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.

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)

Low pressure and high temperature, because particles are spread farther apart and moving faster, so the intermolecular forces of attraction are weaker

Explanation:

At low pressure and high temperature, gas molecules are separated and do not associate appreciably.

Recall that the the kinetic theory of gases holds that there is only a negligible intermolecular interaction between gases hence particles of a gas are relatively free and move about randomly at maximum velocities and kinetic energy.

This is only possible at low pressure and high temperature. At low pressure, gas particles are separated from each other and do not interact appreciably. High temperature imparts maximum kinetic energy to gas particles. Hence the answer.