Suppose you have a mole of water at $25°\mathrm{C}$and atmospheric pressure. Use the data at the back of this book to determine what happens to its Gibbs free energy if you raise the temperature to$30°\mathrm{C}$. To compensate for this change, you could increase the pressure on the water. How much pressure would be required?

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

$\Delta \mathrm{G}=-\mathrm{S}\Delta \mathrm{T}+\mathrm{V}\Delta \mathrm{P}+\mu \Delta {\mathrm{P}}^{\mathrm{P}}$

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 $\Delta \mathrm{P}=0$

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

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

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

Substitute

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

$5\mathrm{K}$for $\Delta T$

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:

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

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

Change in Gibbs free energy $\Delta G=0$

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

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