Question

Enter your answer in the provided box A stockroom supervisor measured the contents of a partially filled 25.0-gallon acetone drum on a day when the temperature was 18.0°C and atmospheric pressure was 756 mmHg, and found that 15.4 gallons of the solvent remained. After tightly sealing the drum, an assistant dropped the drum while carrying it upstairs to the organic laboratory. The drum was dented and its internal volume was decreased to 20.4 gallons. What is the total pressure inside the drum after the accident? The vapor pressure of acetone at 18.0°C is 400.0 mmHg. mm Hg

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Answer #1

the original gas V seems to be 25.0 - 15.4 = 9.6
and after the dent 20.4 - 15.4 = 5.0
then
V1P1 = V2P2
9.6(756) = 5.0(P2)
P2 = 1406 mm

I have assumed that the amount of acetone compressed back into liquid by the drop will be so small that it won't materially affect liquid volume, at least not to the extent that it will show up in tenths of gallons

or

The vapor pressure of acetone is told to us to be 400. torm that is independent of the volume because as the volume decreases some of the acetone will retun to the liquid phase. The total pressure before the accident is 750. tor, that means that the rest of the vapor is ordinary air, which we will treat as an ideal gas. The volume of the vapor (because the liquid is not compressible) has decreased from (25.0-15.4) = 9.6 gallons to (20.5-15.4) = 5.1 gallons; since PV = nRT we can assume PV isa constant for the ordinary air, and if the volume is decreased from 9.6 to 5.1 (whatever) units, the pressure due to the ordinary air must increase from 350. torr to 350. x 659 tom. Thus the total pressure in the cylinder will be 659+400 1059 torr. 5.1

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