Thermodynamics and Thermochemistry - Result Question 100
####25. For the process, $H _2 O(l) \longrightarrow H _2 O(g)$ at $T=100^{\circ} C$ and 1 atmosphere pressure, the correct choice is
(a) $\Delta S _{\text {system }}>0$ and $\Delta S _{\text {surrounding }}>0$
(2014 Adv.)
(b) $\Delta S _{\text {system }}>0$ and $\Delta S _{\text {surrounding }}<0$
(c) $\Delta S _{\text {system }}<0$ and $\Delta S _{\text {surrounding }}>0$
(d) $\Delta S _{\text {system }}<0$ and $\Delta S _{\text {surrounding }}<0$
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Solution:
- PLAN This problem is based on assumption that total entropy change of universe is zero.
At $100^{\circ} C$ and 1 atmosphere pressure,
$$ \begin{aligned} & \qquad H _2 O(l) \rightleftharpoons H _2 O(g) \text { is at equilibrium. } \ & \text { For equilibrium, } \quad \Delta S _{\text {total }}=0 \ & \text { and } \Delta S _{\text {system }}+\Delta S _{\text {surrounding }}=0 \end{aligned} $$
As we know during conversion of liquid to gas entropy of system increases, in a similar manner entropy of surrounding decreases.
$\therefore \quad \Delta S _{\text {system }}>0$ and $\Delta S _{\text {surrounding }}<0$
