A stockroom supervisor measured the contents of a 25.0 gal drum partially filled
ID: 1050464 • Letter: A
Question
A stockroom supervisor measured the contents of a 25.0 gal drum partially filled with acetone on a day when the temperature was 18.0 degree C and the atmospheric pressure was 750 mmHg, and found that 15.4 gal of the solvent remained. After tightly sealing the drum, an assistant dropped the drum while carrying it upstairs to an organic laboratory. The drum was dented, and its internal volume was decreased to 20.4 gal. What is the total pressure inside the drum after the accident? The vapor pressure of acetone at 18.0 degree C is 400 mmHg.Explanation / Answer
pressure of acetone + pressure of air =760 mm Hg
400 + pressure of air = 760 mmHg
pressure of air = (760-400)mm Hg = 350mm Hg
V.P of acetone at 180 C = 400 mmHg
This value is independent of the volume because as the volume decreases,
certain amount acetone which is in vapor state might return to liquid state
The total pressure in the drum, before the accident will be = 750 mm Hg
This indicates that the rest of the acetone saturated into the air can be
considered as ordinary air. Therefore consider this as an Ideal gas.
The volume of vapor before accident = 25 gal - 15.4 gal = 9.6 gal
The volume of vapor before accident = 20.4 gal - 15.4 gal = 5gal
Therefore the volume of vapor decreased from 9.6 gal to 5 gal.
PV=constant for air, If the volume decreases from 9.6 gal to 5 gal
then pressure increases from 350 mm Hg to 350*(9.6 / 5) mmHg
that is 672mm Hg (Boyle's law)
Therefore the total pressure inside the cylinder will be =672 +400 =1072 mm Hg
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