Question

Short answer

The number of air molecules is $1.119\times {10}^{27}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}$, and the number of moles is $1858.8\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}$.

Step-by-step solution

Given data

The length of the room is$L=6\mathrm{m}$.

The width of the room is$w=3\mathrm{m}$.

The height of the room is$h=2.5\mathrm{m}$.

The temperature is $T={22}^{\circ }\mathrm{C}$.

Understanding the number of moles

In order to calculate the number of moles in a certain gas, divide the value of the number of molecules by Avogadro’s number.

Calculation of the number of molecules

The relation from the ideal gas law can be written as:

$\begin{array}{rl}cPV& =nkT\\ n& =\frac{PV}{kT}\end{array}$

Here, Pis the atmospheric pressure, Vis the volume, kis the Boltzmann constant, and n is the number of molecules.

On plugging the values in the above relation, you get:

$\begin{array}{rl}ln& =\left(\frac{\left(1\mathrm{a}\mathrm{t}\mathrm{m}\times \frac{1.013\times {10}^5\mathrm{P}\mathrm{a}}{1\mathrm{a}\mathrm{t}\mathrm{m}}\right)\left(6\mathrm{m}\times 3\mathrm{m}\times 2.5\mathrm{m}\right)}{\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left({22}^{\circ }\mathrm{C}+273\right)\mathrm{K}}\right)\\ n& =1.119\times {10}^{27}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}\end{array}$

Thus, $n=1.119\times {10}^{27}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}$ is the number of molecules.

Calculation of the number of moles

The relation to find the number of moles can be written as:

$N=\left(n\times \frac{1\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}}{N_{\mathrm{A}}}\right)$

Here,$N_{\mathrm{A}}$is Avogadro’s number.

On plugging the values in the above relation, you get:

$\begin{array}{rl}lN& =\left(1.119\times {10}^{27}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}\times \frac{1\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}}{6.02\times {10}^{23}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}}\right)\\ N& =1858.8\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\end{array}$

Thus, $N=1858.8\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}$ is the required number of moles.