Water standing in the open at 33.0°C evaporates because of the escape of some of the surface molecules. The heat of vaporization (557 cal/g) is approximately equal to εn, where ε is the average energy of the escaping molecules and n is the number of molecules per gram. (a) Find ε. (b) What is the ratio of ε to the average kinetic energy of H2O molecules, assuming the latter is related to temperature in the same way as it is for gases?

a) ε = 6.961 * 10⁻²⁰Joules

b) The ratio of ϵ to the average kinetic energy of H2O molecules = 10.642

Explanation:

a) The formula to be used is given below as:

Heat of Vapourisation (Lv) = εn

Where: ε = is the average energy of the escaping molecules and

n = is the number of molecules per gram

The first step is to convert 557 cal/g to joules/kilogram (j/kg)

1 cal/g = 4186.8j/kg

557cal/g = ?

We cross multiply

557cal/g * 4186.8j/kg

= 2332047.6j/kg

Therefore, 557 cal/g = 2332047.6j/kg

ε = Lv/n

ε = LvM/n

Where Lv = 2332047.6j/kg

M = 0.018kg/mol

n = 6.03 * 10²³mol

ε = (2332047.6j/kg * 0.018kg/mol) : 6.03 * 10²³mol

= 6.961336119 * 10⁻²⁰Joules

Approximately, ε = 6.961 * 10⁻²⁰Joules

b) Kinetic energy = (3/2) KT

The ratio of ε to Kinetic energy = ε / (3/2) kT = 2ε / 3kT

Where ε = 6.961 * 10⁻²⁰Joules

k = 1.38 * 10⁻²³ Joules/kelvin

T = 33°C, which will be converted to kelvin as

33°C + 273K

= 306K

The ratio of ε to Kinetic energy will be calculated as

2ε / 3kT

= (2*6.961 * 10⁻²⁰ Joules) : (3 * 1.38 * 10⁻²³Joules/kelvin * 306K)

= 10.642

Hence, The ratio of ϵ to the average kinetic energy of H2O molecules = 10.642