Question

The lower flammability limit of methanol in air is 6 mole %. Calculate the temperature at which a saturated methanol-air mixture at 1 atm would have a composition corresponding to the LFL.

Answer 1

LFL =6 mole %

Basis : 1 mole of air methanol mixture

Moles of methanol= 6/100 =0.06, moles of air = 1-0.06= 0.94

mole fractioof methanol= 0.06, partial pressure of methanol= mole fraction* total pressure = 0.06*1= 0.06 atm= 0.06*760 =45.6 mmHg

this is also the vapor pressure of methanol when satruated.

from Antoine equation for methanol

log P= 8.08097- 1582.27/(239.7+t)

P is in mm Hg and t in deg.c

log(45.6)= 8.08097- !582.27/(239.7+t)

1.658965= 8.08097- 1582.27/(239.7+t)

1582.27/(239.7+t)= 6.432

239.7+t= 1582.27/6.432=6.29 deg.c

This is the temperature at which the mixture is saturated.