Ice at –5ºC is heated to become vapour with temperature of 110ºC at atmospheric pressure. The entrop — JEE Mains Chemistry Past Papers Chemistry Question
Question
Ice at –5ºC is heated to become vapour with temperature of 110ºC at atmospheric pressure. The entropy change associated with this process can be obtained from (A) K K m , p dT C + f m T fusion , H + b m T on vaporisati , H + K K m , p dT C + K K m , p dT C (B) K K m , p dT T C + f m T fusion , H + b m T on vaporisati , H + K K m , p T dT C + K K m , p T dT C (C) K K p dT C + T qrev (D) K K p dT C + Hmelting + Hboiling
💡 Solution & Explanation
S = ? H2O(s) Ice –5ºC H2O(g) 110ºC Solid H2O(s) 268 K () H2O(s) 273 K Fusion () H2O() 273 K Liq. () H2O() 373 K Vaporisation (V) H2O(g) 373 K gas (V) H2O(g) 383 K STotal = S + S +S +SV +SV = K K m , p dT T C + f m T fusion , H + K K m , p T dT C + b m T on vaporisati , H + K K m , p T dT C