Chapter 5, Problem 5.148QP

Interpretation Introduction

Interpretation:

The distance between molecules of water at given conditions has to be estimated and commented on results.

Concept Introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T. An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

$\text{PV = nRT}$

Explanation of Solution

To calculate the moles of water vapour per liter using the ideal gas equation

$\begin{array}{l}\text{P}\text{=}\frac{\text{nRT}}{\text{V}}\\ \\ \frac{\text{n}}{\text{V}}\text{=}\frac{\text{P}}{\text{RT}}\text{=}\frac{\text{1}\text{.0}\text{atm}}{\left(\text{0}\text{.08206}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(100+273)K}}\text{=}\text{0}\text{.033}\text{mol/L}\end{array}$

Eventually want to find the distance between molecules. Therefore, let’s convert moles to molecules, and convert liters to a volume unit that will allow us to get to distance (${\text{m}}^{\text{3}}$).

$\left(\frac{\text{0}\text{.033}\text{mol}}{\text{1}\text{L}}\right)\left(\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}}{\text{1}\text{mol}}\right)\left(\frac{\text{1000}\text{L}}{\text{1}{\text{m}}^{\text{3}}}\right)\text{=2}{\text{.0×10}}^{\text{25}}{\text{molecules/m}}^{\text{3}}$

This can be the number of ideal gas molecules in a cube that is 1 m on each side.  Assuming an equal distribution of molecules along the three mutually perpendicular directions defined by the cube, a linear density in one direction may be found:

${\left(\frac{\text{2}\text{.0}\text{×}{\text{10}}^{\text{25}}\text{molecules}}{\text{1}{\text{m}}^{\text{3}}}\right)}^{\frac{\text{1}}{\text{3}}}\text{=2}\text{.7}{\text{10}}^{\text{8}}{\text{molecules/m}}^{\text{3}}$

This can be the number of molecules on a line 1 m in length. The distance between each vapour molecule is given by:

$\frac{\text{1}\text{m}}{\text{2}\text{.70}\text{×}{\text{10}}^{\text{8}}}\text{=}\text{3}\text{.7}\text{×}{\text{10}}^{\text{-9}}\text{m}\text{=3}\text{.7}\text{nm}$

Assuming a water molecule to be a sphere with a diameter of 0.3 nm, the water molecules are separated by over 12 times their diameter,

$\frac{\text{3}\text{.7}\text{nm}}{\text{0}\text{.3}\text{nm}}\approx \text{12}\text{times}$

A similar calculation can be done for liquid water. Starting with density, we convert to molecules per cubic meter.

$\frac{\text{0}\text{.96}\text{g}}{\text{1}{\text{cm}}^{\text{3}}}\text{×}\frac{\text{1}\text{mol}\text{H}\text{O}}{\text{18}\text{.02}\text{g}{\text{H}}_{\text{2}}\text{O}}\text{×}\frac{\text{6}\text{.022×}{\text{10}}^{\text{23}}}{\text{1}\text{mol}{\text{H}}_{\text{2}}\text{O}}\text{×}{\left(\frac{\text{100}\text{cm}}{\text{1}\text{m}}\right)}^{\text{3}}\text{=}\text{3}\text{.2}\text{×}{\text{10}}^{\text{28}}{\text{molecules/m}}^{\text{3}}$

This is the number of liquid water molecules in 1 m³. From this point, the calculation is the same as that for water vapor, and the space between liquid molecules is found using the same assumptions.

${\left(\frac{\text{3}\text{.2}\text{×}{\text{10}}^{\text{28}}\text{molecules}}{\text{1}{\text{m}}^{\text{3}}}\right)}^{\frac{\text{1}}{\text{3}}}\text{=3}\text{.2}{\text{10}}^{\text{9}}\text{molecules/m}$

$\frac{\text{1}\text{m}}{\text{3}\text{.2}\text{×}{\text{10}}^{\text{9}}}\text{=}\text{3}\text{.1}\text{×}{\text{10}}^{\text{10}}\text{m}\text{=}\text{0}\text{.31}\text{nm}$

Assuming a water molecule to be a sphere with a diameter of 0.3 nm, to one significant figure, water molecules are packed very closely together in the liquid, but much farther apart in the steam.

Conclusion

The number density of water molecules and the number of molecules in one direction was calculated.