Question

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?

Short answer

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

Step-by-step solution

Explanation

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

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

Calculation

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.