Question

For the process $H_{2}O\left(l\right)\to H_{2}O(1bar,373K)$, the correct set of thermodynamic parameters is:

• A. $△G=0,△S=+ve$
• B. $△G=0,△S=-ve$
• C. $△G=+ve,△S=0$
• D. $△G=-ve,△S=+ve$

Answer 1

Answer: (A)

$△G=0,△S=+ve$

Due to equilibrium between liquid and gaseous state of $H_{2}O$ on $373K$, $\Delta G=0$

As the state changes from liquid to gas, the entropy also increases, $\therefore \Delta S=$ positive