Question

For the process:
$H_2{O}_{\left(l\right)}\text{(1 bar, 373 K)}\rightleftharpoons H_2{O}_{\left(g\right)}\text{(1 bar, 373 K)}$
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

The correct option is A $\Delta G=0,\Delta S=+ve$

$\underset{\text{(1 bar, 373 K)}}{H_{2}O\left(l\right)}\rightleftharpoons \underset{\text{(1 bar, 373 K)}}{H_{2}O\left(g\right)}$
At $100{}^{\circ }\text{C}$, $H_{2}O\left(l\right)$ is in equilibrium with $H_{2}O\left(g\right)$ therefore, $\Delta G=0$. Because liquid molecules are converting into gaseous molecules, therefore randomness increases and hence, $\Delta S=+ve$