Question

It is hard to imagine that there can be enough air between a book and a table so that there is a net upward (buoyant) force on the book despite the large downward on the top of the book. About how many air molecules are there between a textbook and a table if there is an average distance of 0.1 mm between the uneven surfaces of the book and table?

Short answer

The number of molecules between textbook and table is 1.61x10¹⁹.

Step-by-step solution

Identification of the given data

The given data is listed below,

The distance between uneven surfaces of book and table is, $t=0.1mm\times \frac{1m}{1000mm}=0.0001m$

Assume the other data as follows,

• The dimensions of the book are, $I\times b=20cm\times 30cm$
• The air is assumed as an ideal gas. The molar Volume of the gas is$V_m=0.224m^3/mol$

Significance of the number of moles of air

The number of moles of air between two surfaces depends on the space between them and the volume of objects.

The moles of air can be obtained by dividing the volume of the object by the molar volume of an ideal gas.

Determination of the number of moles of air

Write the expression for the volume of air between the surfaces.

$V=I\times b\times t$

Here, I is the length of the book, b is the breadth of the book, and t is the distance between uneven surfaces of book and table.

Substitute all the values in the above expression.

$\begin{array}{rcl}V& =& 20cm\times 30cm\times 0.0001m\\ & =& \left(20cm\times \frac{1m}{100cm}\right)\times \left(30cm\times \frac{1m}{100cm}\right)\times 0.0001m\\ & =& 0.2m\times 0.3m\times 0.0001m\\ & =& 6\times {10}^{-5}{m}^3\end{array}$

The expression for the moles of air is given as,

$n=\frac{V}{V_m}$

Here, V volume of air between the surfaces, and V_m is the molar volume of air.

Substitute all the values in the above expression.

$\begin{array}{rcl}n& =& \frac{6\times {10}^{-5}{m}^3}{0.224m^3/mol}\\ & =& 2.678\times {10}^{-5}mol\end{array}$

Determination of the number of molecules of air

The expression for the number of molecules between textbook and table is give as,

N = nN_a

Here, n is the number of moles of air, and N_a is Avogadro’s number with the value $(6.022\times {10}^{23}mo{l}^{-1})$.

Substitute all the values in the above expression.

$\begin{array}{rcl}N& =& 2.678\times {10}^{-5}mol\times (6.022\times {10}^{23}mo{l}^{-1})\\ & =& 1.61\times {10}^{19}\end{array}$

Thus, the number of molecules between textbook and table is $1.61\times {10}^{19}$.