Rešenje
Rešenje
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<strong>Rešenje</strong>:• Na početkukondenzacije x A ’=0,4tj.:pA000,4 = Ptot= pAxA+ (1 − xA)pB= 1,2 − 0,8xAxAPtotP tot= 1 ,2 − 0,8 ⋅0,67= 0, 664atm=0,671 1,2 − 0,8 ⋅ x Ax ==A0,25
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