A mole of steam is condensed at `100^(@)` C, the water is cooled to `0^(@)`C and frozen to i.e . What is the difference in entropies of the stem and ice? The heat of vaporization and fusion are `540 cal" "gm^(-1)` and `80 cal" "gm ^(-1)` respectively . Use the average heat capacity of liquild water as `1 cal" "gm^(-1)` `degree^(-1)`.

Answer

B

Explanation

`DeltaS_("condensation")= (-DeltaH_("vap"))/(T)=(-540 xx 18)/(373)`
`DeltaS_("cooling")= + nC_(p)ln .((T_(2))/(T_(2))) = 18 ln. ((273)/(373))`
`DeltaS_("fusion") = (-80 xx 18)/(273)`
`= - [18((80)/(273)+ (540)/(373)) + (18ln.(273)/(373))]`
`= - 18((80)/(273) +(540)/(373)+ ln.(373)/(273))" "rArr 36.95 cal//^(@)`