Prove that raoult's law follows as special case of henry law
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The Henry's law is given by:
p=k^h×x
where p is the partial pressure and x is the mole fraction.
kH is the proportionality constant (Henry's constant)
The Raoul'ts law is given by:
p=p*x
where p is the partial pressure and x is the mole fraction.
p* is the vapor pressure of the pure component.
So, Raoul'ts law is a special case of Henry's law when kH = p* .
The Henry's constant is approximately equal to the vapor pressure only for 'ideal' mixtures (mixtures obeying the Raoult's law over the entire range of composition) which doesn't entirely exist but closely related substances like benzene and toluene can form a close-to-ideal-mixture where this case applies.
In real-time scenario, the Raoult's law and the Henry's law are obeyed in opposite composition ranges. That is probably why, it's Raoul'ts law is a "special" case of Henry's law.
p=k^h×x
where p is the partial pressure and x is the mole fraction.
kH is the proportionality constant (Henry's constant)
The Raoul'ts law is given by:
p=p*x
where p is the partial pressure and x is the mole fraction.
p* is the vapor pressure of the pure component.
So, Raoul'ts law is a special case of Henry's law when kH = p* .
The Henry's constant is approximately equal to the vapor pressure only for 'ideal' mixtures (mixtures obeying the Raoult's law over the entire range of composition) which doesn't entirely exist but closely related substances like benzene and toluene can form a close-to-ideal-mixture where this case applies.
In real-time scenario, the Raoult's law and the Henry's law are obeyed in opposite composition ranges. That is probably why, it's Raoul'ts law is a "special" case of Henry's law.
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