Introduction
There is a field of physics that studies the mechanical behavior of rigid objects, The objects of study were found in a solid state, they were rigid (of fixed shape and volume), and they were acted on by forces, In this lesson, we will study different phases: in particular, liquid and gas phases, And we will define basic concepts such as density and pressure, We will also study why some objects float and others don't, and how pressure spreads in a fluid.
Phases of Matter
The previous chapters analyzed one of the states that matter can be found in—the solid phase, At an atomic level, the solid state is defined by a clear structure where atoms and ions occupy rather fixed positions. Solid objects have a clear shape and volume. Although they are characterized by some elastic properties, it takes different strengths to change the shape of a solid object. This strength depends on the internal bonding of the particles.
We will now look at some basic characteristics of two other phases of matter—the liquid phase and gas phase. Liquids and gases together are also known as fluids.
Pressure
Analogy is a powerful tool in the world of science. Remember, we defined forces that deal with mechanical interaction and cause static and dynamic effects. We also defined torque as the anaJog of force, in cases where rotational motion (instead of linear motion) was a possible outcome of interaction. The objects of interaction were solid objects of definite shape. In the case of liquids, the shape is no longer a constraint: Depending on the container, a liquid can change its shape and its effect on the surrounding medium. What do we mean by that? Imagine this case: two liters of water (remember, liter is a unit for volume) divided equally into two containers of shapes as shown in Figure 10.1. Do you think they have the same interaction with the surface of support?
The answer is no because although each liter of water will act with the same weight on the surface of support, but the area is different in the two cases. If a person wears sneakers and then changes into high heels, the mark left on a soft asphalt will be different. It will be deeper when the person is on heels even though the weight of the person did not change. Although the weight was the same, the surface area of support is smaller with the heels, and therefore, the effect on the support surface area is larger. The quantity measuring the force and the effect on the surface on which it acts is called pressure.
There are many units for pressure, but the SI unit is 
Different fields of application deal with different ranges of pressure and with different measuring instruments. Therefore, applications have imposed different scales. Some of the most used scales are the pascal (Pa), the milibar (mb), the inch of mercury or milimeter of mercury (mmHg), and the pound per square inch (PSI).
Pressure
Pressure measures the force exerted perpendicularly per unit area.
If the force is due to the weight of the liquid, then the definition above becomes:
Units of Pressure
1 atm = 101,325 N/m2
1 mb = 1 hPa (hecto pascal) = 100 Pa
1 in Hg = 33.8639 mb = 3,386.39 Pa
1 mmHg = 133.32 Pa
1 PSI = 6.895 kPa
· 
Example
At sea level, the average atmospheric pressure is 760 mmHg. Find the SI value of this pressure and in inch Hg.
Solution
This problem requires us to convert pressure to SI units.
Density
As pressure takes the place of the simple forces acting on an area, what is similar to mass in the case of fluids? Mass is a quantity associated with matter, but at the same amount of mass, some objects occupy a lot of space. In fact, they will take all that you give them in the case of gases; some other objects will take very little space. The quantity that measures the amount of mass relative to the space occupied is called density.
As with pressure, density can be measured in many different units. The SI unit is kg/m3. Other units for density are Vcm3, g/m3, lb/ft3, oz/in3, and Ib/gal. The device used to measure density is a specific gravity bottle, called a pyknometer. The most commonly known density is for water at standard atmospheric pressure and temperature of 4° C: 1,000 kg/m3, or 1 kilogram of water occupying a volume of 1 liter.
Atomically, the explanation of different densities associated with different materials and phases is the arrangement of atoms in the structure of the material. Solids are more compact, presenting a clear structure, and their atoms and ions are closer to each other amounting to a larger mass in a given volume. This structure is mostly missing in liquids and is not existent in gases, where particles are freely moving, hence, they can occupy large volumes.
Density
Density measures the mass per unit volume.
Example
The Malibu Maxx's cargo space, according to GM data from 2003, is about 50% larger than other midsize sedans, and it is calculated to be 22.8 cubic feet. If the air density is 1.2 kg/m3, what is the quantity of air occupying that space?
Solution
The first step is to convert all data into SI.
V = 22.8 ft3 = 22.8 ft3 · 0.0283168 m3/1 ft3 = 0.65 m3
p = 1.2 kg/m3
m = ?
p = m/V
1.2 kg/m3 = m/0.65 m3
m = 1.2 kg/m3 · 0.65 m3 = 0.78 kg
m = 0.78 kg
Pressure and Depth
Let's begin with the definition of pressure and then isolate a cylindrical volume inside a container filled with a liquid as shown in Figure 10.2. The liquid is at rest and is occupying a space between levels h1 and h2 inside the water. The volume considered has a certain mass, and the pressure at the bottom of the column is due to the weight of the liquid. Hence, we can calculate the pressure of this volume of liquid.
If we consider a pressure exerted upon a fluid (air pressure at the surface of a lake for instance), then the variation of the pressure with depth is:
Where pois the pressure of the air and the second term, p · (h2 - h1) · g, is the contribution of the liquid.
In addition to this depth dependence, we have to add that at the same level in a fluid, the pressure is the same in all directions as shown in Figure 10.3. For example, an object floating in a liquid will be exerted with a pressure p in all directions.
Pressure
The pressure exerted by a fluid at rest is proportional to the height or depth of the fluid column and the density of the liquid.
Example
Lake Baikal in Russia has a depth of 5,371 ft. Find the pressure exerted on the bottom of the lake by the water layer.
Solution
Convert first into 51 units and then solve for the unknown: pressure.
h = 5,371 ft
p = 1,000 kg/m3
p = ?
h = 5,371 ft = 5,371 ft · 0.3048 m/1 ft = 1,637 m
p = p · h · g = 1,000 kg/m3 · 1,637 m · 9.8 m/s2 = 16,040 N/m2 = 1.604 · 104N/m2
p = 1.604 · 107N/m2
Archimedes's Principle
Recall that in mechanics, we defined the condition of equilibrium as referring to the net force on the object being zero. For an object to be at rest on a table, gravity has to be counteracted by another force, an equal and opposite force, and we call that reaction force. In a similar way, you can relate to the new concept we will discuss here: buoyancy. When an object is immersed in a fluid and floats at some level, the weight will act as if weighted down and there is no other force to be noted, or is there? Well, a piece of wood will not sink. And that is because the wood is pushed up by a force. We call this phenomena buoyancy. How large is the force of buoyancy? This is the subject of Archimedes's principle.
The force of buoyancy is usually symbolized by B and according to the principle, the expression is:
B = pfluid · Vfluid · g
Because in some cases, the object floats inside the liquid, the volume displaced is equal to the object's volume: Vliquid = Vobject. In other cases, the object floats at the surface of the liquid, and only part of the volume of the object will have to be taken into account.
Archimedes's Principle
The upward force exerted by a fluid on an object immersed in the fluid is equal to the weight of the displaced fluid.
Example
A totally submerged object floats in water as shown in Figure 10.4. Find the ratio of the object's density to the density of water.
Solution
First, let's show the forces acting on this object. The horizontal forces determined by the surrounding pressure will cancel each other out because at equal levels, the pressure is the same, and therefore, the forces will be the same. The object floats inside the liquid, so the weight and the buoyancy force are also equal. Using Archimedes's principle, we can write the equation for the vertical forces.
W – B=0
W – B = m · g – pliquid · Vliquid · g = 0
m = pobject · Vobject
0 = pobject · Vobject g – pliquid · Vliquid · g
Vobject = Vliquid
0 = (pobject – pliquid) · Vliquid · g
0 = (pobject – pliquid)
Pascal's Principle
As we mentioned at the beginning of this lesson, pressure is exerted in all directions in the same manner. Therefore, when the pressure changes, it will change in all directions and is also transmitted through the liquid. The manner of transmission constitutes the subject of Pascal's principle.
This phenomenon is very important in the way hydraulic presses function: A liquid transmits pressure between two mobile pistons reducing, the forces necessary to counteract a weight (see the following example).
Pascal's Principle
Any pressure applied to a static fluid is transmitted equally in all directions through the liquid.
Example
The diagram in Figure 10.5 shows a hydraulic press used in a car shop to lift cars. An automobile is placed on the large area of the press. The automobile has a mass of 1,540 kg, and the two sides of the press have a diameter of 25 and 15 cm, respectively. Find how much force will need to be applied on the right piston to counteract the weight of the car.
Solution
First, we convert the data into S1 units. Next, complete the diagram with the forces and pressure acting on the liquid, and then solve the problem.
25 cm = 25 cm · 1 m/l00 cm = 0.25 m
15 cm = 15 cm · 1 m/100 cm = 0.15 cm
m = 1,540 kg
F = ?
According to Pascal's principle, the pressure exerted by the automobile spreads equally in the liquid, and it will reach the piston on the left-hand side of the press.
p1 = p2
The pressure exerted by the car is:
The platform on which the car rests pushes down on a cylindrical piston of radius 0.25 m:
Al = π · r2 = π · (0.25 m)2 = 0.196 m2
The piston on which the counteraction is applied is also a cylindrical piston of radius 0.15 m:
A2 = π · r2 = π · (0.15 m)2 = 0.071m2
The pressures exerted on the pistons are:
And now we can solve for the unknown force.
Compare this force with the weight of the car:
W = 1,540 kg · 9.8 m/s2 = 15,092 N
W = 1,540 kg · 9.8 m/s2 = 15,092 N
Practice problems of this concept can be found at: Properties of Matter Practice Questions
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From Physics Success in 20 Minutes A Day. Copyright © 2006 by LearningExpress, LLC. All Rights Reserved.
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