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Question 15.8.3: Calculate ∆G for the conversion of 1 mole of water at 100 °C......

Calculate ∆G for the conversion of 1 mole of water at 100 °C and 1 atm to steam at 100 °C and 0.5 atm pressures.

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∆G = (∆G for the conversion of H_2O (l) at 100 °C and 1 atm to steam at 100 °C and 1 atm).

+ (∆G for the transition of steam at 100 °C from 1 atm to 0.5 atm).

The first term on the RHS is zero, as H_2O (l) and H_2O (g) are in equilibrium at 100 °C and 1 atm pressure.

The second term is =2.303 \mathrm{~nRT}  \log \frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}

\begin{array}{l}\Delta G=2.303 \times 1 \times 1.987 \times 373 \times \log _{10} \frac{0.5}{1}\\  \\=2.303 \times 1 \times 1.987 \times 373 \times(-0.3010) \\  \\=-513.82 \mathrm{~cal}\end{array}

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