Question

A mixture containing 15.0 mol% hexane balance N2 flows through a pipe at 6.53 kmol/h. The pressure is 2.00 atm (abs) and the temperature is 100°C. If I want to begin to condense the hexane, what temperature do I need cool to? If I want to condense 80% of the hexane, what temperature do I need to cool to?

Answer 1

moles of hexane in 6.53 kmo/h of mixture= 6.53*0.15=0.9795 kmol/hr

moles of N2= 0.85*6.53=5.55 Kmol/hr

Partial pressure of nitrogen= 0.85*2= 1.7 atm

when 80% of the hexane condenses, hexane remaining= 0.9795*0.2=0.1959 Kmol/hr

Moles of nitrogen remain the same during condensation of hexane

after condensation

Moles of Hexane/ Moles of nitrogen = partial pressure of hexane/partial pressure of nitrogen

0.1959/5.5= partial pressure of hexane/ 1.7

partial pressure of hexane= 1.7*0.1959/5.5=0.06 atm

The temperatue to which the mixture needs to be cooled to corresponds a vapor pressure of hexane which is equal to 0.06 atm= 0.06*760=45.6 mm Hg

Antoine Equation for hexane can be written as

logP (mmHg)= A-B/C+T

A=7.01051 , B = 1246.33, C= 232.988 T is in deg.c

log (45.6)= 7.01051- 1246.33/232.988+T

1.66= 7.01051- 1246.33/232.988 +T

1246.33/(232.988+T)= 5.535

T+232.988= 1246.33/5.535=232.9589

T= 232.9589=232.988=-.03 deg.c

The mixture will have to be cooled to -0.03 deg.c