Problem 2 For the same system in Problem 1, the process is run until 50 mol% of the benzene originally in the still-pot has been vaporised. Determine a) the amount of o-xylene remaining in the still pot b) the amount and composition of the distillate c) which of the runs takes longer
Problem 1
A simple (i.e. single equilibrium stage) batch still is being used to separate benzene
from o-xylene; a system which may be assumed to have a constant relative volatility
of 6.7. The feed to the still is 1000 mol of 60 mol% benzene. The process is run until
the instantaneous distillate composition is 70 mol% benzene. Determine:
a) the composition and amount of the residue remaining in the still pot
b) the amount and average composition of the distillate
c) the time required for the process to run if the boil-up rate is 50 mol/h
Problem 2
For the same system in Problem 1, the process is run until 50 mol% of the benzene
originally in the still-pot has been vaporised. Determine
a) the amount of o-xylene remaining in the still pot
b) the amount and composition of the distillate
c) which of the runs takes longer
plz solve question 2 and explain every step
Step by step
Solved in 3 steps
Amount of benzene in Still pot:
ABW=W·xwABW=200 mol
Can you explain why did you write 200 mol instead of 300 mol because 50% of the benzene has evaporated so the remaining should be 300 mol not 200 mol
by Bartleby ExpertAmount of benzene in Still pot:
ABW=W·xwABW=200 mol
how did this become 200 when 300 mol of benzene has evaporated
600 -300 = 200 ?
by Bartleby ExpertAmount of benzene in Still pot:
ABW=W·xwABW=200 mol
should not this be 300 mol as 50% has evaporated? (300+300 =600)?
AOFAOWα=ABFABW400AOW6.7=600200AOW=339.506 mol
and what does this step mean? what is AOFAOW and ABFABW400AOW6.7
by Bartleby ExpertAmount of benzene in Still pot:
ABW=W·xwABW=200 mol
Amount of o-xylene in feed:
AOF=F×1-xFAOF=1000×0.4AOF=400 mol
Can you explain these two steps plz
by Bartleby Expert
