Water standing in the open at$\mathbf{32}\mathbf{.}\mathbf{0}\mathbf{°}\mathit{C}$evaporates because of the escape of some of the surface molecules. The heat of vaporization () is approximately equal to$\mathit{\epsilon }\mathit{n}$, where$\mathit{\epsilon }$is the average energy of the escaping molecules and is the number of molecules per gram.

1. Find$\mathit{\epsilon }$
2. What is the ratio of$\mathit{\epsilon }$to the average kinetic energy of${\mathit{H}}_{\mathbf{2}}\mathit{O}$molecules, assuming the latter is related to temperature in the same way as it is for gases?

• Temperature is $32°\text{C}=305\text{\hspace{0.17em}K}$
• Heat of vaporization is$539\text{\hspace{0.17em}cal}/\text{g}$

The expression for the average kinetic energy is given by,

$K_{avg}=\frac{3}{2}KT$

Here is the Boltzmann constant, T is the temperature, $K_{avg}$is the average kinetic energy.

The value of K is equal to the$1.38\times {10}^{-23}J/molecule\cdot K$

First, we have to find number of molecules.

The molar mass of water is as follows:

$\text{m}=2\left(\text{molarmassofH}\right)+\left(\text{molarmassofo}\right)$

$\begin{array}{c}\text{m}=2\left(1\frac{\text{g}}{\text{mole}}\right)+16\frac{\text{g}}{\text{mole}}\\ \text{m}=18\frac{\text{g}}{\text{mole}}\end{array}$

Now, number of molecules:

$\begin{array}{c}\text{n}=\frac{6.023\times {10}^{23}}{18}\\ \Rightarrow \text{n}=3.3461\times {10}^{22}\text{molecules}/\text{g}\end{array}$

Now, the average escape energy,

$\Delta =\frac{L}{n}$

Substitute $539\text{\hspace{0.17em}cal}/\text{g}$for L and $3.3461\times {10}^{22}\text{molecules}/\text{g}$for n into the above equation,

$\begin{array}{c}\Delta =\frac{539}{3.3461\times {10}^{22}}\\ =1.6108\times {10}^{-20}\text{cal}\\ =1.61\times {10}^{-20}\times 4.186\text{J}\\ =6.74\times {10}^{-20}\text{J}\end{array}$

Therefore the value of $\epsilon$is $6.74\times {10}^{-20}\text{J}$.

The translational kinetic energy is given by,

$K_{avg}=\frac{3}{2}kT$

Substitute $1.38\times {10}^{-23}J/molecule\cdot K$for k and $305K$for T into the above equation,

$\begin{array}{c}{\text{K}}_{avg}=\frac{3}{2}\times 1.38\times {10}^{-23}\times 305\\ =6.313\times {10}^{-21}\text{J}\\ \end{array}$

$\begin{array}{c}\frac{\Delta }{K_{avg}}=\frac{6.74\times {10}^{-20}}{6.313\times {10}^{-21}}\\ =10.67\\ \approx 10.7\end{array}$

Therefore the ratio of average escaping energy to average kinetic energy is$10.7$ .