Question

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?

Answer 1

a) ε = 6.961 × 10⁻²⁰Joules

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

Step-by-step 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