Question

For a mixture of n-butane (1) + n-pentane (2) at 25 degrees C, 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

For n-butane

LogP(mmHg)= 6.80776- 935.77/(238.789+T)

At 25 deg.c, logP1sat(mmHg)= 6.80776- 935.77/(238.789+25)=1821.132 mm Hg

For n-pentane

Log P(mm Hg)= A-B/(T+C)

Where A= 6.876 B= 1064.8 and C=233.01   T is in deg.c

Logp2sat(mmHg)= 6.876- 1064.8/ (233.01+25) at 25 deg.c

P(mmHg)= 561.04mm Hg

For 20% n –butane mixture

x1= 0.2 and x2= 0.8

Bubble point pressure = x1P1sat+x2P2sat=0.2*1821.132+0.8*561.04=813.0584 mm Hg

Raoults' law can be applied to dilute solutions. In this case, 20 mole % n-butane can be considered as dilute soluition.