For the vaporisation of a liquid at its normal boiling point, Delta_vapH = 38.6 kJ mol^-1 and Delta_vapS = 104 J K^-1mol^-1. The normal boiling point is about

JEE Main levelSingle correct optionDifficulty 2 of 3td102

For the vaporisation of a liquid at its normal boiling point, $Δ_{v a p} H = 38.6 kJ mo l^{- 1}$ and $Δ_{v a p} S = 104 J K^{- 1} mo l^{- 1}$ . The normal boiling point is about

At the boiling point liquid and vapour are in equilibrium, so

Δ G = 0

and

T_{b} = Δ H /Δ S = 38600/104 = 371.2 K

. The value

98

is the result in

^{\circ} C

, not in kelvin.