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# 5.1 Pressure and Temperature

When we heat an object, we speed up the mind-bogglingly complex random motion of its molecules. One method for taming complexity is the conservation laws, since they tell us that certain things must remain constant regardless of what process is going on. Indeed, the law of conservation of energy is also known as the first law of thermodynamics.

But as alluded to in the introduction to this chapter, conservation of energy by itself is not powerful enough to explain certain empirical facts about heat. A second way to sidestep the complexity of heat is to ignore heat's atomic nature and concentrate on quantities like temperature and pressure that tell us about a system's properties as a whole. This approach is called macroscopic in contrast to the microscopic method of attack. Pressure and temperature were fairly well understood in the age of Newton and Galileo, hundreds of years before there was any firm evidence that atoms and molecules even existed.

Unlike the conserved quantities such as mass, energy, momentum, and angular momentum, neither pressure nor temperature is additive. Two cups of coffee have twice the heat energy of a single cup, but they do not have twice the temperature. Likewise, the painful pressure on your eardrums at the bottom of a pool is not affected if you insert or remove a partition between the two halves of the pool.

We restrict ourselves to a discussion of pressure in fluids at rest and in equilibrium. In physics, the term “fluid” is used to mean either a gas or a liquid. The important feature of a fluid can be demonstrated by comparing with a cube of jello on a plate. The jello is a solid. If you shake the plate from side to side, the jello will respond by shearing, i.e., by slanting its sides, but it will tend to spring back into its original shape. A solid can sustain shear forces, but a fluid cannot. A fluid does not resist a change in shape unless it involves a change in volume.

5.1.1 Pressure

If you're at the bottom of a pool, you can't relieve the pain in your ears by turning your head. The water's force on your eardrum is always the same, and is always perpendicular to the surface where the eardrum contacts the water. If your ear is on the east side of your head, the water's force is to the west. If you keep your ear in the same spot while turning around so your ear is on the north, the force will still be the same in magnitude, and it will change its direction so that it is still perpendicular to the eardrum: south. This shows that pressure has no direction in space, i.e., it is a scalar. The direction of the force is determined by the orientation of the surface on which the pressure acts, not by the pressure itself. A fluid flowing over a surface can also exert frictional forces, which are parallel to the surface, but the present discussion is restricted to fluids at rest.

Experiments also show that a fluid's force on a surface is proportional to the surface area. The vast force of the water behind a dam, for example, in proportion to the dam's great surface area. (The bottom of the dam experiences a higher proportion of its force.)

Based on these experimental results, it appears that the useful way to define pressure is as follows. The pressure of a fluid at a given point is defined as $$F_\perp/A$$, where $$A$$ is the area of a small surface inserted in the fluid at that point, and $$F_\perp$$ is the component of the fluid's force on the surface which is perpendicular to the surface. (In the case of a moving fluid, fluid friction forces can act parallel to the surface, but we're only dealing with stationary fluids, so there is only an $$F_\perp$$.)

This is essentially how a pressure gauge works. The reason that the surface must be small is so that there will not be any significant difference in pressure between one part of it and another part. The SI units of pressure are evidently $$\text{N}/\text{m}^2$$, and this combination can be abbreviated as the pascal, 1 Pa=1 $$\text{N}/\text{m}^2$$. The pascal turns out to be an inconveniently small unit, so car tires, for example, normally have pressures imprinted on them in units of kilopascals.

a / A simple pressure gauge consists of a cylinder open at one end, with a piston and a spring inside. The depth to which the spring is depressed is a measure of the pressure. To determine the absolute pressure, the air needs to be pumped out of the interior of the gauge, so that there is no air pressure acting outward on the piston. In many practical gauges, the back of the piston is open to the atmosphere, so the pressure the gauge registers equals the pressure of the fluid minus the pressure of the atmosphere.

Example 1: Pressure in U.S. units

In U.S. units, the unit of force is the pound, and the unit of distance is the inch. The unit of pressure is therefore pounds per square inch, or p.s.i. (Note that the pound is not a unit of mass.)

Example 2: Atmospheric pressure in U.S. and metric units

$$\triangleright$$ A figure that many people in the U.S. remember is that atmospheric pressure is about 15 pounds per square inch. What is this in metric units?

$$\triangleright$$

\begin{align*} (\text{15 lb})/(\text{1 in}^2) &= \frac{68\ \text{N}}{(0.0254\ \text{m})^2}\\ &= 1.0\times10^5\ \text{N}/\text{m}^2 \\ &= 100\ \text{kPa} \end{align*}

#### Only pressure differences are normally significant.

If you spend enough time on an airplane, the pain in your ears subsides. This is because your body has gradually been able to admit more air into the cavity behind the eardrum. Once the pressure inside is equalized with the pressure outside, the inward and outward forces on your eardrums cancel out, and there is no physical sensation to tell you that anything unusual is going on. For this reason, it is normally only pressure differences that have any physical significance. Thus deep-sea fish are perfectly healthy in their habitat because their bodies have enough internal pressure to cancel the pressure from the water in which they live; if they are caught in a net and brought to the surface rapidly, they explode because their internal pressure is so much greater than the low pressure outside.

Example 3: Getting killed by a pool pump

$$\triangleright$$ My house has a pool, which I maintain myself. A pool always needs to have its water circulated through a filter for several hours a day in order to keep it clean. The filter is a large barrel with a strong clamp that holds the top and bottom halves together. My filter has a prominent warning label that warns me not to try to open the clamps while the pump is on, and it shows a cartoon of a person being struck by the top half of the pump. The cross-sectional area of the filter barrel is 0.25 $$\text{m}^2$$. Like most pressure gauges, the one on my pool pump actually reads the difference in pressure between the pressure inside the pump and atmospheric pressure. The gauge reads 90 kPa. What is the force that is trying to pop open the filter?

$$\triangleright$$ If the gauge told us the absolute pressure of the water inside, we'd have to find the force of the water pushing outward and the force of the air pushing inward, and subtract in order to find the total force. Since air surrounds us all the time, we would have to do such a subtraction every time we wanted to calculate anything useful based on the gauge's reading. The manufacturers of the gauge decided to save us from all this work by making it read the difference in pressure between inside and outside, so all we have to do is multiply the gauge reading by the cross-sectional area of the filter:

\begin{align*} F &= PA\\ &= (90\times10^3\ \text{N}/\text{m}^2)( 0.25\ \text{m}^2)\ = 22000\ \text{N} \end{align*}

That's a lot of force!

The word “suction” and other related words contain a hidden misunderstanding related to this point about pressure differences. When you suck water up through a straw, there is nothing in your mouth that is attracting the water upward. The force that lifts the water is from the pressure of the water in the cup. By creating a partial vacuum in your mouth, you decreased the air's downward force on the water so that it no longer exactly canceled the upward force.

#### Variation of pressure with depth

b / This doesn't happen. If pressure could vary horizontally in equilibrium, the cube of water would accelerate horizontally. This is a contradiction, since we assumed the fluid was in equilibrium.

c / The pressure is the same at all the points marked with dots.

d / This does happen. The sum of the forces from the surrounding parts of the fluid is upward, canceling the downward force of gravity.

The pressure within a fluid in equilibrium can only depend on depth, due to gravity. If the pressure could vary from side to side, then a piece of the fluid in between, b, would be subject to unequal forces from the parts of the fluid on its two sides. Since fluids do not exhibit shear forces, there would be no other force that could keep this piece of fluid from accelerating. This contradicts the assumption that the fluid was in equilibrium.

self-check:

How does this proof fail for solids? (answer in the back of the PDF version of the book)

To find the variation with depth, we consider the vertical forces acting on a tiny, imaginary cube of the fluid having infinitesimal height $$dy$$ and areas $$dA$$ on the top and bottom. Using positive numbers for upward forces, we have

\begin{align*} P_{bottom}dA - P_{top}dA - F_g = 0 . \end{align*}

The weight of the fluid is $$F_g = mg = \rho Vg = \rho dAdy\:g$$, where $$\rho$$ is the density of the fluid, so the difference in pressure is

$\begin{multline*} dP = -\rho g dy . \\ {\text{[variation in pressure with depth for}}\\ \\ {\text{a fluid of density \rho in equilibrium;}} \ {\text{positive y is up.]}} \end{multline*}$

A more elegant way of writing this is in terms of a dot product, $$dP = \rho\mathbf{g}\cdot d\mathbf{y}$$, which automatically takes care of the plus or minus sign, depending on the relative directions of the g and $$d\mathbf{y}$$ vectors, and avoids any requirements about the coordinate system.

The factor of $$\rho$$ explains why we notice the difference in pressure when diving 3 m down in a pool, but not when going down 3 m of stairs. The equation only tells us the difference in pressure, not the absolute pressure. The pressure at the surface of a swimming pool equals the atmospheric pressure, not zero, even though the depth is zero at the surface. The blood in your body does not even have an upper surface.

In cases where $$g$$ and $$\rho$$ are independent of depth, we can integrate both sides of the equation to get everything in terms of finite differences rather than differentials: $$\Delta P = -\rho g \Delta y$$.

self-check:

In which of the following situations is the equation $$\Delta P = -\rho g \Delta y$$ valid? Why?\ (1) difference in pressure between a tabletop and the feet (i.e., predicting the pressure of the feet on the floor)\ (2) difference in air pressure between the top and bottom of a tall building\ (3) difference in air pressure between the top and bottom of Mt. Everest\ (4) difference in pressure between the top of the earth's mantle and the center of the earth\ (5) difference in pressure between the top and bottom of an airplane's wing

Example 4: Pressure of lava underneath a volcano

$$\triangleright$$ A volcano has just finished erupting, and a pool of molten lava is lying at rest in the crater. The lava has come up through an opening inside the volcano that connects to the earth's molten mantle. The density of the lava is 4.1 $$\text{g}/\text{cm}^3$$. What is the pressure in the lava underneath the base of the volcano, 3000 m below the surface of the pool?

$$\triangleright$$

\begin{align*} \Delta P &= \rho g\Delta y\\ &= ( 4.1\ \text{g}/\text{cm}^3)( 9.8\ \text{m}/\text{s}^2)(3000\ \text{m})\\ &= ( 4.1\times10^6\ \text{g}/\text{m}^3) ( 9.8\ \text{m}/\text{s}^2)(3000\ \text{m})\\ &= ( 4.1\times10^3\ \text{kg}/\text{m}^3) ( 9.8\ \text{m}/\text{s}^2)(3000\ \text{m})\\ &= 1.2\times10^8\ \text{N}/\text{m}^2 \\ &= 1.2\times10^8\ \text{Pa} \end{align*}

This is the difference between the pressure we want to find and atmospheric pressure at the surface. The latter, however, is tiny compared to the $$\Delta P$$ we just calculated, so what we've found is essentially the pressure, $$P$$.

Example 5: Atmospheric pressure

Gases, unlike liquids, are quite compressible, and at a given temperature, the density of a gas is approximately proportional to the pressure. The proportionality constant is discussed on page 306, but for now let's just call it $$k$$, $$\rho= kP$$. Using this fact, we can find the variation of atmospheric pressure with altitude, assuming constant temperature:

\begin{align*} dP &= -\rho gdy\\ dP &= - kPgdy\\ \frac{dP}{ P} &= - kgdy\\ \text{ln}\: P &= - kgy+\text{constant} \ \text{[integrating both sides]}\\ P &= (\text{constant}) e^{- kgy} \ \text{[exponentiating both sides]} \end{align*}

Pressure falls off exponentially with height. There is no sharp cutoff to the atmosphere, but the exponential factor gets extremely small by the time you're ten or a hundred miles up.

## 5.1.2 Temperature

#### Thermal equilibrium

f / Thermal equilibrium can be prevented. Otters have a coat of fur that traps air bubbles for insulation. If a swimming otter was in thermal equilibrium with cold water, it would be dead. Heat is still conducted from the otter's body to the water, but much more slowly than it would be in a warm-blooded animal that didn't have this special adaptation.

We use the term temperature casually, but what is it exactly? Roughly speaking, temperature is a measure of how concentrated the heat energy is in an object. A large, massive object with very little heat energy in it has a low temperature.

But physics deals with operational definitions, i.e., definitions of how to measure the thing in question. How do we measure temperature? One common feature of all temperature-measuring devices is that they must be left for a while in contact with the thing whose temperature is being measured. When you take your temperature with a fever thermometer, you are waiting for the mercury inside to come up to the same temperature as your body. The thermometer actually tells you the temperature of its own working fluid (in this case the mercury). In general, the idea of temperature depends on the concept of thermal equilibrium. When you mix cold eggs from the refrigerator with flour that has been at room temperature, they rapidly reach a compromise temperature. What determines this compromise temperature is conservation of energy, and the amount of energy required to heat or cool each substance by one degree. But without even having constructed a temperature scale, we can see that the important point is the phenomenon of thermal equilibrium itself: two objects left in contact will approach the same temperature. We also assume that if object A is at the same temperature as object B, and B is at the same temperature as C, then A is at the same temperature as C. This statement is sometimes known as the zeroth law of thermodynamics, so called because after the first, second, and third laws had been developed, it was realized that there was another law that was even more fundamental.

e / We have to wait for the thermometer to equilibrate its temperature with the temperature of Irene's armpit.

#### Thermal expansion

g / A hot air balloon is inflated. Because of thermal expansion, the hot air is less dense than the surrounding cold air, and therefore floats as the cold air drops underneath it and pushes it up out of the way.

The familiar mercury thermometer operates on the principle that the mercury, its working fluid, expands when heated and contracts when cooled. In general, all substances expand and contract with changes in temperature. The zeroth law of thermodynamics guarantees that we can construct a comparative scale of temperatures that is independent of what type of thermometer we use. If a thermometer gives a certain reading when it's in thermal equilibrium with object A, and also gives the same reading for object B, then A and B must be the same temperature, regardless of the details of how the thermometers works.

i / The volume of 1 kg of neon gas as a function of temperature (at standard pressure). Although neon would actually condense into a liquid at some point, extrapolating the graph gives to zero volume gives the same temperature as for any other gas: absolute zero.

What about constructing a temperature scale in which every degree represents an equal step in temperature? The Celsius scale has 0 as the freezing point of water and 100 as its boiling point. The hidden assumption behind all this is that since two points define a line, any two thermometers that agree at two points must agree at all other points. In reality if we calibrate a mercury thermometer and an alcohol thermometer in this way, we will find that a graph of one thermometer's reading versus the other is not a perfectly straight $$y=x$$ line. The subtle inconsistency becomes a drastic one when we try to extend the temperature scale through the points where mercury and alcohol boil or freeze. Gases, however, are much more consistent among themselves in their thermal expansion than solids or liquids, and the noble gases like helium and neon are more consistent with each other than gases in general. Continuing to search for consistency, we find that noble gases are more consistent with each other when their pressure is very low.

h / A simplified version of an ideal gas thermometer. The whole instrument is allowed to come into thermal equilibrium with the substance whose temperature is to be measured, and the mouth of the cylinder is left open to standard pressure. The volume of the noble gas gives an indication of temperature.

As an idealization, we imagine a gas in which the atoms interact only with the sides of the container, not with each other. Such a gas is perfectly nonreactive (as the noble gases very nearly are), and never condenses to a liquid (as the noble gases do only at extremely low temperatures). Its atoms take up a negligible fraction of the available volume. Any gas can be made to behave very much like this if the pressure is extremely low, so that the atoms hardly ever encounter each other. Such a gas is called an ideal gas, and we define the Celsius scale in terms of the volume of the gas in a thermometer whose working substance is an ideal gas maintained at a fixed (very low) pressure, and which is calibrated at 0 and 100 degrees according to the melting and boiling points of water. The Celsius scale is not just a comparative scale but an additive one as well: every step in temperature is equal, and it makes sense to say that the difference in temperature between 18 and $$28°\text{C}$$ is the same as the difference between 48 and 58.

#### Absolute zero and the kelvin scale

We find that if we extrapolate a graph of volume versus temperature, the volume becomes zero at nearly the same temperature for all gases: -273°\textup{C}. Real gases will all condense into liquids at some temperature above this, but an ideal gas would achieve zero volume at this temperature, known as absolute zero. The most useful temperature scale in scientific work is one whose zero is defined by absolute zero, rather than by some arbitrary standard like the melting point of water. The temperature scale used universally in scientific work, called the Kelvin scale, is the same as the Celsius scale, but shifted by 273 degrees to make its zero coincide with absolute zero. Scientists use the Celsius scale only for comparisons or when a change in temperature is all that is required for a calculation. Only on the Kelvin scale does it make sense to discuss ratios of temperatures, e.g., to say that one temperature is twice as hot as another.

Example 6: Which temperature scale to use

$$\triangleright$$ You open an astronomy book and encounter the equation

$\begin{equation*} (\text{light emitted}) = (\text{constant}) \times T^ 4 \end{equation*}$

for the light emitted by a star as a function of its surface temperature. What temperature scale is implied?

$$\triangleright$$ The equation tells us that doubling the temperature results in the emission of 16 times as much light. Such a ratio only makes sense if the Kelvin scale is used.

Although we can achieve as good an approximation to an ideal gas as we wish by making the pressure very low, it seems nevertheless that there should be some more fundamental way to define temperature. We will construct a more fundamental scale of temperature in section 5.4.

##### Discussion Questions

j / Discussion questions A-C.

◊ Figure j/1 shows objects 1 and 2, each with a certain temperature $$T$$ and a certain amount of thermal energy $$E$$. They are connected by a thin rod, so that eventually they will reach thermal equilibrium. We expect that the rate at which heat is transferred into object 1 will be given by some equation $$dE_1/dt=k(...)$$, where $$k$$ is a positive constant of proportionality and “$$...$$” is some expression that depends on the temperatures. Suppose that the following six forms are proposed for “...:”

1. $$T_1$$
2. $$T_2$$
3. $$T_1-T_2$$
4. $$T_2-T_1$$
5. $$T_1/T_2$$
6. $$T_2/T_1$$

Give physical reasons why five of these are not possible.

◊ How should the rate of heat conduction in j/2 compare with the rate in j/1?

◊ The example in j/3 is different from the preceding ones because when we add the third object in the middle, we don't necessarily know the intermediate temperature. We could in fact set up this third object with any desired initial temperature. Suppose, however, that the flow of heat is steady. For example, the $$36°$$ object could be a human body, the $$0°$$ object could be the air on a cold day, and the object in between could be a simplified physical model of the insulation provided by clothing or body fat. Under this assumption, what is the intermediate temperature? How does the rate of heat conduction compare in the two cases?

◊ Based on the conclusions of questions A-C, how should the rate of heat conduction through an object depend on its length and cross-sectional area? If all the linear dimensions of the object are doubled, what happens to the rate of heat conduction through it? How would this apply if we compare an elephant to a shrew?