Question

Question 41:(II) The lowest attainablepressure using the best available vacuum techniques is about $\mathbf{1}{\mathbf{0}}^{\mathbf{-}\mathbf{12}}\mathbf{N}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$. At such a pressure, how many molecules are there per cm³ at 0°C?

Short answer

There are $$300\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}/\mathrm{c}{\mathrm{m}}^3$$at $0^{\circ }\mathrm{C}$ for the lowest attainable pressure.

Step-by-step solution

Given Data 

The pressure of the gas is $P={10}^{-12}\mathrm{N}/{\mathrm{m}}^2$.

The temperature of the gas is $T=0^{\circ }\mathrm{C}=273\mathrm{K}$

The volumeis$V=1\mathrm{c}{\mathrm{m}}^3$.

Understanding of Ideal Gas Law  

The ideal gas law is the equation of state for an ideal gas. It gives the relation between the pressure (P), volume (V), and temperature (T) forn moles of the ideal gas by using the following expression:

$PV=nRT$

Here, R is the universal gas constant.

Determination of the number of molecules 

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

$\begin{array}{c}PV=\frac{N}{N_{\mathrm{A}}}RT\\ PV=NkT\\ N=\frac{PV}{kT}\end{array}$

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

Substitute the values into the above equation.

\(\begin{array}{c}N = \frac{{{{10}^{ - 12}}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}} \times \left( {1\;{\rm{c}}{{\rm{m}}^{\rm{3}}} \times \frac{{1\;{{\rm{m}}^3}}}{{{{10}^6}\;{\rm{c}}{{\rm{m}}^{\rm{3}}}}}} \right)}}{{1.38 \times {{10}^{23}}\;{\rm{J/K}} \times 273\;{\rm{K}}}}\ = 2.65 \times

{10^2}\;{\rm{molecules/c}}{{\rm{m}}^{\rm{3}}}\ \approx 300\;{\rm{molecules/c}}{{\rm{m}}^{\rm{3}}}\end{array}\)

Thus, there are $$300\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}/\mathrm{c}{\mathrm{m}}^3$$at $0^{\circ }\mathrm{C}$ for the lowest attainable pressure.