Question

The vapour pressure of liquid mercury at ${\mathbf{27}}^{\mathbf{o}}\mathbf{C} \mathbf{is} \mathbf{2}\mathbf{.}\mathbf{87}\times {\mathbf{10}}^{\mathbf{-}\mathbf{6} } \mathbf{atm}$ Calculate the number of molecules of Hg atoms per cubic centimeter in the “empty space”above the top of the mercury column in a barometer at ${\mathbf{27}}^{\mathbf{o}}\mathbf{C}\mathbf{ }\mathbf{ }$

Short answer

The number of molecules of Hg is $7\times {10}^{13}\mathrm{atoms}$.

Step-by-step solution

Given data

$$\begin{gathered} P=2.87\times {10}^{-6}\text{atm}\times \frac{101325\text{Pa}}{1\text{atm}}=0.291\text{Pa} \\ \text{V}=1{\text{cm}}^3={10}^{-6}{\text{m}}^3 \\ \text{T}=27°\text{C}+273=300\text{K} \end{gathered}$$

Concept of Ideal gas equation

The ideal gas law shows the relation between various parameters of the gas, pressure, volume, moles and temperature. The mathematical expression of ideal gas: PV = nRT , Where

$$\begin{gathered} \text{P}=\text{pressure.} \\ \text{V = volume} \\ \text{n = number} \text{of} \text{moles} \\ \text{R =}\left(\text{8.314}\frac{\text{J}}{\text{K}\times \text{mol}}\right) \\ \text{T = temperature.} \end{gathered}$$

Calculation of  number of moles

Putting the given values in the ideal gas equation and calculate

$\begin{array}{rcl}\text{n}& =& \frac{\text{P}·\text{V}}{\text{R}·\text{T}}\\ & =& \frac{0.291\text{Pa}·1·{10}^{-6}{\text{m}}^3}{8.314\frac{\text{J}}{\text{K}·\text{mol}}·300\text{K}}\\ & =& 1.17\times {10}^{-10}\mathrm{mol}\end{array}$

So, of Hg =$1.17\times {10}^{-10}\text{mol}$

Calculation of number of molecules of Hg atoms

We can use the below equation to calculate number of molecules

$\mathrm{n}=\frac{\mathrm{N}}{{\mathrm{N}}_{\mathrm{A}}}$

N = number of molecules and ${\text{N}}_{\text{A}}$ Avogadro constant

Putting the values of number of moles and Avogadro constant = 6×10²³

$\begin{array}{rcl}& & \text{N = n}\times {\text{N}}_{\text{A}}\\ & & \text{= 1.17}\times {\text{10}}^{\text{- 10}}\times \text{6}\times {\text{10}}^{\text{23}}\\ & & \text{= 7}\times {\text{10}}^{\text{13}} \text{Hg atoms}\end{array}$