Entropy change involved in conversion of one mole of liquid water at 373 K to vapour at the same temperature (latent heat of vaporization of water =2.257 kJ g -1 ).

Question

Entropy change involved in conversion of one mole of liquid water at 373 K to vapour at the same temperature (latent heat of vaporization of water =2.257 kJ g -1 ). (A) 30.7 JK -1 mol -1 (B) 60.3 JK -1 mol -1 (C) 90.8 JK -1 mol -1 (D) 108.9 JK -1 mol -1

Tardigrade Admin · Accepted Answer

Δ H text vap =2.257 kJ / g =(2.257 × 18) kJ mol -1=40.626 kJ mol -1 Δ S text vap =(40.626/373)=0.1089 kJ K -1 mol -1

Q. Entropy change involved in conversion of one mole of liquid water at $373 K$ to vapour at the same temperature (latent heat of vaporization of water $= 2.257 k J g^{- 1}$ ).

$30.7 J K^{- 1} m o l^{- 1}$

$60.3 J K^{- 1} m o l^{- 1}$

$90.8 J K^{- 1} m o l^{- 1}$

$108.9 J K^{- 1} m o l^{- 1}$

$Δ H_{vap} = 2.257 k J / g = (2.257 \times 18) k J m o l^{- 1} = 40.626 k J m o l^{- 1}$

$Δ S_{vap} = \frac{40.626}{373} = 0.1089 k J K^{- 1} m o l^{- 1}$

Q. Entropy change involved in conversion of one mole of liquid water at 373K to vapour at the same temperature (latent heat of vaporization of water =2.257kJg−1 ).

30.7JK−1mol−1

60.3JK−1mol−1

90.8JK−1mol−1

108.9JK−1mol−1