91Ó°ÊÓ

Convert the units of p and T to be in Pascal and kelvin respectively

p = l atm = 1.013 x 10⁵ Pa

T = 27C° + 273 = 300K

Calculate the number of nitrogen molecules N in the lecture room, and the number of molecules is given by

$\mathbf{N}\mathbf{=}{\mathbf{nN}}_{\mathbf{A}}$

where n is the number of moles and N_A is the Avogadro's number and equals 6.022 x 10²³ molecules/mol.

Calculate N we want to calculate the number of moles n. It can be calculated by Ideal gas law

$\mathbf{pV}\mathbf{=}\mathbf{nRT}$

now sub with all given values of p, V, R and T in the equation and solve for n

$\begin{array}{r}n=\frac{pV}{RT}\\ =\frac{1.013×{10}^5\mathrm{Pa}×216{\mathrm{m}}^3}{(8.314\mathrm{J}/\mathrm{mol}.\mathrm{K})×\left(300\mathrm{K}\right)}\\ =8773\mathrm{mol}\end{array}$

Now after calculating the number of moles n we can sub with its value in equation (1) to find the number of molecules N, where

$\begin{array}{r}N=n{N}_A\\ =8773\mathrm{mol}×6.022×{10}^{23}\text{molecule}/\mathrm{mol}\\ =52.8×{10}^{28}\text{molecule}\end{array}$

The physics lecture room with volume 216 m³ contain 52.8 x 10²⁶ molecules of Nitrogen gas.

The particle density is given by

$$\begin{gathered} \mathit{P}\mathit{a}\mathit{r}\mathit{t}\mathit{i}\mathit{c}\mathit{l}\mathit{e}\mathbf{}\mathit{d}\mathit{e}\mathit{n}\mathit{s}\mathit{i}\mathit{t}\mathit{y}\mathbf{=}\frac{\mathbf{N}}{\mathbf{V}} \\ Particledensity=\frac{52.8×{10}^{26}molecules}{216m^3×{\displaystyle \frac{1c{m}^3}{{10}^{-6}{m}^3}}} \\ =2.45×{10}^{19}molecules/c{m}^3 \end{gathered}$$

The particle density is $2.45×{10}^{19}molecules/c{m}^3$

The mass m of the nitrogen air in the room could be calculated by the relation between the number of moles n and the molar mass M of nitrogen gas

$\begin{array}{r}\mathbf{m}\mathbf{=}\mathbf{nM}\\ =8773\mathrm{mol}×29.014\mathrm{g}/\mathrm{mol}\\ =245.77×{10}^3\mathrm{g}\\ =245.77\mathrm{kg}\end{array}$

The mass of N in the physics lecture room equals 245.77 kg.