Question

For the process :

$H_{2}O\left(l\right)$ ($1$ bar, $373$ $K$) $\to H_{2}O\left(g\right)$ ( $1$ bar, $373K$),

the correct set of thermodynamic parameters is

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

Answer 1

Answer: (A)

$\Delta G=0,\Delta S=+ve$

$H_{2}O\left(l\right)$ ( 1 bar, 373 $K$) $\to H_{2}O\left(g\right)$ ( 1 bar, 373 )

As this process is in equilibrium so $\Delta G=0$

Also, here water changes from liquid to more random state gas so entropy of system increases.

$\Delta S=+$ ve