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

Short Answer

Expert verified

There are \[300\;{\rm{molecules/c}}{{\rm{m}}^{\rm{3}}}\]at \(0^\circ {\rm{C}}\) for the lowest attainable pressure.

Step by step solution

01

Given Data 

The pressure of the gas is \(P = {10^{ - 12}}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\).

The temperature of the gas is \(T = 0^\circ {\rm{C}} = 273\;{\rm{K}}\)

The volumeis\(V = 1\;{\rm{c}}{{\rm{m}}^{\rm{3}}}\).

02

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.

03

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:

\(\begin{array}{c}PV = \frac{N}{{{N_{\rm{A}}}}}RT\\PV = NkT\\N = \frac{{PV}}{{kT}}\end{array}\)

Here, k is the Boltzmann constant,\({N_{\rm{A}}}\) 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 \[300\;{\rm{molecules/c}}{{\rm{m}}^{\rm{3}}}\]at \(0^\circ {\rm{C}}\) for the lowest attainable pressure.

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Most popular questions from this chapter

Consider a container of oxygen gas at a temperature of 23°C that is 1.00 m tall. Compare the gravitational potential energy of a molecule at the top of the container (assuming the potential energy to be zero at the bottom) with the average kinetic energy of the molecules. Is it reasonable to neglect the potential energy?

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