A stockroom supervisor measured the contents of a partially filled 25.0 gallon acetone drum on a day when the temperature was 18^(@)C and atmospheric pressure was 780 mm Hg and found that 15.4 gallon of the solvent remained. After tightly sealing the drum a student 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 gallon. What is the total pressure inside the drum volume was decreased to 20.4 gallon. What is the total pressure inside the drum after the accident ? The vapour pressure of acetone at 18^(@)C is 400 mm Hg.

A stockroom supervisor measured the contents of a partially filled 25.

1.	A stockroom supervisor measured the contents of a partially filled 25.0 gallon acetone drum on a day when the temperature was 18^(@)C and atmospheric pressure was 780 mm Hg and found that 15.4 gallon of the solvent remained. After tightly sealing the drum a student 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 gallon. What is the total pressure inside the drum volume was decreased to 20.4 gallon. What is the total pressure inside the drum after the accident ? The vapour pressure of acetone at 18^(@)C is 400 mm Hg.
Answer» Solution :At the TIME the drum was sealed, the pressure inside the drum which is equal to the sum of the pressures of air and acetone, is equal to the atmospheric pressure. Thus, `p_("air") = 780 - 400 = 380 mm` Mole of air `= (pV)/(RT) = ((380)(25-15.4))/(RT)` After the ACCIDENT, mole of air will remain the same. `p_("air") = (nRT)/(V)` `= ((380) xx 9.6)/(RT) xx (RT)/((20.4 - 15.4))` = 729.6 mm. `therefore` total pressure `= p_("air") + p_("acetone")` `= 729.6 + 400` `= 1129.6 mm`.