Answer
(1) As represented by Figure 19-9, the conversion of $(H_2O$ $l, 1 atm)$ into $(H_2O$$g, 1 atm)$ is a process whose both $\Delta H$ and $\Delta S$ > 0.
(2) Below 100$^{\circ} C$, ($H_2O$ $g, 1 atm)$ is condensed.
At 100$^{\circ}C$ , $\Delta G=0$; condensation and vaporization processes are at equilibrium.
(3) At 100 $^{\circ}C$, vaporization occurs, $\Delta G<0$, and $T$ $\Delta S$ > $\Delta H$.
Work Step by Step
(1) As represented by Figure 19-9, the conversion of $(H_2O$ $l, 1 atm)$ into $(H_2O$$g, 1 atm)$ is a process whose both $\Delta H$ and $\Delta S$ > 0.
(2) Below 100$^{\circ} C$, ($H_2O$ $g, 1 atm)$ is condensed.
At 100$^{\circ}C$ , $\Delta G=0$; condensation and vaporization processes are at equilibrium.
(3) At 100 $^{\circ}C$, vaporization occurs, $\Delta G<0$, and $T$ $\Delta S$ > $\Delta H$.
