Question

The tungsten filament in an incandescent light bulb ordinarily operates operates at a temperature of about ${\mathbf{2500}}^{\mathbf{o}}\mathbf{C}$. At this temperature, the vapour pressure of solid tungsten is $\mathbf{7}\mathbf{.}\mathbf{0}\times {\mathbf{10}}^{\mathbf{-}\mathbf{9}} \mathbf{atm}$ .Estimate the number of tungsten atoms per cubic centimeter under these conditions.

Short answer

The number of molecules of gaseous tungsten is $1.84\times {10}^{10} \text{atoms}$

Step-by-step solution

Given data

$$\begin{gathered} P=7·{10}^{-9}\text{atm}\times \frac{101325\text{Pa}}{1\text{atm}}=7.09\times {10}^{-4}\text{Pa} \\ {\text{V = 1cm}}^{\text{3}}\text{= 1}\times {\text{10}}^{\text{- 6}}{\text{m}}^{\text{3}} \\ \text{T = 2500}°\text{C + 273 = 27}73\text{K} \end{gathered}$$

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

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

 Step3: Calculation of number of moles

Putting the given values in the ideal gas equation and calculating the number of moles of tungsten:

$\begin{array}{rcl}\text{n}& =& \frac{\text{P}·\text{V}}{\text{R}·\text{T}}\\ & =& \frac{7.09·{10}^{-4}\text{Pa}·1\times {10}^{-6}{\text{m}}^3}{8.314\frac{\text{J}}{\text{K}·\text{mol}}·2773\text{K}}\\ & =& 3.08\times {10}^{-14}\text{mol}\end{array}$

So, number of moles of tungsten atoms$=3.08\times {10}^{-14}\text{mol}$

Calculation of number of molecules of tungsten atoms

Using the below equation to calculate number of molecules

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

N is the number of molecules and ${\text{N}}_{\text{A}}$ is Avogadro number

Putting the values of number of moles and Avogadro number

$$\begin{gathered} \text{N}=\text{n}·{\text{N}}_{\text{A}} \\ =3.08\times {10}^{-14}\times 6\times {10}^{23} \\ =1.84·{10}^{10}\text{atoms} \end{gathered}$$