Example 2

Assuming a paperback book has a mass of 2.00 kg, a length of 27.0 cm, a width of 21.0 cm, and a thickness of 4.5 cm, what pressure does it exert on a surface if it is

1. lying flat?
2. standing on edge in a bookcase?

Given: mass and dimensions of object

Asked for: pressure

Strategy:

A Calculate the force exerted by the book and then compute the area that is in contact with a surface.

B Substitute these two values into Equation 10.1 to find the pressure exerted on the surface in each orientation.

Solution:

The force exerted by the book does not depend on its orientation. Recall that the force exerted by an object is F = ma, where m is its mass and a is its acceleration. In Earth’s gravitational field, the acceleration is due to gravity (9.8067 m/s² at Earth’s surface). In SI units, the force exerted by the book is therefore

F = ma = (2.00 kg)(9.8067 m/s²) = 19.6 (kg·m)/s² = 19.6 N

1. A We calculated the force as 19.6 N. When the book is lying flat, the area is (0.270 m)(0.210 m) = 0.0567 m². B The pressure exerted by the text lying flat is thus

   $$P=\frac{19.6\text{N}}{0.0567{\text{m}}^2}=3.46\times {10}^2\text{Pa}$$

2. A If the book is standing on its end, the force remains the same, but the area decreases:

   (21.0 cm)(4.5 cm) = (0.210 m)(0.045 m) = 9.5 × 10⁻³ m²

   B The pressure exerted by the book in this position is thus

   $$P=\frac{19.6\text{N}}{9.5\times {10}^{-3}{\text{m}}^2}=2.1\times {10}^3\text{Pa}$$

   Thus the pressure exerted by the book varies by a factor of about six depending on its orientation, although the force exerted by the book does not vary.

Exercise

What pressure does a 60.0 kg student exert on the floor

1. when standing flat-footed in the laboratory in a pair of tennis shoes (the surface area of the soles is approximately 180 cm²)?
2. as she steps heel-first onto a dance floor wearing high-heeled shoes (the area of the heel = 1.0 cm²)?

Answers:

1. 3.27 × 10⁴ Pa (4.74 lb/in.²)
2. 5.9 × 10⁶ Pa (8.5 × 10² lb/in.²)

Example 3

One of the authors visited Rocky Mountain National Park several years ago. After departing from an airport at sea level in the eastern United States, he arrived in Denver (altitude 5280 ft), rented a car, and drove to the top of the highway outside Estes Park (elevation 14,000 ft). He noticed that even slight exertion was very difficult at this altitude, where the atmospheric pressure is only 454 mmHg. Convert this pressure to

1. atmospheres.
2. kilopascals.

Given: pressure in millimeters of mercury

Asked for: pressure in atmospheres and kilopascals

Strategy:

Use the conversion factors in Equation 10.4 to convert from millimeters of mercury to atmospheres and kilopascals.

Solution:

From Equation 10.4, we have 1 atm = 760 mmHg = 101.325 kPa. The pressure at 14,000 ft in atm is thus

$$P=(454\cancel{\text{mmHg}})\left(\frac{1\text{atm}}{760\cancel{\text{mmHg}}}\right)=0.597\text{atm}$$

The pressure in kPa is given by

$$P=(0.597\cancel{\text{atm}})\left(\frac{101.325\text{kPa}}{1\cancel{\text{atm}}}\right)=60.5\text{kPa}$$

Exercise

Mt. Everest, at 29,028 ft above sea level, is the world’s tallest mountain. The normal atmospheric pressure at this altitude is about 0.308 atm. Convert this pressure to

1. millimeters of mercury.
2. kilopascals.

Answer: a. 234 mmHg; b. 31.2 kPa

Example 4

Suppose you want to construct a closed-end manometer to measure gas pressures in the range 0.000–0.200 atm. Because of the toxicity of mercury, you decide to use water rather than mercury. How tall a column of water do you need? (At 25°C, the density of water is 0.9970 g/cm³; the density of mercury is 13.53 g/cm³.)

Given: pressure range and densities of water and mercury

Asked for: column height

Strategy:

A Calculate the height of a column of mercury corresponding to 0.200 atm in millimeters of mercury. This is the height needed for a mercury-filled column.

B From the given densities, use a proportion to compute the height needed for a water-filled column.

Solution:

A In millimeters of mercury, a gas pressure of 0.200 atm is

$$P=(0.200\cancel{\text{atm}})\left(\frac{760\text{mmHg}}{1\cancel{\text{atm}}}\right)=152\text{mmHg}$$

Using a mercury manometer, you would need a mercury column at least 152 mm high.

B Because water is less dense than mercury, you need a taller column of water to achieve the same pressure as a given column of mercury. The height needed for a water-filled column corresponding to a pressure of 0.200 atm is proportional to the ratio of the density of mercury $(d_{\text{Hg}})$ to the density of water $(d_{{\text{H}}_{\text{2}}\text{O}})$:

$$\begin{array}{c}\left({\text{height}}_{{\text{H}}_{\text{2}}\text{O}}\right)\left(d_{{\text{H}}_{\text{2}}\text{O}}\right)=\left({\text{height}}_{\text{Hg}}\right)\left(d_{\text{Hg}}\right)\\ {\text{height}}_{{\text{H}}_2\text{O}}=\left({\text{height}}_{\text{Hg}}\right)\left(\frac{d_{\text{Hg}}}{d_{{\text{H}}_2\text{O}}}\right)\\ =(152\text{mm})\left(\frac{13.53\cancel{{\text{g/cm}}^3}}{0.9970\cancel{{\text{g/cm}}^3}}\right)\\ =2.06\times {10}^3\text{mm}{\text{H}}_2\text{O}=\text{2}\text{.06 m}{\text{H}}_2\text{O}\end{array}$$

This answer makes sense: it takes a taller column of a less dense liquid to achieve the same pressure.

Exercise

Suppose you want to design a barometer to measure atmospheric pressure in an environment that is always hotter than 30°C. To avoid using mercury, you decide to use gallium, which melts at 29.76°C; the density of liquid gallium at 25°C is 6.114 g/cm³. How tall a column of gallium do you need if P = 1.00 atm?

Answer: 1.68 m