Chapter 1, Problem 1.1P

Interpretation Introduction

Interpretation:

The number of molecules which crosses the circular hole is to be determined.

Concept Introduction:

The formula to calculate number of molecules $\left(N\right)$

that crosses a unit area per unit time in only one direction is:

  $N=\frac{1}{4}n\overline{v}$ ........(1)

Here, $n$ is the number of molecules per unit volume and $\overline{v}$ is the average molecular velocity.

The formula to calculate the speed of sound $\left(c\right)$

in a perfect gas is:

  $c=\sqrt{k{g}_{c}RT}$ ........(2)

Here, $k$ is the ratio of specific heats, $R$ is the specific gas constant for a gas in $\text{kJ}/\left(\text{kg}\cdot \text{K}\right)$ , $g_c$ is the conversion factor in $\left(\text{kg}\cdot \text{m}\right)/\left(\text{N}\cdot {\text{s}}^2\right)$ , and $T$ is temperature in $\text{K}$ .

Answer to Problem 1.1P

  $1.07\times {\text{10}}^{18}\text{molecules}/\text{s}$ .

Explanation of Solution

Given information:

The gas is at standard conditions with $4\times {10}^{20}$ molecules per $\text{in}{\text{.}}^3$ . The average molecular velocity is taken approximately equal to the sound speed in a perfect gas.

The diameter of the circular hole is, $d={10}^{-3}\text{ in}\text{.}$

Assume that the perfect gas is air with the value of $k=1.4$ . For air, the value of R is taken as $R=0.287\text{kJ}/\left(\text{kg}\cdot \text{K}\right)$ . At standard condition, the temperature is taken as, $T=298\text{K}$ .

Use equation (2) to calculate the speed of sound in perfect gas as:

  $\begin{array}{c}c=\sqrt{k{g}_{c}RT}\\ =\sqrt{\left(1.4\right)\left(\frac{\cancel{\text{kg}}\cdot \text{m}}{\cancel{\text{N}}\cdot {\text{s}}^2}\right)\left(\frac{0.287\times {10}^3\cancel{\text{N}}\text{-m}}{\cancel{\text{kg}}\cdot \cancel{\text{K}}}\right)\left(298\cancel{\text{K}}\right)}\\ =346.028\text{m}/\text{s}\end{array}$

Since,

  $1\text{ m =39}\text{.37 in}\text{.}$

Thus,

  $\begin{array}{l}c=346.028\text{ m}\left(\frac{\text{39}\text{.37 in}\text{.}}{1\text{ m}}\right)\text{/s}\\ =1.36\times {10}^4\text{in}\text{.}/\text{s}\end{array}$

It is given that the sound speed in the perfect gas is nearly equal to the average molecular velocity.

Thus,

  $\overline{v}=c=1.36\times {10}^4\text{in}\text{.}/\text{s}$

Area of the circular hole is calculated as:

  $\begin{array}{c}A=\frac{\pi }{4}{d}^2\\ =\frac{3.14}{4}{\left({10}^{-3}\text{ in}\text{.}\right)}^2\\ =7.85\times {10}^{-7}\text{ in}{\text{.}}^2\end{array}$

Equation (1) gives the number of molecules that crosses a unit area per unit time in one direction. To calculate the number of molecules that crosses a particular area per unit time, multiply equation (1) by A as:

  $\begin{array}{c}NA=\frac{1}{4}n\overline{v}A\\ =\frac{1}{4}\left(\frac{4\times {10}^{20}\text{ molecules}}{\cancel{\text{in}{\text{.}}^3}}\right)\left(\frac{1.36\times {10}^4\cancel{\text{in}\text{.}}}{\text{s}}\right)\left(7.85\times {10}^{-7}\cancel{\text{in}{\text{.}}^2}\right)\\ =1.07\times {\text{10}}^{18}\text{molecules}/\text{s}\end{array}$

Conclusion

Thus, the number of molecules which crosses the given circular hole is, $1.07\times {\text{10}}^{18}\text{molecules}/\text{s}$ .