About 1 dm³ gastric juice are produced per day in the human body from other body fluids such as blood, whose pH is 7.6021 [[H⁺] = 10-^pH] at 53°C. Calculate the minimum work (in kJ) required to produce1 dm³ . p^H  of gastric juice may be taken as unity.

The production of gastric juice from body fluids is a slow process hence, the change may be considered as reversible one.

$\mathrm{W}=-2.303\mathrm{nRTlog}\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}$

$\text{ Concentration }\left[{\mathrm{H}}^{+}\right]\frac{\mathrm{mol}}{\mathrm{L}}\left(\propto \frac{1}{\mathrm{V}}\right)$

Thus, $\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}=\frac{{\mathrm{C}}_1}{{\mathrm{C}}_2}$

$\mathrm{W}=-2.303\mathrm{nRT} \log \frac{{\mathrm{C}}_1}{{\mathrm{C}}_2}$

$\begin{array}{l}\mathrm{n}=\text{ moles of }{\mathrm{H}}^{+}\text{ions present in }0.1\mathrm{M}\\ (\mathrm{pH}\left.=1,\left[{\mathrm{H}}^{+}\right]={10}^{-1}\mathrm{M}\right)\text{ of }1{\mathrm{dm}}^3\text{ juice }=0.1\\ {\mathrm{C}}_1={\left[{\mathrm{H}}^{+}\right]}_1={10}^{-\mathrm{pH}}={10}^{-7.6021}=2.5\times {10}^{-8}\mathrm{M}\\ {\mathrm{C}}_2={\left[{\mathrm{H}}^{+}\right]}_2={10}^{-\mathrm{pH}}={10}^{-1}=0.1\mathrm{M}\\ \mathrm{R}=8.3143\times {10}^{-3}\mathrm{kJ}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1},\\ \mathrm{T}=326\mathrm{K}\\ \mathrm{W}=-2.303\times 0.1\times 8.314\times {10}^{-3}\times 326K \log \frac{2.5\times {10}^{-8}}{0.1}\\ =+1.74\mathrm{kJ}\end{array}$