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>> Raoults Law
According to this law, the partial pressure of any volatile constituent of a solution at a constant temperature is equal to the vapour pressure of pure constituent multiplied by the mole fraction of that constituent in the solution.
Let a mixture (solution) be prepared by mixing nA moles of liquid A and nB moles of liquid B. Let pA and pB be the partial pressures of two constituents A and B in solution and pA0 and pB0 the vapour pressures in pure state respectively.
Thus, according to Raoult’s law,
pA = nA/nA+nB pA0 = mole fraction of A × pA0 = XApA0
And pB = nB/nA+nB pA0 = mole fraction of B × pB0 = XBpB0
If the total pressure be P, then
P = pA + pB
= nA/nA+nB pA0 + nB/nA+nB pA0
= XAPA0 + XBPB0
This law, in fact, is the major deciding factor, whether a solution will be ideal or non-ideal. Ideal solutions obey Raoult’s law at every range of concentration. Non-ideal solutions do not obey Raoult’s law. They show either positive or negative deviation from Raoult’s law. for comparison between ideal and non-ideal solutions a table has been given on next page. (Only binary combinations of miscible liquids have been considered.)
Relation between Dalton’s Law and Raoult’s Law:
The composition of the vapour in equilibrium with the solution can be calculated applying Dalton’s law of partial pressures. Let the mole fractions of vapours A and B be YA and YB respectively. Let PA and PB be the partial pressures of vapours A and B respectively and total pressure P.
pA = YAP ..(i)
pB = YBP ..(ii)
pA = XAPA0 ..(iii)
pB = XBPB0 ..(iv)
Equating (i) and (iii)
YAP = XAPA0
or YA = XAPA0/P = pA/P
Similarly, equation (iii) and (iv)
YB = XBPB0/P = pB/P
Thus, in case of ideal solution the vapour phase is richer with more volatile component i.e., the one having relatively greater vapour pressure.