Question

For the process ${H_2}O(l) \to {H_2}O(g)$, at $T = {100^o}C$ and 1 atmosphere, the correct choice is:
A) $\Delta {S_{system}} > 0{\text{ and }}\Delta {S_{surrounding}} > 0$
B) $\Delta {S_{system}} > 0{\text{ and }}\Delta {S_{surrounding}} < 0$
C) $\Delta {S_{system}} < 0{\text{ and }}\Delta {S_{surrounding}} < 0$
D) $\Delta {S_{system}} < 0{\text{ and }}\Delta {S_{surrounding}} < 0$

Answer 1

Hint: In the given process, water from liquid state is converted into gaseous state. This conversion is an equilibrium reaction and for equilibrium reactions, total entropy change is zero i.e., $\Delta {S_{total}} = 0$.
And, $\Delta {S_{total}} = \Delta {S_{system}} + \Delta {S_{surrounding}}$
Also, you should know that gas molecules are more randomly distributed than liquid molecules.

Complete step by step answer:
The symbol $S$ is denoted for a thermodynamic term called entropy. Entropy can be defined as a measure of degree of randomness or disorder in a system. It can also be thought of as a relative distribution of the energy within the system.
Now, let us understand the entropy of the process or system given in question. The given process is:
${H_2}O(l) \to {H_2}O(g)$
Here in this system, water molecules first are in liquid state i.e., ${H_2}O(l)$ and then they start converting in the gaseous state, i.e., ${H_2}O(g)$

The gaseous state of any substance possesses large positive values of entropy because gas molecules are more chaotically and randomly distributed than molecules of the liquid state. Then, entropy change for the given system ($\Delta {S_{system}}$) will be positive or we can write, $\Delta {S_{system}} > 0$ because there is an increase in entropy from liquid state to gaseous state.
And, total entropy change for the system, $\Delta {S_{total}} = \Delta {S_{system}} + \Delta {S_{surrounding}}$
But, $\Delta {S_{total}} = 0$ will be zero for the given process. Hence, $\Delta {S_{surrounding}} = - \Delta {S_{system}}$. Therefore, if entropy change for the system is positive, $\Delta {S_{system}} > 0$ then, entropy change for the surrounding $(\Delta {S_{surrounding}})$ will be negative or less than zero.
Or, $\Delta {S_{surrounding}} < 0$
So, the correct answer is “Option B”.

Note: The greater is the disorder or randomness in a system, the higher is the entropy. If the structure of the products is very much disordered than that of the reactants, then there will be a resultant increase in entropy as we also see in the given process of conversion of state of water. For any substance, the solid state is the state of lowest entropy and gaseous state is the state of highest entropy.