Answer
From Chapter $19 \text { (see Eq. } 19-9)$, we have
$$\frac{N}{V}=\frac{p}{k T}=\frac{1.01 \times 10^{5} \mathrm{Pa}}{\left|1.38 \times 10^{-23} \mathrm{J} / \mathrm{K} | \ | 273 \mathrm{K} \ | \right.}=2.68 \times 10^{25} \mathrm{m}^{-3}$$
or an ideal gas under "standard conditions.' Thus,
$$\frac{n_{p}}{|N / V|}=\frac{3.14 \times 10^{31} \mathrm{m}^{-3}}{2.44 \times 10^{25} \mathrm{m}^{-3}}=1.2 \times 10^{6}$$
Work Step by Step
From Chapter $19 \text { (see Eq. } 19-9)$, we have
$$\frac{N}{V}=\frac{p}{k T}=\frac{1.01 \times 10^{5} \mathrm{Pa}}{\left|1.38 \times 10^{-23} \mathrm{J} / \mathrm{K} | \ | 273 \mathrm{K} \ | \right.}=2.68 \times 10^{25} \mathrm{m}^{-3}$$
or an ideal gas under "standard conditions.' Thus,
$$\frac{n_{p}}{|N / V|}=\frac{3.14 \times 10^{31} \mathrm{m}^{-3}}{2.44 \times 10^{25} \mathrm{m}^{-3}}=1.2 \times 10^{6}$$
