Answer
See answer.
Work Step by Step
Find the density of water vapor using the ideal gas law.
$$\frac{n}{V}=\frac{P}{RT}$$
$$\rho=\frac{m}{V}=\frac{Mn}{V}=M\frac{P}{RT}$$
$$=(0.0180\;kg/mol)\frac{1.013\times10^5 Pa}{(8.314)(273.15+100\;K)}=0.588\;kg/m^3$$
This is slightly off from the density value for steam in Table 10–1, $0.598\;kg/m^3$.
Because steam at this temperature is very close to condensing into liquid water, it’s not surprising that the ideal gas model doesn’t predict steam’s properties that well. The molecules do not interact solely via elastic collisions; there must also be an attractive force between them since they are starting to condense, i.e., stick together.