The air in a jar of home-canned green beans is heated to the boiling point of
water at 100oC and the jar is sealed at that temperature. Assume that the
pressure is 1.00 atm when the jar is sealed. What will be the pressure in the
jar when it is cooled to room temperature at 20.0oC?
2007-10-26 09:01:46 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Using a combined Charles/Boyles Laws.
PV/T = pv/t
assuming the volume does not alter at 1cm^3, then
1 atm x 1cm^3 /100oC = p x 1 cm^3 / 20oC
p = 1 atm x 1 cm^3 x 20oC / 1 cm^3 x 100oC
p = 0.2 atm.
2007-10-26 09:29:11 · answer #1 · answered by lenpol7 7 · 0⤊ 0⤋
At 100 deg C the vapor pressure of water is 1 atm. The headspace will be filled with water vapor. On sealing and cooling the head space will contain only water vapor. The vapor pressure of water at 20.0 deg C is 17.5 mm Hg (0.023 atm).
2007-10-26 09:18:00 · answer #2 · answered by skipper 7 · 0⤊ 0⤋