derive expression for Raoult's law.
Answers
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Raoult's Law: According to this law 'The vapour pressure of solution containing non-volatile solute is directly proportional to the mole fraction of the solvent'.
For a solution of two components A (Volatile solvent) and B (non-volatile solute)
Vapour pressure of solution = Vapour pressure of solvent ∝ Mole fraction of solvent
Or p=p
A
∝X
a
Pr P
A
=KX
A
....(i)
Where, k= Proportionality constant
Apply equation (i) for pure solvent for this purpose, put (i) X
a
=1 and instead of p
A
put p
A
0
where P
A
0
= vapour pressure of pure solvent
∴ P
A
0
=KX
1
Or P
A
0
=K ....(ii)
According to equation (ii) the value of proportionality constant k is equal to the vapour pressure of pure solvent by putting the value of k in equation (i) we get
P
A
=P
A
0
X
A
Or p solution =p pure solvent × mole fraction of solvent.
Derive it mathematically: Mathematically, if P is the vapour pressure of the pure solvent and P
S
, that of the solution, X
S
is the mole fraction of the solvent in the solution, n are number of moles of the solute and N are the number of moles of the solvent in the solution, then
P
S
∝X
S
or P
S
=KX
S
(Where X
S
=
n+N
N
and K is constant ....(i)
In case of pure solvent n=0 hence
Mole fraction of solvent,
X
S
=
n+N
N
=
0+N
N
=1 and P
S
=p
P=k×1=k
From equation (i) P
S
=P×X
S
...(ii)
Thus, the vapour pressure of the solvent in the solution is equal to the product of its mole fraction and its vapour pressure in the pure state at the same temperature.