Chemistry Textbook

• Define gas pressure
• Define and convert among the units of pressure measurements

Conversion of pressure Units

The United States National Weather Service reports pressure in both inches of Hg and millibars. Convert a pressure of 29.2 in. Hg into:

(a) torr

(b) atm

(c) kPa

(d) mbar

This is a unit conversion problem. The relationships between the various pressure units are given in Table 6.1.

(a) $29.2\sout{\text{in Hg}}\times \frac{\text{25.4}\sout{\text{mm}}}{1\sout{\text{in}}}\times \frac{\text{1}}{\text{torr}}=\text{742 torr}$

(b) $742\sout{\text{torr}}\times \frac{\text{1}}{}\frac{}{\text{atm}}=\text{0.976}$

(c) $742\sout{\text{torr}}\times \frac{\text{101.325}}{\text{kPa}}=\text{98.9 kPa}$

(d) $98.9\sout{\text{kPa}}\times \frac{1000\sout{\text{Pa}}}{1\sout{\text{kPa}}}\times \frac{1\sout{\text{bar}}}{\mathrm{100,}}\frac{000\sout{\text{Pa}}}{}\times \frac{\text{1000}}{}\frac{}{\text{mbar}}=\text{989 mbar}$

Check Your Learning:

A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, in kilopascals, and in bar?

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?

acceleration due to gravity (g) = 9.8 m/s²

1.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

When the book is lying flat, the area is (0.270 m)(0.210 m) = 0.0567 m²

2. If the book is standing on its end, the force remains the same, but the area decreases: $\left(21.0cm\right)\left(4.5cm\right)=\left(0.210m\right)\left(0.045m\right)=9.5\times <!--\; \times \; -->\times {10}^{-<!--\; -\; -->3}{m}^2$ The pressure exerted by the book in this position is thus: $P={\displaystyle \frac{19.6N}{9.5\times <!--\; \times \; -->{10}^{-<!--\; -\; -->3}{m}^2}}=2.1\times <!--\; \times \; -->{10}^{3}Pa$ 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.

What pressure does a 60.0 kg student exert on the floor when standing flat-footed in the laboratory in a pair of tennis shoes (the surface area of the soles is approximately 180 cm²)? as she steps heel-first onto a dance floor wearing high-heeled shoes (the area of the heel = 1.0 cm²)?