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A cook puts 7.20 g of water in a 2.00-L pressure cooker that is then warmed to 530°C. What is the pressure inside the container? (Assume room temperature is 20°C.)

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Josie

  1. Note: I used PV=nRT to solve (there are 0.400 mol), but the answer I got was incorrect; the hint I received was that all the water has evapourated.

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    Josie

  2. (2.00 L) x (273 / (20 + 273)) / (22.414 L/mol) = 0.083139 mol air in the "empty" container
    (7.20 g H2O) / (18.01532 g H2O/mol) = 0.39966 mol H2O added
    Supposing all the water vaporized:
    PV = nRT
    P = nRT / V = (0.083139 mol + 0.39966 mol) x (0.08205746 L atm/K mol) x
    (530 + 273 K) / (2.00 L) = 15.9 atm
    15.9atm x 101.3kPa = 1611.31

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    Aris

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