Question

For a mixture of n-butane (1) + n-pentane(2) at 298.15 K, what would be the predicted bubble-point pressure using Raoult’s Law if your mixture was 20% n-butane by mole? Would you expect Raoult’s law to be a good model for this system? Why or why not

Answer 1

at 298.15

Saturation pressures : Butane : (P1sat) 2.43 bar n: penatane (P2sat): 0.68 bar

mole fraction of n butane x1=0.2 and mole fraction of n-pentane= 1-0.2= 0.8

From Raoults' law y1P= = x1P1sat (1) and y2P= x2P2sat (2) where x1,x2 are mole fractinos of n-butane and n-pentane in the liquid phase while y1 and y2 are mole fractions of n-butane and n-pentane in the vapor phase.

Addiing Eq.1 1 ans Eq.2 gives P=x1p1sat+x2p2sat= 0.2*2.43+0.8*0.68=1.03 bar

since the concentration of n-butane is only 20%, the solution can be considered dilute ( more of n-pentane ) and hence they nearly form ideal solution

Bubble point pressure=