For the process \[ \mathrm{H}_{2} \mathrm{O}(I) \rightarrow \mathrm{H}_{2} \mathrm{O}(g) \] at $T=100^{\circ} \mathrm{C}$ and 1 atmosphere pressure, the correct choice is

(B) $\Delta \mathrm{S}_{\text {system }}>0$ and $\Delta \mathrm{S}_{\text {surroundings }}<0$

(A) $\Delta \mathrm{S}_{\text {system }}>0$ and $\Delta \mathrm{S}_{\text {surroundings }}>0$

(C) $\Delta \mathrm{S}_{\text {system }} 0$

(D) $\Delta \mathrm{S}_{\text {system }}<0$ and $\Delta \mathrm{S}_{\text {surroundings }}<0$

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Chemical Thermodynamics

For the process

H_{2} O (I) \to H_{2} O (g)

T = 10 0^{\circ} C

and 1 atmosphere pressure, the correct choice is

Answer: (B)

For the vaporization of water at its boiling point (100°C, 1 atm):

**System entropy (

Δ S_{system}

)**: When liquid water converts to vapor, the disorder increases significantly, so

Δ S_{system} > 0

.

**Surroundings entropy (

Δ S_{surroundings}

)**: The process is endothermic (heat is absorbed from surroundings), so

Δ S_{surroundings} = - \frac{Δ H _{vap}}{T} < 0

.

At the boiling point, the process is at equilibrium, so

Δ G = 0

, which means

Δ S_{total} = Δ S_{system} + Δ S_{surroundings} = 0

. However, individually,

Δ S_{system} > 0

and

Δ S_{surroundings} < 0

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option