Entropy of vaporisation of water at 100°C, if molar heat of vaporisation is 9710 cal mol-1 will be

Question

Entropy of vaporisation of water at 100°C, if molar heat of vaporisation is 9710 cal mol-1 will be (A) 20 cal mol-1K-1 (B) 26 cal mol-1K-1 (C) 24 cal mol-1K-1 (D) 28 cal mol-1K-1

Tardigrade Admin · Accepted Answer

Δ Svap = Δ Hvap /Tb = 9710/373 = 26.032 cal K-1 mol-1

Q. Entropy of vaporisation of water at $10 0^{\circ}$ C, if molar heat of vaporisation is $9710 c a l$ $m o l^{- 1}$ will be

$20 c a l m o l^{- 1} K^{- 1}$

$26 c a l m o l^{- 1} K^{- 1}$

$24 c a l m o l^{- 1} K^{- 1}$

$28 c a l m o l^{- 1} K^{- 1}$

$Δ S_{v a p} = Δ H_{v a p} / T_{b}$ = 9710/373
= 26.032 cal $K^{- 1} m o l^{- 1}$

Q. Entropy of vaporisation of water at 100∘C, if molar heat of vaporisation is 9710cal mol−1 will be

20calmol−1K−1

26calmol−1K−1

24calmol−1K−1

28calmol−1K−1