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The heat of melting of ice at 1 atmosphere pressure and 0°C is 1.4363 kcal/mol. The volume of ice is 0.0196 liter and the volume of water is 0.018 liter. If 1 mole of ice is melted under these conditions, what will be a) The work done in cal? b) The change in internal energy in cal? c) The change in entropy in cal? Note that 1 literxatmosphere is equal to 101.325 joule

The heat of melting of ice at 1 atmosphere pressure and 0°C is 1.4363 kcal/mol. The volume of ice is 0.0196 liter and the volume of water is 0.018 liter. If 1 mole of ice is melted under these conditions, what will be a) The work done in cal? b) The change in internal energy in cal? c) The change in entropy in cal? Note that 1 literxatmosphere is equal to 101.325 joule

General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter18: Thermodynamics And Equilibrium
Section: Chapter Questions
Problem 18.98QP: When 1.000 g of ethylene glycol, C2H6O2, is burned at 25C and 1.00 atmosphere pressure, H2O(l) and...
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Sekton) The heat of melting of ice at 1 atmosphere pressure and 0°C is 1.4363 kcal/mol. The volume of ice is 0.0196 liter and the volume of water is 0.018 liter. If 1 mole of ice is melted under these conditions, what will be a) The work done in cal? b) The change in internal energy in cal? c) The change in entropy in cal? Note that 1 literxatmosphere is equal to 101.325 joule Given: C 1-4363 Kcal/male тся = 0.0196 liter Val of ice Vel of water = 0·018 liter (a) Work done W Pdv - = 1 (0-018-0-0196) W = 0.0016 cal (b) Change in internal energy AE = Q - W = 1-4363 X 10³-0-0016 AE = 1436-29 cal (C) Change in Entrepy (AS) = 2 AS = 1436-30 273 AS = 5-261 Liter xatm=101.325 Joule J= J⇒cal what??
Transcribed Image Text:Sekton) The heat of melting of ice at 1 atmosphere pressure and 0°C is 1.4363 kcal/mol. The volume of ice is 0.0196 liter and the volume of water is 0.018 liter. If 1 mole of ice is melted under these conditions, what will be a) The work done in cal? b) The change in internal energy in cal? c) The change in entropy in cal? Note that 1 literxatmosphere is equal to 101.325 joule Given: C 1-4363 Kcal/male тся = 0.0196 liter Val of ice Vel of water = 0·018 liter (a) Work done W Pdv - = 1 (0-018-0-0196) W = 0.0016 cal (b) Change in internal energy AE = Q - W = 1-4363 X 10³-0-0016 AE = 1436-29 cal (C) Change in Entrepy (AS) = 2 AS = 1436-30 273 AS = 5-261 Liter xatm=101.325 Joule J= J⇒cal what??
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