mathematical expression for lowering vapour of pressure
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Raoult established that the lowering of vapour pressure depends only on the concentration of the solute particles and it is independent of their identity.
establishes a relation between vapour pressure ofthe solution, mole fraction and vapour pressure of the solvent, i.e.
Suppose 1 is solvent 2 is solute
P
solution
=
P1
=
P0
1
X1
Psolution = P1 = P01X1
Lowering of vapour pressure
=
ΔP =
P0
1
X1
∆P = P01X1
Lower of vapour pressure
= ΔP =
P0
1
−
P1
=
P0
1
(1−
X1
)
= ∆P = P01-P1= P01(1-X1)
ΔP =
P0
1
.
X2
∆P = P01. X2
or
ΔP
P1
0
=
X2
.
∆PP10 = X2.
2nd..
For an ideal solution, the equilibrium vapor pressure is given by Raoult's law as
p= pa*xa + pb*xb + ….,,
where pi* is the vapor pressure of the pure component (i= A, B, ...) and xi is the mole fraction of the component in the solution
For a solution with a solvent (A) and one non-volatile solute (B), pb*=0 and p= pa*xa
The vapor pressure lowering relative to pure solvent is
Δp= pa*- p= pa*(1-xa) = pa*xb,
which is proportional to the mole fraction of solute.
establishes a relation between vapour pressure ofthe solution, mole fraction and vapour pressure of the solvent, i.e.
Suppose 1 is solvent 2 is solute
P
solution
=
P1
=
P0
1
X1
Psolution = P1 = P01X1
Lowering of vapour pressure
=
ΔP =
P0
1
X1
∆P = P01X1
Lower of vapour pressure
= ΔP =
P0
1
−
P1
=
P0
1
(1−
X1
)
= ∆P = P01-P1= P01(1-X1)
ΔP =
P0
1
.
X2
∆P = P01. X2
or
ΔP
P1
0
=
X2
.
∆PP10 = X2.
2nd..
For an ideal solution, the equilibrium vapor pressure is given by Raoult's law as
p= pa*xa + pb*xb + ….,,
where pi* is the vapor pressure of the pure component (i= A, B, ...) and xi is the mole fraction of the component in the solution
For a solution with a solvent (A) and one non-volatile solute (B), pb*=0 and p= pa*xa
The vapor pressure lowering relative to pure solvent is
Δp= pa*- p= pa*(1-xa) = pa*xb,
which is proportional to the mole fraction of solute.
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