A swimmer coming out from a pool with a film of water weighing about 18 g .how much heat must be supplied to evapourite this water?also calculaye internal energy of vapiurisation at 100℃ given Enthalpy of vapiurisation at 373k is 40.66kj/mole

Asked by ashutosharnold1998 | 11th Aug, 2019, 12:37: AM

Expert Answer:

Given:
 
Mass of water = 18 gm
 
ΔvapH2O = 40.66 kJ/mol
 
R = 8.314×10−3 kJ/mol
 
T = 373 K
 
Evaporation of water is given as,
 
straight H subscript 2 straight O subscript open parentheses straight l close parentheses end subscript space space space rightwards arrow with Vaporisation on top space space straight H subscript 2 straight O subscript open parentheses straight g space close parentheses space end subscript space space space increment subscript space vap end subscript straight H space equals space 40.66 space kJ divided by mol
18 space gm space space space space space space space space space space space space space space space space space space space space space space space space 18 space gm space
1 space mol space space space space space space space space space space space space space space space space space space space space space space space space space space space space 1 space mol
 
ΔvapU = ΔH − PΔV
 
         =ΔH −ΔngRT
 
Δng = 1−0 
      = 1 mol
 
Therefore
 
ΔvapU = 40.66 − (1)(8.314×10−3)(373)
 
          = 40.66 − 3.10
 
ΔvapU = 37.55 kJ/mol
 
Internal energy of vaporisation is 37.55 kJ/mol

Answered by Varsha | 12th Aug, 2019, 11:19: AM

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