Question

Calculate the number of molecules/m³in an ideal gas at STP.

Short answer

There are $2.69\times {10}^{25}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}/{\mathrm{m}}^3$in an ideal gas at STP.

Step-by-step solution

Step 1:Statement of the ideal gas law in terms of molecules 

The ideal gas law gives the relation between the pressure (P), volume (V), and temperature (T) of n moles of the ideal gas by the following expression:

$PV=nRT$

Here,R is the universal gas constant whose value is$8.14\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K}$.

In terms of the number of molecules (N) of the gas, the ideal gas law is written as

$PV=\frac{N}{N_{\mathrm{A}}}RT$.

Here, $N_{\mathrm{A}}$is the number of molecules in one mole of asubstance and is known as Avogadro’s number. Its value is$6.02\times {10}^{23}$.

Determination of the number of molecules in one cubic meter ofan ideal gas at STP

The volume of 1 mole of an ideal gas at STPis $V=22.4\mathrm{L}=22.4\times {10}^{-3}{\mathrm{m}}^3$.

The value of the Avogadro number is$N_{\mathrm{A}}=6.02\times {10}^{23}$.

Avogadro’s number is the number of molecules in volume $V{\mathrm{m}}^3$. Thus, the number of molecules in $1{\mathrm{m}}^3$ is

$\begin{array}{c}\frac{N_{\mathrm{A}}}{V}=\frac{6.02\times {10}^{23}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}}{22.4\times {10}^{-3}{\mathrm{m}}^3}\\ =0.269\times {10}^{26}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}/{\mathrm{m}}^3\\ =2.69\times {10}^{25}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}/{\mathrm{m}}^3.\end{array}$

Thus, there are $2.69\times {10}^{25}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}/{\mathrm{m}}^3$in an ideal gas at STP.