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When water is boiled under a pressure of 2.00atm, the heat of vaporization is 2.20×106j/kg and the boiling point is 120∘ c. at this pressure, 1.00kg of water has a volume of 1.00×10−3m3, and 1.00 kg of
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When water is boiled under a pressure of 2.00atm, the heat of vaporization is 2.20×106j/kg and the boiling point is 120∘

c. at this pressure, 1.00kg of water has a volume of 1.00×10−3m3, and 1.00 kg of steam has a volume of 0.824m3.?

User Assen Kolov
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If we are asked for the work done in forming 1 kg of steam at this pressure and temperature, we use the formula.
W = -P(V2 - V1)
W = 2 atm (101325 Pa/1 atm) (0.824 m3 - 1.00x10-3 m3)
W = 166,780.95 J or 166.78kJ
User Ludlow
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