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I need help with this problem

A 0.75–L bottle is cleaned, dried, and closed in a room where the air is 22oC and 44% relative humidity (that is, the water vapour in the air is 0.44 of the equilibrium vapour pressure at 22oC). The bottle is brought outside and stored at 0.0oC.

b) Would liquid water condense at 10oC?

I don't get the letter b part. I know how to get the mass of the H20 i am just stuck with letter B

  • Chemistry - ,

    vapor pressure H2O @ 22C = 19.8 mm Hg.
    vapor pressure H2O @ 0C = 4.6 mm Hg.

    How much water is trapped in the bottle at 22C (295K)?
    19.8mm x 0.44 = 8.71 mm
    n = PV/RT = (8.71 x 0.75/760*0.08206*295) = about 0.00035 mols but you should redo it more accurately.

    How many mols vapor can there be in the bottle after cooling to 0C?
    n = (4.6*0.75/760*0.08206*273) = about 0.0002 mols.
    You have more mols than that; therefore, some of it will condense. How much. That will be about
    0.00035 - 0.0002 = about 0.00015 and that x molar mass H2O = about 0.003 g (about 3 mg).

  • Chemistry - ,

    it is not deal with any delta H vapor? because i saw it changes from liquid to gas and have vapor pressure?

  • Chemistry - ,

    and if you use: pv/rt. p must be atm.

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