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One mole of H2O(g) at 1.00 atm and 100.°C occupies a volume of 30.6 L. When one mole of H2O(g) is condensed to one mole of H2O(l) at 1.00 atm and 100.°C, 40.66 kJ of heat is released. If the density of H2O(l) at this temperature and pressure is 0.996 g/cm3, calculate ΔE for the condensation of one mole of water at 1.00 atm and 100.°C.

Respuesta :

Answer:

ΔE (of water)= -37,56 KJ

Explanation:

Using the first law of thermodynamics:

ΔE= Q - W = ΔH - P(Vf - Vi)

- The heat released by water in its condensation diminishes its internal energy --> Negative contribution

- The work done to compress the water should increase its internal energy --> positive contribution (Vf< Vi) , Vf= mf/Df

W= P(Vf - Vi) = 1 atm * (101325 Pa/atm) * [  1/(0,996gr/cm3)* 18 gr/mole*1 mole - 30,6 L * 1m3/1000L]] * 1 KJ/1000J = -3,01 KJ

ΔE= Q - W = (-40,66 KJ/mole)*1 mole - (-3,01 KJ) = -40,66 KJ + 3,01 KJ = (-37,56 KJ)

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