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What will be the partial vapor pressure of hexane at 68oC above the mixture of cyclohexane - hexane that contains 34 % of hexane by volume if it is known that the vapor pressure of pure hexane at the given temperature is 760 mm Hg

Respuesta :

Answer:

the vapor pressure will be pV=  298.4 mm Hg

Explanation:

Assuming ideal behaviour of liquid mixture of the cyclohexane - hexane, then the Raoult equation applies:

pL= p⁰*x

where pL = partial pressure of the liquid in the mixture  , p⁰= partial pressure of pure hexane  , and  x= partial molar fraction of hexane in the mixture =  0.34

(if we assume that they have approximately the same density , 34 (V/V)% = 0.34 of partial molar fraction of hexane )

Then since the liquid mixture is at equilibrium with the vapor , the partial vapor pressure pV should be equal to the liquid vapor pressure. Thus

pV=pL

pV= p⁰*x = 760 mm Hg* 0.34 = 298.4 mm Hg

pV=  298.4 mm Hg

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