JEE Mains Chemistry Past PapershardMCQ SINGLE

Ice at –5ºC is heated to become vapour with temperature of 110ºC at atmospheric pressure. The entropJEE 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

Answer: B

💡 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 +SV +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

💬
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes
Ask on WhatsApp →

Practice 22,000+ questions like this

AI-adaptive practice, video lectures, and full JEE Mains Chemistry content — all in one place.

JEE Advanced · JEE Mains · NEET · IChO · AP Chemistry