Water standing in the open at 29.1°C evaporates because of the escape of some of

Question

Water standing in the open at 29.1°C evaporates because of the escape of some of the surface molecules. The heat of vaporization (523 cal/g) is approximately equal to n, where is the average energy of the escaping molecules and n is the number of molecules per gram. (a) Find . (b) What is the ratio of to the average kinetic energy of H2O molecules, assuming the latter is related to temperature in the same way as it is for gases?

I have the answer for part A at 157E-22 Cal, but cannot figure out part B.

Explanation / Answer

(b)

As, Eav = 3/2 kT, therefore the ratio of to the average kinetic energy Eav of H2O molecule will be,

/ Eav = / (3/2 kT)

           = 2 / 3 kT

To obtain the ratio of to the average kinetic energy Eav of H2O molecule, substitute 157xE-22 J for , 1.38×10-23J/K for k and 32° C for T in the equation / Eav= 2 / 3 kT, we get,

/ Eav= 2 / 3 kT

           = 2(157xE-22)/3(1.38×10-23 J/K)( 29.1° C)

           = 2(157xE-22)/3(1.38×10-23 J/K)( 29.1+273) K)

          =2(157xE-22)/3(1.38×10-23 J/K)( 302.1K)

          = 2.510606111

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