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w = -P∆V
P = 1.0 atm
To determine ∆V, use V1/T1 = V2/T2 and assume V1 = 1 liter and T2 = 373K and solve for V2
V2 = 1.4 L and ∆V = 0.4 L
w = -P∆V = -(1.0 atm)(0.4L) = -0.4 L-atm
-0.4 L-atm x 101.325 J/L-atm = -179 J of work done (note - sign as work is done by the gas on the surrounding)

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Q: Calculate the work done when 1.0 mole of water at 273 k vaporizes against as atmospheric pressure of 1.0 atmosphere. assume ideal gas behaviour.?
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