State and derive raoult's law

State and derive raoult's law
  • 1 answers

Himanshu Gautam 1 year, 4 months ago

Suppose a mixture contains 2 completely miseble volatile components A and B having mole fraction xA and xB. Therefore partial pressure of each component is pA=xA p°A eq1 pB=xB p°B eq2 Therefore P=pA+pB P=xA p°A+xB p°B eq3 We know xA+xB=1 xA=1-xB Putting value of xA in eq3 P=(1-xB)p°A+xB p°B P=p°A-xB p°A+xB p°B P-p°A=xB(p°B-p°A) xB=(P-p°A)÷(p°B-p°A) eq4 xB= mole fraction of solute xA= mole fraction of solvent P= vapour pressure of solution p°A= vapour pressure of pure solvent p°B= vapour pressure of pure solute We know xA=1-xB xA=1-(eq4) We get xA=(p°B-P)÷(p°B-p°A) Definition- In a solution, the vapour pressure of a component at a given temperature is equal to the mole fraction of that component in the solution multiplied by the vapour pressure of that component in the pure state.

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