Question

Water standing in the open at 29.2°C evaporates because of the escape of some of the surface molecules. The heat of vaporization (548 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

Answer:

a) ε = 6.86x$10^{-20}$ J

b) ε/Kavg = 10.98

Explanation:

a) The heat of evaporation (Lv) is

Lv = εn

ε = Lv/n

n is the number of molecules in a gram of the substance. The molar mass of water (H2O) is 2x1 g of H + 16 g of O = 18 g/mol.

By the Avogadros' number(NA), we know that 1 mol = 6.02x$10^{23}$ molecules, so n will be:

n = NA/ molar mass

n = $\frac{6.02x10^{23} }{18}$

n = 3.34x$10^{22}$ molecules/g

So,

ε = $\frac{548}{3.34x10^{22} }$

ε = 1.64x$10^{-20}$ cal

As 1 cal = 4.18 J, ε = 6.86x$10^{-20}$ J

b) For ideal gases, the average of the kinetic energy (the internal energy of the molecules) is given by :

Kavg = $\frac{3}{2}$kT

Where Kavg is the average of the kinetic energy, k is the Boltzmann's constant and T is the temperature in Kelvin.

Knowing that k = 1.38x$10^{-23}$ J/K and T = 29.2ºC + 273 = 302.2 K

Kavg = $\frac{3}{2}$x1.38x$10^{-23}$x302.2

Kavg = 6.25x$10^{-21}$ J

So the ratio of ε to the average kinetic energy of H2O molecules is

ε/Kavg = $\frac{6.86x10^{-20} }{6.25x10^{-21} }$

ε/Kavg = 10.98