JEE Challenger
More from Thermodynamics
ChemEasyAdv 2007 • Paper 2

Thermodynamic Parameters for Water-Vapor Equilibrium

For the process H2O(l) (1 bar,373 K)→H2O(g) (1 bar,373 K)\text{H}_2\text{O}(l)\ (1\text{ bar}, 373\text{ K}) \rightarrow \text{H}_2\text{O}(g)\ (1\text{ bar}, 373\text{ K}), the correct set of thermodynamic parameters is

Options

A

ΔG=0,ΔS=+ve\Delta G = 0, \Delta S = \text{+ve}

Correct
B

ΔG=0,ΔS=−ve\Delta G = 0, \Delta S = -\text{ve}

C

ΔG=+ve,ΔS=0\Delta G = \text{+ve}, \Delta S = 0

D

ΔG=−ve,ΔS=+ve\Delta G = -\text{ve}, \Delta S = \text{+ve}

Topics & Concepts

ThermodynamicsThermodynamicsFree EnergyEntropy of VaporizationPhase Equilibrium

Step-by-Step Solution

At 1 bar1\text{ bar} pressure and 373 K373\text{ K} (100∘C100^\circ\text{C}), liquid water and water vapor are at dynamic phase equilibrium (the normal boiling point).

  1. For any system in phase equilibrium at constant temperature and pressure:
    ΔG=0\Delta G = 0

  2. The transition from liquid to gaseous state involves a large increase in randomness and disorder, so the entropy change of vaporization is positive:
    ΔS=Sgas−Sliquid>0(ΔS=+ve)\Delta S = S_{\text{gas}} - S_{\text{liquid}} > 0 \quad (\Delta S = \text{+ve})

Hence, ΔG=0,ΔS=+ve\Delta G = 0, \Delta S = \text{+ve}.

Correct Answer: A

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