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Liquid water in a closed container at 25 degrees C is in equilibrium with water vapor...

Liquid water in a closed container at 25 degrees C is in equilibrium with water vapor at a pressure of 0.0313 atmospheres. Consider a puddle on a sidewalk at 25 degrees C. If the relative humidity is 38% (i.e., the H2O(g) partial pressure is 0.012 atmospheres) what is the free energy change for evaporation of water from the puddle?

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Answer #1

So, here we have to find out the equilibrium constant between the water at the puddle and the atmosphere.

Now, the equilibrium constant is given by, Kp = pressure of water at puddle/ atmospheric pressure = 0.0313/0.012 = 2.608

Thus, the free energy change is obtained from  -RTlnK, AG=RTIn Kp

AG 8.314 x 298 x 1n2.608 -2374.96J/mol

This is the free energy change of vaporization.

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