Question

a) For an ideal mixture of two liquids A and B, show that the pressure p! at which the mole fractions of A in the gas and liquid phases are most different from one another is given by the geometric mean of the partial pressures of pure A and B.

Answer 1

& for an ideal mixture of two liquids A&B show that the pressure P at which the mole fractions of A in the liquid phases are most different from one another is given by the geometric mean of the partial pressure A & B. Ang. The relation that shows between Pressure & mole fraction of A in the galoous Phase is given by a P = PA PA+PB P = PXA Pa partial press POXA + Pg XB PRE Partial presse here g liquid A. 3 of liquid B. PA+PB PA= Pure vapour pressure PoXA + Po XB B = Pure vapoure Pressure of B How For two liquid mixture, (volatile). Raoulty law is applicable Alc to Raoultry law, partial pressure of any volatile liquid A is equal to the Product betuween mole fraction of A & pure rapour pressure I PA = Poxa el Po= P8 X 8 )

Ale to Daltonn law of partial pressure, Total Pressure= sumoy all partial pressure Protal = PA + PB = sum of partial pressure of two volatile liquid = PAXA + PBXB = PxH + Po (17XA) = PAXA + P - PETA IY = ctma Straight line havine positive & slope C Posit intercept I Total Pressun Now, Total pressure For Volatile & non- volatile liquid Pr = PATPR 1 → X (mole fraction) (let liquid Bin (non-volatile (Pg=0 = PAXA + PgX.

(intel CORE 13 Du - mole fraction ho hraph for Ideal & non ideal solution With respect to mole fraction. s This graph pro Pra PatPB represents & K lo Pp Ideal 801MB ецаБои XA=1 mole fraction XA=0 XB=1 for Хg=o Non ideal 8014 O Positive Deviation 10 Negative deviation - x =1 mole x mole faction faction B-1 XBT0 XA=0