx}1\mathrm{mol}=18\mathrm{g}$The Density of the water is $1000\mathrm{kg}/{\mathrm{m}}^3$

Hence, the volume of the water molecule will be:

$V=\frac{18\mathrm{r}}{110\mathrm{k},/{\mathrm{m}}^2}×\frac{{\mathrm{Hin}}^2\mathrm{km}}{\lg }=18×{10}^{-6}{\mathrm{m}}^3$

Consider the pressure is increased and the Gibbs free energy remains constant

Change in Gibbs free energy $\mathrm{Δ}G=0$

If the change in pressure is $\mathrm{Δ}\mathrm{P}$

So, to bring out the required Gibb's free energy, increase the pressure by 194atm.