(II) Compare the value for the density of water vapor at exactly 100°C and 1 atm (Table 10–1) with the value predicted from the ideal gas law. Why would you expect a difference?

The ideal gas law is \(PV = RT\) for one mole of ideal gas.

Here, you have to use the equation\(\rho = \frac{M}{V}\).

The temperature of the water vapor is \[\].

The molar mass of the water is \(M = 18\;{\rm{g}} = 0.018\;{\rm{kg}}\).

Now you know, the density of the water is

\(\begin{array}{c}\rho = \frac{M}{V}\\V = \frac{M}{\rho }\end{array}\).

Now, substituting the values of the volume into the ideal gas equation, you getthe following:

\(\begin{array}{c}P\frac{M}{\rho } = RT\\\rho = \frac{{PM}}{{RT}}\\\rho = \frac{{\left( {1.013 \times {{10}^5}\;Pa} \right) \times \left( {0.018\;{\rm{kg/mol}}} \right)}}{{\left( {8.314\;{\rm{J/molK}}} \right) \times \left( {373\;{\rm{K}}} \right)}}\\\rho = 0.588\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\end{array}\)

The calculated value of the density of the water is lower than the given value in the table. At \({100^ \circ }{\rm{C}}\), you are assuming that the water vapor is an ideal gas, but there is a natural attraction between the molecules,so they don’t follow the ideal gas equation perfectly. Also, at that time, some water vapor may condense and create water droplets. This is the reason for the difference between the two values.