At what temperature liquid water will be in equilibrium with water vapour? Δ Hvap - 40.73 kJ mol-1, Δ Svap = 0.109 kJ K-1 mol-1

Question

At what temperature liquid water will be in equilibrium with water vapour? Δ Hvap - 40.73 kJ mol-1, Δ Svap = 0.109 kJ K-1 mol-1 (A) 282.4 K (B) 373.6 K (C) 100 K (D) 400 K

Tardigrade Admin · Accepted Answer

At equilibrium Δ G = 0 for Δ G = Δ H - TΔ S or T = (Δ H/Δ S) = (40.73/0.109) = 373.6 K

Q. At what temperature liquid water will be in equilibrium with water vapour?
$Δ H_{v a p} - 40.73 k J m o l^{- 1}$ , $Δ S_{v a p} = 0.109 k J K^{- 1} m o l^{- 1}$

$282.4 K$

$373.6 K$

$100 K$

$400 K$

At equilibrium $Δ G = 0$ for $Δ G = Δ H - T Δ S$
or $T = \frac{Δ H}{Δ S} = \frac{40.73}{0.109}$
$= 373.6 K$

Q. At what temperature liquid water will be in equilibrium with water vapour? ΔHvap​−40.73kJmol−1, ΔSvap​=0.109kJK−1mol−1

282.4K

373.6K

100K

400K