Question

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

Answer 1

The partial vapor pressure of hexane at 68°C above the mixture of cyclohexane-hexane can be calculated using Raoult's Law.

Step-by-step explanation:

The partial vapor pressure of hexane at 68°C above the mixture of cyclohexane-hexane that contains 34% of hexane by volume can be calculated using Raoult's Law.

Raoult's Law states that the partial vapor pressure of a component in a mixture is equal to the product of its mole fraction in the mixture and the vapor pressure of the pure component at that temperature.

In this case, the mole fraction of hexane is 0.34 and the vapor pressure of pure hexane at 68°C is 760 mm Hg. Therefore, the partial vapor pressure of hexane can be calculated as:

Partial vapor pressure of hexane = (0.34)(760 mm Hg) = 258.4 mm Hg

Answer 2

the vapor pressure will be pV= 298.4 mm Hg

Step-by-step 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