ThermodynamicsHard
Question
An amount of 5 mole H2O(l) at 100o C and 1 atm is converted into H2O(g) at 100°C and 5 atm. ΔG for the process is
Options
A.zero
B.1865 ln 5 cal
C.3730 ln 5 cal
D.−3730 ln 5 cal
Solution
$H_{2}O\left( l,1\text{ atm,10}\text{0}^{o}C \right)\overset{\quad(1)\quad}{\rightarrow}H_{2}O\left( g,1\text{ atm, 10}\text{0}^{ο}C \right)\overset{\quad(2)\quad}{\rightarrow}H_{2}O\left( g,5\text{ atm, 10}\text{0}^{o}C \right)$
$\Delta G_{1} = 0\text{ and }\Delta G_{2} = nRT\ln\frac{P_{2}}{P_{1}} = 5 \times 2 \times 373 \times \ln\frac{5}{1} = 3730\ln 5\text{ cal}$
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