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Internal energy and heat of vaporization

  1. Apr 30, 2015 #1
    1. One gram of water occupies a volume of 1 cm3 at atmospheric pressure. When this amount of water is boiled, it becomes 1671 cm3 of steam. Calculate the change in internal energy for this vaporization process. The heat of vaporization for water is 2.26 x 106 J/kg.

    2. w = -PΔV, E = q + w

    3. w = -(1 atm)x(1.671 L - 0.001 L) x (101.3 J/L atm) = -169.2 J. q = ΔHvap = 2.26 x 106 J/kg x 0.001 kg = 2.26 x 103 J. E = 2.26 x 103 J - 169.2 J = 2090.8 J.

    For some reason I feel like something is missing, but there are no temperature changes given so I don't know what other steps there could be.
     
  2. jcsd
  3. May 1, 2015 #2

    ehild

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    I do not think there is any problem with your solutions. There are some little things, however. The energy equation is ΔE = q + w. The result should be given by 4 significant digits. Round up that 2090.8 J.
     
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