Chapter 20, Problem 6A

(a)

To determine

The mass of a column of air.

Answer to Problem 6A

The mass of a column of air is $0.1\text{ kg}$ .

Explanation of Solution

Given info:

The cross sectional area of the column of air is $A=1{\text{ cm}}^2$ .

Formula used:

Show the expression for pressure as follows:

  $P=\frac{F}{A}$

Here, $F$ is the force and $A$ is the cross sectional area.

Show the expression for force as follows:

  $F=mg$

Here, $m$ is the mass and $g$ is the acceleration due to gravity.

Calculation:

Consider the average atmospheric pressure at sea level is $P=101.3\text{ kPa}$ .

Consider the acceleration due to gravity as $g=9.8\text{m}/{\text{s}}^2$ .

Find the mass of a column of air as follows:

  $\begin{array}{l}P=\frac{F}{A}\\ P=\frac{mg}{A}\\ 101.3\text{ kPa}=\frac{m\times \left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}{\left({\text{1 cm}}^2\times {\left(\frac{1\text{ m}}{100\text{ cm}}\right)}^2\right)}\\ m=1.0\text{ kg}\end{array}$

Conclusion:

Thus, the mass of a column of air is $1.0\text{kg}$ .

(b)

To determine

The weight of the air column.

Answer to Problem 6A

The weight of the air column is $1.0\text{ N}$ .

Explanation of Solution

Given:

The cross sectional area of the column of air is $A=1{\text{ cm}}^2$ .

Formula used:

The expression for weight is,

  $W=mg$

Here, $m$ is the mass and $g$ is the acceleration due to gravity.

Calculation:

From part (a), the mass of a column of air is $m=1.0\text{ kg}$ .

Consider the acceleration due to gravity as $g=9.8\text{m}/{\text{s}}^2$ .

The weight of the air column is,

  $\begin{array}{c}W=mg\\ W=\left(\text{1}\text{.0 kg}\right)\times \left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ =9.8\text{N}\\ W\approx 10\text{N}\end{array}$

Conclusion:

Thus, the weight of the air column is $10\text{ N}$ .

(c)

To determine

The pressure at the bottom of a column of air.

Answer to Problem 6A

The pressure at the bottom of a column of air is $10\text{N}/{\text{cm}}^2$ .

Explanation of Solution

Given:

The cross sectional area of the column of air is $A=1{\text{ cm}}^2$ .

Formula used:

The expression for pressure is,

  $P=\frac{F}{A}$

Here, $F$ is the force and $A$ is the cross sectional area.

Calculation:

From part (b), the weight of the air column is $F=10\text{ N}$ .

The pressure at the bottom of a column of air is,

  $\begin{array}{l}P=\frac{F}{A}\\ P=\frac{\text{10 N}}{{\text{1 cm}}^2}\\ P=10\text{N}/{\text{cm}}^2\end{array}$

Conclusion:

Thus, the pressure at the bottom of a column of air is $10\text{N}/{\text{cm}}^2$ .