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

part b )

Kavg = 3*k*T/2

T = 291. C = 302.1 K

k = boltzman constant = 1.38 x 10^-23 J/K

Kavg = 4.16898 x 10^-21 J

in part A ) e = 157 x 10^-22 cal = 659.086 x 10^-22 J

e/Kavg = 15.809

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