04 Hess'sLaw - When water vaporizes at 100 oC and 1 atm...

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1 1 When water vaporizes at 100 o C and 1 atm pressure, 2.26 kJ/g of energy are transferred from the surroundings to the water. At 100 o C the density of liquid water is 0.958 g/mL, and the density of water vapor (steam) is 5.98x10 -4 g/mL. What is the enthalpy change, H vap , when 1 mole of water vaporizes? (A) 2.26 kJ (B) 40.7 kJ (C) 3.04 kJ H = E + P V 101.3 J = 1 L atm H 2 O = 18.0 g/mol 2 When 1 mol water vaporizes at 100 o C and 1 atm pressure, 40.7 kJ of energy are transferred from the surroundings to the water. At 100 o C the density of liquid water is 0.958 g/mL, and the density of water vapor (steam) is 5.98x10 -4 g/mL. How much work in joules is done in pushing back the atmosphere when mole of liquid water at 100 o C is turned into steam at 100 o C? H = E + P V 101.3 J = 1 L atm
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2 3 Asking for work done, I know work w = - P V And V = V final – V initial , and P = 1 atm is given. V initial = volume of 1 mol liquid water at 100 o C 18.0 g / (0.958 g/mL) = 18.79 mL = 18.7 x 10 -3 L V final = volume of 1 mol of steam at 100 o C 18.0 g /(5.98x10 -4 g/mL) = 3.010 x 10 4 mL 3.010 x 10 4 x 10 -3 L = 30.10 L So w = - 1 atm (30.10 L – 0.0187 L) = - 30.08 L atm Asks for work in joules so - 30.08 L atm (101.3 J / L atm) = - 3.05 kJ 4 When 1 mol water vaporizes at 100 o C and 1 atm pressure, 40.7 kJ of energy are transferred from the surroundings to the water and – 3.05 kJ of work is done in pushing back the atmosphere. The negative sign means work is done by the
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04 Hess'sLaw - When water vaporizes at 100 oC and 1 atm...

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