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Two liquids A and B are miscibnle over the whole range of compsition and may be treated as ideal. At 350 K, the vapour pressure of pure liquid A is 24.0 kPa and that of pure liquid B is 12.0 kPa. A miozture of 60% of A and 40% of B is distilled at this temperature. What is the pressure in a colse contaiunbiner from which arir is excluded ? A small amount of distillate is collected and redistilled at the same temperature. What is the compositon os the second distillate ? |
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Answer» `"For liquid A":""x_(A)=0.60" and "P_(A)^(@)=24.0 kP_(a)` `"For liquid B":""x_(B)=0.40" and "P_(B)^(@)=12.0 kP_(a)`. `"Partial pressure of A in the liquid mixture "(P_(A))=P_(A)^(@)P_(A)=(24.0kP_(a))xx(0.60)=14.4 kP_(a)` `"Partial pressure of A in the liquid mixture "(P_(B))=P_(B)^(@)x_(B)=(12.0kP_(a))xx(0.40)=4.8 kP_(a)` `"Total pressure in the flask "=14.4+4.8=19.2 kP_(a)` Step II. Calculation of the composition of the second distillate. `"Form the second distillate, only liquid A will distil since it is LOW boiling as is evident from its higher vapour pressure "(24kP_(a)) "COMPAREDTO the liquid B "(12 kP_(a))`. `"Mole fraction of A in the vapoure state "(x'A)=((14.4kP_(a)))/((19.2kP_(a)))=0.75` `"Mole fraction of B in the vapoure state "(x'B)=(1-0.75)=0.25` `"Partial pressure A in the maxiture "(P'_(A))=P_(A)^(@)x_(A)=24xx0.75=18kP_(a).` `"Partial pressure B in the maxiture "(P'_(B))=P_(B)^(@)x_(B)=12xx0.25=3 kP_(a).` `"Mole fraction of A in the second distilatte"=P_(A)^(@)/(P'_(A)+P'_(B))=(18kkP_(a))/(21 kP_(a))=0.857` Thus, the second distillate will contain only 85.7% of A. |
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