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Chapter 11. Pressures

Useful Calculations

Useful Calculations

The purpose of this work is to promote fluency in the language of mathematics among students who have had little experience (or perhaps an unpleasant experience) with standard math courses. Every educated person needs to become "numerate", not only to solve practical problems, but also to develop paths of communication with scientists, physicians, and engineers, whose work requires that they think in a quantitative manner.

Atmospheric Pressure

Everyone is under pressure - atmospheric pressure outside, and blood pressure inside. Pressures of work and family life are not readily quantified: see your psychotherapist about them.

Atmospheric pressure is due to the weight of the air, which varies a little from day to day, and decreases with altitude. Ordinarily noticed only by mariners and weather people, it is measured by a barometer. The simplest type of barometer is a tube of some liquid inverted, mouth down, in an open dish of the same liquid. The column falls, leaving a vacuum at the top of the tube. It remains high because of the atmospheric pressure on the surface of the liquid in the dish. The favorite liquid is mercury, a metallic element that has a very high density, and is in the liquid state at ordinary environmental temperatures. ('υδραργυροσ' = quicksilver = mercury = Hg).

The height of the column of mercury supported by the weight of the atmosphere is usually just over ¾ meter, or 760 mm (Figure 11-1). This apparatus is so simple that it is tempting to express all pressures in mm Hg, but this is a naïve, unscientific way of thinking about pressures.  Automobile tires are marked with maximum inflation pressures in psi (pounds per square inch) and kPa (kilopascals), but not in mm Hg. The relation of these units is worth examining. If the tire pressure is 44 psi, the first step is to convert one pound into kilograms:

1lb/ 2.20462 lbs = x/ 1 kg,   x = 0.453592909 kg

Pressure is a force, and a force is the product of mass and acceleration:

f = ma.

Thank you, Sir Isaac.

The acceleration of gravity a = 9.80665 m / s2. Therefore:

f = 0.453592909  kg X  9.80665 m / s2

=  4.448226905 kg- m / s2 = 4.448 Newtons

The next step is to find the area subject to this force:

1 inch / 39.37 in/m = 0.0254 m, 1 in2 = (0.0254 m)2 = 0.000645162 m2

4.448 N / 0.000645162 m2 = 6894.744119 N/ m2

The unit of force with these SI dimensions is called the Pascal, after a Frenchman who studied the meaning of pressure, among other things. This unit is too small for most measuring devices, so the usual practice is to deal with the unit 1 000 times larger, the kiloPascal (kPa). The result above can be expressed as 1 psi = 6.8947 kPa. Thus 44 psi = 303.3668 kPa. On the tire, this is rounded off to 300 kPa.

Exercise 11-1.

(a) Atmospheric pressure is 760 mm Hg or 14.696 psi. Convert psi to kPa.

(b) Use 760 mm,  density of mercury = 13.595 g/cm3, and the acceleration of gravity above, and calculate 1 atmosphere in kPa. (This is rather challenging because of the mixture of units, but give it a try.)

The answer to both calculations should be the same:

1 atm = 760 mm Hg = 14.695 psi = 101.32 kPa.

Tire inflation pressures and blood pressures are actually the amounts that they exceed atmospheric pressure.

Exercise 11-2.

Typical blood pressures are still given as 120 systolic and 80 diastolic. To agree with modern practice, convert these mm Hg into kPa.

Pressure of Gasses

Breathing is accomplished by changing pressure on the air in the lungs, causing inhaling and exhaling. Gases expand and contract when pressure is varied. This is evident in hot-air balloons, tanks of compressed gases, and compression in the cylinders of gasoline and diesel engines.  The behavior of gases in response to changes in pressure and temperature is described by a beautiful equation:

pV = nRT.

n is number of moles;  to simplify calculation, n is taken as 1 mole, and p is counted in atmospheres rather than kPa. V is volume, T is temperature in degrees Kelvin, and R is the ideal gas constant = 0.0821026 L-atm / mol-deg.  Standard  conditions are 1 atm and 273oK = 0o C. Therefore the volume of 1 mole of any gas is:

V = RT / p = 0.0821026  X 273  = 22.414 L.

One mole of nitrogen (N2) weighs 28 grams, O2 - 32 g, CO2 - 44 g , H2 only 2 g. All occupy the same space. In trying to picture this volume, think of a cube 28.2 cm on each edge: 28.23 = 22 426 cm3, about the size of a bread- box, or the box containing the chips and circuits of a computer, measuring

32 X 39 X 18 = 22 464 cm3.

Equations for distance - time - speed, and voltage across a resistance, like the ideal gas equation, have the simple form y = a x. As Cromer wrote, “None of the linear relations of elementary physics is exactly true.”  At very high pressures, the volume decrease is less than predicted by V = RT / p.

In order to make the equation match the data of experiments, a parameter is introduced:

p (V - b) = nRT.

For example, 1 mole of hydrogen under 500 atm  proved to have volume  0.060524 L. The expected volume was: 22.414 L / 500 atm  = 0.044828 L.

The difference observed sets the value of the parameter b.

b = Vobs – Vexp =  0.060524 – 0.044828 = 0.015696 L

Exercise 11-3.

A mad physicist found that his machine could produce a pressure of 1000 atm on a gas. When he subjected 1 mole of  hydrogen to this pressure, its volume was reduced to 0.038255 L. Calculate b.

This parameter represents volume occupied by the molecules themselves, called the volume of exclusion or incompressible volume. Deviations from “ideal gas” behavior also occur due to attraction between gas molecules, which tends to make volumes a little less than expected, and requires an additional parameter. As a talk-show host said, “Anyone who uses the word parameter in ordinary conversation should not be invited to a dinner party.”