1

BAYERO UNIVERSITY, KANO

PHY1230: BEHAVIOR OF MATTER (BoM)

PREPARED

BY

T. H. DARMA, PhD

AND

PRESENTED

BY

A. A. BISU, MSc. (MIEEE, MACM, MIET)

2012/2013 ACADEMIC SESSION

2

“REMEMBER THAT; THERE IS NO ROYAL PATH TO EDUCATION (LEARNING)”

READ HARD, PRAY HARD AND WORK HARD, THEN HAVE FUN! A WORD TO A WISE!

Structure of Matter

Atoms are composed of electrons, protons, and neutrons. Electron and protons are negative and

positive charges of the same magnitude, 1.6 × 10-19

Coulombs.

The mass of the electron is negligible with respect to those of the proton and the neutron, which

form the nucleus of the atom. The unit of mass is an atomic mass unit (amu) = 1.66 × 10-27

kg,

and equals 1/12 the mass of a carbon atom. The Carbon nucleus has Z=6, and A=6, where Z is

the number of protons, and A the number of neutrons. Neutrons and protons have very similar

masses, roughly equal to 1 amu. A neutral atom has the same number of electrons and protons,

Z.

A mole is the amount of matter that has a mass in grams equal to the atomic mass in amu of the

atoms. Thus, a mole of carbon has a mass of 12 grams. The number of atoms in a mole is called

the Avogadro number, Nav = 6.023 × 1023

. Note that Nav = 1 gram/1 amu.

Electrons in Atoms

The forces in the atom are repulsions between electrons and attraction between electrons and

protons. The electrons form a cloud around the nucleus moving in orbits. According to quantum

mechanics, only certain orbits are allowed. The orbits are identified by a principal quantum

number n, which can be related to the size, n = 0 is the smallest; n = 1, 2 .. are larger. (They are

"quantized" or discrete, being specified by integers). The angular momentum l is quantized, and

so is the projection in a specific direction m. The structure of the atom is determined by the Pauli

exclusion principle, only two electrons can be placed in an orbit with a given n, l, m – one for

each spin. The ability to gain or lose electrons is termed electronegativity or electropositivity, an

important factor in ionic bonds.

Bonding Forces and Energies

The Coulomb forces are simple: attractive between electrons and nuclei, repulsive between

electrons and between nuclei. The force between atoms is given by a sum of all the individual

forces, and the fact that the electrons are located outside the atom and the nucleus in the center.

When two atoms come very close, the force between them is always repulsive, because the

electrons stay outside and the nuclei repel each other. Unless both atoms are ions of the same

charge (e.g., both negative) the forces between atoms is always attractive at large internuclear

distances r. Since the force is repulsive at small r, and attractive at small r, there is a distance at

which the force is zero. This is the equilibrium distance at which the atoms prefer to stay.

3

The interaction energy is the potential energy between the atoms. It is negative if the atoms are

bound and positive if they can move away from each other. The interaction energy is the integral

of the force over the separation distance, so these two quantities are directly related. The

interaction energy is a minimum at the equilibrium position. This value of the energy is called the

bond energy, and is the energy needed to separate completely to infinity (the work that needs to

be done to overcome the attractive force.) The strongest the bond energy, the hardest is to move

the atoms, for instance the hardest it is to melt the solid, or to evaporate its atoms.

Ionic Bonding

This is the type of bond in which one of the atoms is negative (has an extra electron) and another

is positive (has lost an electron). Then there is a strong, direct Coulomb attraction. An example is

NaCl. In the molecule, there are more electrons around Cl, forming Cl- and less around Na,

forming Na+. Ionic bonds are strong. In real solids, ionic bonding is usually combined with

covalent bonding.

Covalent Bonding

In covalent bonding, electrons are shared between the molecules, to saturate the valency. The

simplest example is the H2 molecule, where the electrons spend more time in between the nuclei

than outside, thus producing bonding.

Metallic Bonding

In metals, the atoms are ionized, loosing some electrons from the valence band. Those electrons

form a electron sea, which binds the charged nuclei in place, in a similar way that the electrons

in between the H atoms in the H2 molecule bind the protons.

Secondary Bonding (Van der Waals)

Fluctuating Induced Dipole Bonds

Since the electrons may be on one side of the atom or the other, a dipole is formed: the + nucleus

at the center, and the electron outside. Since the electron moves, the dipole fluctuates. This

fluctuation in atom A produces a fluctuating electric field that is felt by the electrons of an

adjacent atom, B. Atom B then polarizes so that its outer electrons are on the side of the atom

closest to the + side (or opposite to the – side) of the dipole in A. This bond is called van der

Waals bonding.

Polar Molecule-Induced Dipole Bonds

A polar molecule like H2O (Hs are partially +, O is partially –), will induce a dipole in a nearby

atom, leading to bonding.

4

Permanent Dipole Bonds

This is the case of the hydrogen bond in ice. The H end of the molecule is positively charged and

can bond to the negative side of another dipolar molecule, like the O side of the H2O dipole.

GAS LAWS:

Boyle’s Law:- Boyle's law states that the volume of a given mass of gas is inversely

proportional to its pressure, if the temperature remains constant. Conversely, it shows that, at

constant temperature, the product of an ideal gas pressure and volume is always constant.

(1.1)

As a mathematical equation, Boyle's law is:

(1.2)

where P is the pressure (Pa), V the volume (m3) of a gas.

Charles's Law:- Charle’s law states that, for an ideal gas at constant pressure, the volume is

directly proportional to the absolute temperature (in kelvin).

(1.3)

Gay-Lussac's law:- Gay-Lussac’s law, or the pressure law, states that the pressure exerted on a

container's sides by an ideal gas is proportional to the absolute temperature.

(1.4)

Avogadro's law:- Avogadro’s law states that the volume occupied by an ideal gas is proportional

to the number of moles (or molecules) present in the container. This gives rise to the molar

volume of a gas, which at standard temperature and pressure (STP) is 22.4 dm3. The relation is

given by

(1.5)

where n is equal to the number of moles of gas (the number of molecules divided by Avogadro’s

number).

5

Combined gas law and ideal gas law:- The combined gas law or general gas equation is formed

by the combination of the three laws, and shows the relationship between the pressure, volume,

and temperature for a fixed mass of gas:

(1.6)

With the addition of Avogadro’s law, the combined gas law develops into the ideal gas law:

(1.7)

where P is pressure, , V is volume, n is the number of moles, R is the universal gas constant, T is

temperature (K)

An equivalent formulation of this law is:

(1.8)

where P is the absolute pressure, V is the volume, N is the number of gas molecules, k is the

Boltzmann constant (1.381×10−23

J·K−1

in SI units), T is the temperature (K)

Other Gas Laws:

Grahams law:- or law of diffusion states that the rate at which gas molecules diffuse is inversely

proportional to the square root of its density. Combined with Avogadro's law (i.e. since equal

volumes have equal number of molecules) this is the same as being inversely proportional to the

root of the molecular weight.

Dalton’s law of partial pressures:- states that the pressure of a mixture of gases is simply the

sum of the partial pressure’s of the individual components. Dalton's Law is stated

mathematically as follows:

, (1.9)

TUTORIAL

1. Using the gas equations, derive an expression for the general gas equation (equation of

state).

2. Water of mass 9.00 g is placed in a 2.00-L pressure cooker and heated to 500°C. What is

the pressure inside the container

3. One mole of oxygen gas is at a pressure of 6.00 atm and a temperature of 27.0°C. (a) If

the gas is heated at constant volume until the pressure triples, what is the final

temperature? (b) If the gas is heated until both the pressure and volume are doubled, what

is the final temperature?

6

4. A spray can containing a propellant gas at twice atmospheric pressure (202 kPa) and

having a volume of 125.00 cm3 is at 22°C. It is then tossed into an open fire. When the

temperature of the gas in the can reaches 195°C, what is the pressure inside the can?

Assume any change in the volume of the can is negligible.

HEAT & TEMPERATURE

Heat is a form of energy which flows from a region of higher temperature to a region of low

temperature. Temperature is a measure of hotness or coldness of a body. Heat flow occurs either

by conduction, convection or radiation. Conduction occurs in solid bodies in which the heat is

transferred through the vibration of the molecules of the body. Convection occurs in liquids and

gasses and involves the actual motion of the molecules from one point to another. Radiation

involves heat transfer through empty space i.e. it does not require a material body (slid, liquid or

gas).

Two objects are said to be in thermal contact if energy can be can be exchanged between them

due to a temperature difference. Thermal equilibrium is a situation in which two objects would

not exchange energy by heat or electromagnetic radiation if they were placed in thermal contact.

Thermometers and Temperature scales Thermometers are devices that are used to measure the temperature of a system. All

thermometers are based on the principle that some physical property of a system changes as the

system’s temperature changes. Some physical properties that change with temperature are (1) the

volume of a liquid, (2) the dimensions of a solid, (3) the pressure of a gas at constant volume, (4)

the volume of a gas at constant pressure, (5) the electric resistance of a conductor, and (6) the

color of an object. A temperature scale can be established on the basis of any one of these

physical properties.

A common thermometer in everyday use consists of a mass of liquid—usually mercury or

alcohol—that expands into a glass capillary tube when heated. In this case the physical property

that changes is the volume of a liquid. Any temperature change in the range of the thermometer

can be defined as being proportional to the change in length of the liquid column. The

thermometer can be calibrated by placing it in thermal contact with some natural systems that

remain at constant temperature. One such system is a mixture of water and ice in thermal

equilibrium at atmospheric pressure. On the Celsius temperature scale, this mixture is defined to

have a temperature of zero degrees Celsius, which is written as 0°C; this temperature is called

the ice point of water. Another commonly used system is a mixture of water and steam in

thermal equilibrium at atmospheric pressure; its temperature is 100°C, which is the steam point

of water. Once the liquid levels in the thermometer have been established at these two points, the

length of the liquid column between the two points is divided into 100 equal segments to create

the Celsius scale. Thus, each segment denotes a change in temperature of one Celsius degree.

7

Constant-Volume Gas thermometer

The physical change exploited in this device is the variation of pressure of a fixed volume of gas

with temperature. The thermometer is calibrated by using the ice and steam points of water as

follows.

Fig. 1 Constant-Volume gas thermometer

The flask is immersed in an ice-water bath, and mercury reservoir B is raised or lowered until the

top of the mercury in column A is at the zero point on the scale. The height h, the difference

between the mercury levels in reservoir B and column A, indicates the pressure in the flask at

0°C.

The flask is then immersed in water at the steam point, and reservoir B is readjusted until the top

of the mercury in column A is again at zero on the scale; this ensured that the gas’s volume is the

same as it was when the flask was in the ice bath. This adjustment of reservoir B gave a value for

the gas pressure at 100°C.

When these two pressure and temperature values are plotted, as shown in Figure 2, the line

connecting the two points serves as a calibration curve for unknown temperatures.

8

Fig. 2. Typical graph of pressure versus temperature taken with a constant-volume gas

thermometer.

If temperatures are measured with gas thermometers containing different gases at different initial

pressures, experiments show that the thermometer readings are nearly independent of the type of

gas used, as long as the gas pressure is low and the temperature is well above the point at which

the gas liquefies.

Fig. 3 Pressure versus temperature for experimental trials in which gases have different pressures

in a constant-volume gas thermometer

If the straight lines in Figure 3 is extended toward negative temperatures, a remarkable result is

found—in every case, the pressure is zero when the temperature is -273.15°C. This is the basis for

the absolute temperature scale, which sets -273.15°C as its zero point.

This temperature is often referred to as absolute zero. The size of a degree on the absolute

temperature scale is chosen to be identical to the size of a degree on the Celsius scale. Thus, the

conversion between these temperatures is

TC = T - 273.15

9

where TC is the Celsius temperature and T is the absolute temperature.

The Celsius, Fahrenheit, and Kelvin Temperature Scales

A common temperature scale in everyday use in the United States is the Fahrenheit Scale, while

most countries in the world uses the Celcius scale. The Fahrenheit scale sets the temperature of

the ice point at 32°F and the temperature of the steam point at 212°F. The relationship between

the Celsius and Fahrenheit temperature scales is

Conversely,

Exercises:

1. A pan of water is heated from 25°C to 80°C. What is the change in its temperature on the

Kelvin scale and on the Fahrenheit scale?

2. On a day when the temperature reaches 50°F, what is the temperature in degrees Celsius and

in Kelvin?

Zeroth Law of Thermodynamics

The zeroth law states that if two systems are separately in thermal equilibrium with a third

system, they are also in thermal equilibrium with each other.

Temperature is a property that determines whether an object is in thermal equilibrium with other

objects. Two objects in thermal equilibrium with each other are at the same temperature.

Conversely, if two objects have different temperatures, then they are not in thermal equilibrium

with each other.

Thermal Expansion of Solids and Liquids The underlining principle of the liquid thermometer makes use of one of the best-known changes

in a substance: as its temperature increases, its volume increases. This phenomenon, known as

thermal expansion, has an important role in numerous engineering applications. For example,

thermal-expansion joints must be included in buildings, concrete highways, railroad tracks, brick

walls, and bridges to compensate for dimensional changes that occur as the temperature changes.

Thermal expansion is a consequence of the change in the average separation between the atoms

in an object.

At ordinary temperatures, the atoms in a solid oscillate about their equilibrium positions. The

average spacing between the atoms is about 10-10

m. As the temperature of the solid increases,

the atoms oscillate with greater amplitudes; as a result, the average separation between them

increases. Consequently, the object expands.

10

If thermal expansion is sufficiently small relative to an object’s initial dimensions, the change in

any dimension is, to a good approximation, proportional to the first power of the temperature

change. Suppose that an object has an initial length Li along some direction at some temperature

and that the length increases by an amount ΔL for a change in temperature ΔT. Considering the

fractional change in length per degree of temperature change, the average coefficient of linear

expansion is defined as;

Experiments show that α is constant for small changes in temperature. For purposes of

calculation, this equation is usually rewritten as;

or as

where Lf is the final length, Ti and Tf are the initial and final temperatures, and the proportionality

constant α is the average coefficient of linear expansion for a given material and has units of

(°C)-1

.

Because the linear dimensions of an object change with temperature, it follows that surface area

and volume change as well. The change in volume is proportional to the initial volume Vi and to

the change in temperature according to the relationship;

where β is the average coefficient of volume expansion. For a solid, the average coefficient

of volume expansion is three times the average linear expansion coefficient. i.e. :

β = 3α

Exercise: Show that for a given body; β = 3α

Each substance has its own characteristic average coefficient of expansion as shown in Table 1

.

Table 1. Average linear and volume expansion coefficients of materials

11

Examples:

1. A segment of steel railroad track has a length of 30.000 m when the temperature is 0.0°C.

What is its length when the temperature is 40.0°C?

The Unusual Behavior of Water

Liquids generally increase in volume with increasing temperature and have average coefficients

of volume expansion about ten times greater than those of solids. Cold water is an exception to

this rule..As the temperature increases from 0°C to 4°C, water contracts and thus its density

increases as shown in the figure 4 below. Above 4°C, water expands with increasing

temperature, and so its density decreases. Thus, the density of water reaches a maximum value of

1.000 g/cm3 at 4°C.

This explains why a pond begins freezing at the surface rather than at the bottom. When the

atmospheric temperature drops from, for example, 7°C to 6°C, the surface water also cools and

consequently decreases in volume. This means that the surface water is denser than the water

below it, which has not cooled and decreased in volume. As a result, the surface water sinks, and

warmer water from below is forced to the surface to be cooled. When the atmospheric

temperature is between 4°C and 0°C, however, the surface water expands as it cools, becoming

less dense than the water below it. The mixing process stops, and eventually the surface water

freezes. As the water freezes, the ice remains on the surface because ice is less dense than water.

The ice continues to build up at the surface, while water near the bottom remains at 4°C.

12

Fig. 4. The variation in the density of water at atmospheric pressure with temperature.

Heat and Internal Energy

Heat is defined as the transfer of energy across the boundary of a system due to a temperature

difference between the system and its surroundings Internal energy is all the energy of a system

that is associated with its microscopic components—atoms and molecules—when viewed from a

reference frame at rest with respect to the center of mass of the system. Internal energy includes

kinetic energy of random translational, rotational, and vibrational motion of molecules, potential

energy within molecules, and potential energy between molecules.

Units of Heat

One of the oldest unit of heat is the calorie (cal), which is defined as the amount of energy

transfer necessary to raise the temperature of 1 g of water from 14.5°C to 15.5°C. Another unit

of energy is the British thermal unit (Btu), which is defined as the amount of energy transfer

required to raise the temperature of 1 lb of water from 63°F to 64°F. In SI unit, heat, work, and

internal energy are usually measured in joules, which is also the unit of energy.

1 cal = 4.186 J

Specific Heat and Calorimetry

When energy is added to a system and there is no change in the kinetic or potential energy of the

system, the temperature of the system usually rises (except in the case in which a system

undergoes a change of state.

If the system consists of a sample of a substance, it is found that the quantity of energy required

to raise the temperature of a given mass of the substance by some amount varies from one

substance to another. The heat capacity C of a particular sample of a substance is defined as the

13

amount of energy needed to raise the temperature of that sample by 1°C. From this definition, it

is seen that if energy Q produces a change ΔT in the temperature of a sample, then

Q = CΔT

The specific heat c of a substance is the heat capacity per unit mass. Thus, if energy Q transfers

to a sample of a substance with mass m and the temperature of the sample changes by ΔT, then

the specific heat of the substance is

Specific heat is essentially a measure of how thermally insensitive a substance is to the addition

of energy. The greater a material’s specific heat, the more energy must be added to a given mass

of the material to cause a particular temperature change. The energy Q transferred between a

sample of mass m of a material and its surroundings to a temperature change ΔT as

Specific heat varies with temperature. However, if temperature intervals are not too great, the

temperature variation can be ignored and c can be treated as a constant. Specific heat capacities

of some substances at 25oC and atmospheric pressure, is given in Table 2.

Conservation of Energy: Calorimetry

One technique for measuring specific heat involves heating a sample to some known temperature

Tx, placing it in a vessel containing water of known mass and temperature Tw <' Tx, and

measuring the temperature of the water after equilibrium has been reached. This technique is

called calorimetry, and devices in which this energy transfer occurs are called calorimeters.

If the system of the sample and the water is isolated, the law of the conservation of energy

requires that the amount of energy that leaves the sample (of unknown specific heat) equal the

amount of energy that enters the water

Qcold = - Qhot The negative sign indicating that the sample has lost energy when transferred into the vessel

containing the water

Example:

1. An ingot of metal of mass 0.050 kg is heated to 200.0°C and then dropped into a beaker

containing 0.400 kg of water initially at 20.0°C. If the final equilibrium temperature of the mixed

system is 22.4°C, find the specific heat of the metal.

14

2. A policeman fires a silver bullet with a muzzle speed of 200 m/s into the wall of a saloon.

Assume that all the internal energy generated by the impact remains with the bullet. What is the

temperature change of the bullet?

Phase change or Change of state: Latent Heat A substance often undergoes a change in temperature when energy is transferred between it and

its surroundings. There are situations, however, in which the transfer of energy does not result in

a change in temperature. This is the case whenever the physical characteristics of the substance

change from one form to another; such a change is commonly referred to as a phase change or

change of state. Two common phase changes are from solid to liquid (melting) and from liquid

to gas (boiling or vaporisation). All such phase changes involve a change in internal energy but

no change in temperature. The increase in internal energy in boiling, for example, is represented

by the breaking of bonds between molecules in the liquid state; this bond breaking allows the

15

molecules to move farther apart in the gaseous state, with a corresponding increase in

intermolecular potential energy.

If a quantity Q of energy transfer is required to change the phase of a mass m of a substance, the

ratio L= Q/m characterizes an important thermal property of that substance. Because this added

or removed energy does not result in a temperature change, the quantity L is called the latent heat

(literally, the ―hidden‖ heat) of the substance. The value of L for a substance depends on the

nature of the phase change, as well as on the properties of the substance. The energy required to

change the phase of a given mass m of a pure substance is thus

Q = ± mL

The positive sign is used when energy enters a system, causing melting or vaporization. The

negative sign corresponds to energy leaving a system, such that the system freezes or condenses.

Latent heat of fusion Lf is the term used when the phase change is from solid to liquid, and latent

heat of vaporization Lv is the term used when the phase change is from liquid to gas (the liquid

―vaporizes‖). Table 3 shows the latent heats of fusion and vaporization of some substances

Table 3. latent heats of fusion and vaporization of some substances

Example: Determine the energy required to convert a 1.00-g cube of ice at -30.0°C to steam at

100.0°C.

The latent heat of fusion (Lf) of a crystalline solid is the quantity of heat required to melt a unit

mass of the solid at constant temperature.

The latent heat of vaporization (Lv) of a liquid is the quantity of heat required to vaporize a unit

mass of the liquid at constant temperature.

The heat of sublimation of a substance is the quantity of heat required to convert a unit mass of

the substance from the solid to the gaseous state at constant temperature.

First Law of Thermodynamics

16

The first law of thermodynamics is a statement of the law of conservation of energy. It states that

if an amount of heat (ΔQ) flows into a system, then this energy must appear as increased internal

energy ΔU for the system and /or work ΔW done by the system on its surroundings.

Mathematically, the First law is written as;

ΔQ = ΔU + ΔW

Work Done by a System

Heat supplied to a system may increase its internal energy and enable it to expand and thereby do

external work. The work done by a system (ΔW) is positive if the system thereby loses energy to

its surroundings (e.g. expansion). When the surroundings do work on the system to give it

energy, ΔW is a negative quantity (e.g. compression). In a small expansion ΔV, a fluid at

constant pressure does work given by

ΔW = P ΔV

Consider a mass of gas enclosed in a cylinder. If the gas expands by moving the piston through a

small distance δx, which is so small that the pressure P remains constant during the expansion,

the external work done δw by the gas against the force f is

δw = f δx = PA δx = P δv

where δv is the increase in volume of the gas.

The total work done W by the gas during finite expansion from V1 to V2 is

If P is constant during the expansion from V1 to V2, then,

W = P(V2 – V1)

Q. A litre of air, initially at 20oC and at 76.0cmHg pressure, is heated at constant pressure until

its volume is doubled. Find (a) the final temperature, (b) the external work done by the air in

expanding.

Sol.

V1 = 1 lit. = 0.001m3, T1 = 20

oC = 293K, P = 76.0cmHg = 1.012 x10

5 N/m

2, V2 = 2V1, T2 = ?, W

= ?

(a) V1/ T1 = V2/ T2 hence, T2 = V2 T1/ V1 = 586K

(b) W = P(V2 – V1) hence, W = 1.012 x105( 2 x 0.001 – 0.001) = 101.2J

17

Specific heat of Gases

Suppose a gas is heated in a closed vessel. If the gas is not allowed, then it does no work and all

the heat goes to increase its internal energy. The heat required to warm unit mass of a gas

through one degree, when its volume is kept constant, is called the specific heat capacity of the

gas at constant volume and is denoted by cv, expressed in JKg-1

K-1

.

If the gas is allowed to expand, then external work is done in addition to the increase in its

internal energy. The work done depends on the increase in volume of the gas which in turn

depends on the way the gas expands. Suppose the pressure remains constant during this

expansion. The heat required to warm unit mass of a gas at constant pressure is called the

specific heat capacity of the gas at constant pressure and is denoted by cp, expressed in the same

unit as cv.

For a temperature rise of 1oC of one mole of the gas, since the internal energy is independent of

volume, the rise in internal energy is cv, and the heat supplied to raise the temperature at

constant pressure is cp. i.e.

δQ = cp δT, and δU = cv δT, while δW = p δV

therefore, the expression for the first law becomes;

cp δT = cv δT + p δV

Applying the ideal gas equation (PV = RT) for one mole of gas to the initial and final stages;

P(V + δV) = R(T + δT) or PδV = RδT

Using these last two equations we have;

Cp = cv + R or cp – cv = R

The internal energy of an ideal monoatomic gas consists entirely of translational kinetic energy

and using the kinetic theory,

If the temperature increase by δT and the internal energy by δU, using δU = cv δT then

U + δU = 3/2R(T + δT) or δU = 3/2R δT

Therefore;

cv = 3/2R and since cp = cv + R, then cp = 5/2R

The ratio cp/cv = γ. For the monoatomic gas γ = 5/3

For diatomic molecule;

U = 5/2RT, hence, cv = 5/2R and cp = 7/2R which means γ = 7/5

For polyatomic molecule;

U = 6/2RT, hence cv = 6/2R and cp = 8/2R which means γ = 8/6

Thermodynamic Processes

Adiabatic process: An adiabatic process is one during which no energy enters or leaves the

system by heat—that is, Q = 0. An adiabatic process can be achieved either by thermally

insulating the walls of the system, or by performing the process rapidly, so that there is

18

negligible time for energy to transfer by heat. Applying the first law of thermodynamics to an

adiabatic process, we see that

U = W (adiabatic process)

From this result, if a gas is compressed adiabatically such that W is positive, then U is positive

and the temperature of the gas increases. Conversely, the temperature of a gas decreases when

the gas expands adiabatically.

In a reversible adiabatic change (an adiabatic change that can be reversed through the same P, V,

and T values, the first law (with δT Q = 0) is written as

cv δT + p δV = 0

To eliminate δT, the general equation of the ideal gas law is used for one mole of gas. i.e.

pV = RT

Since both pressure and volume may change, the above equation can be differentiated as;

p δV + V δp = R δT or

Recall that cv δT + p δV = 0; therefore,

cv

+ p δV = 0 or, cv(p δV + V δp) + Rp δV

but R = cp – cv. Therefore,

cv(p δV + V δp) + (cp – cv)p δV = 0

or,

cvVδp + cpp δV = 0

hence,

Therefore,

Integrating this equation gives;

Or

i.e. piVi ؞γ = pfVf

γ

This is the equation for a reversible adiabatic change.

In order to introduce temperature into this equation, the general gas equation is used as;

pV = RT or p = RT/V

Therefore,

Or,

19

i.e.

Isobaric process: A process that occurs at constant pressure is called an isobaric process. An

isobaric process could be established by allowing the piston to move freely so that it is always in

equilibrium between the net force from the gas pushing upward and the weight of the piston plus

the force due to atmospheric pressure pushing downward.

In such a process, the values of the heat and the work are both usually nonzero. The work done

on the gas in an isobaric process is simply

W = - P(Vf – Vi)

where P is the constant pressure.

Isovolumeric process: A process that takes place at constant volume is called an isovolumetric

process. In such a process, the value of the work done is zero because the volume does not

change. Hence, from the first law,

Q = U

This expression specifies that if energy is added by heat to a system kept at constant volume,

then all of the transferred energy remains in the system as an increase in its internal energy.

Isothermal process: A process that occurs at constant temperature is called an isothermal

process. A plot of P versus V at constant temperature for an ideal gas yields a hyperbolic curve

called an isotherm.

The internal energy of an ideal gas is a function of temperature only. Hence, in an isothermal

process involving an ideal gas, U = 0. For an isothermal process, then, it follows from the first

law that the energy transfer Q must be equal to the negative of the work done on the gas—that is,

Q = - W. Any energy that enters the system by heat is transferred out of the system by work; as a

result, no change in the internal energy of the system occurs in an isothermal process.

Isothermal Expansion of an Ideal Gas

Suppose that an ideal gas is allowed to expand quasi-statically at constant temperature.

This process is described by the PV diagram.

PV diagram for an isothermal expansion of an ideal gas from an initial state i to a final state f.

20

Because the gas is ideal and the process is quasi-static, then PV = nRT for each point on the path.

The work done on the gas during expansion from state i to state f. is;

Since T is constant,

Therefore,

Numerically, this work W equals the negative of the shaded area under the PV curve. Because the

gas expands, Vf > Vi and the value for the work done on the gas is negative, as expected. If the

gas is compressed, then Vf < Vi and the work done on the gas is positive.

Q. A 1.0-mol sample of an ideal gas is kept at 0.0°C during an expansion from 3.0 L to 10.0 L.

How much work is done on the gas during the expansion?

Sol. This is an isothermal expansion. Therefore,

Using the first law and considering that the heat Q is different for each path, specific heats for

two processes that frequently occur can be defined: changes at constant volume and changes at

constant pressure. Because the number of moles is a convenient measure of the amount of gas

the molar specific heats associated with these processes are defined with the following equations:

Q = nCV ΔT (constant volume)

Q = nCP ΔT (constant pressure)

where CV is the molar specific heat at constant volume and CP is the molar specific heat at

constant pressure.

KINETIC THEORY OF GASES

The model shows that the pressure that a gas exerts on the walls of its container is a consequence

of the collisions of the gas molecules with the walls. In developing this model, we make the

following assumptions:

1. The number of molecules in the gas is large, and the average separation between them is large

compared with their dimensions.

2. The molecules obey Newton’s laws of motion, but as a whole they move randomly.

3. The molecules interact only by short-range forces during elastic collisions.

4. The molecules make elastic collisions with the walls

5. The gas under consideration is a pure substance; that is, all molecules are identical.

21

The pressure exerted on the wall is;

This result indicates that the pressure of a gas is proportional to the number of molecules per unit

volume and to the average translational kinetic energy of the molecules.

THERMAL CONDUCTIVITY

Consider a slab of material as shown below. Its thickness is L and its cross-sectional area is A.

The temperature of its two faces are T1 and T2, so the temperature difference between the two

faces is ΔT = T1 – T2. The quantity ΔT/L is called the temperature gradient. It is the rate of

change of temperature with distance.

The quantity of heat transmitted from face 1 to face 2 in time Δt is given by

where k depends on the material of the slab and is called the thermal conductivity and has unit of

Js-1

m-1

K-1

or W/m.K.

Example:

1. Find the amount of heat that is conducted through a 2m2 brick wall 12cm in thickness in

1 hour if the temperature on one side is 8oC and 28

oC on the other side. Given that

thermal conductivity of brick is 0.13Wm-1

K-1

.

Sol. A = 2m2, L = 12cm = 12x10

-2m, t = 1 hour = 3600s, T1 = 8

oC, T2 = 28

oC,

k = 0.13Wm-1

K-1

, Q = ?

Q = kAt x temperature gradient

= 0.13 x 2 x 3600 x 28 – 8/ 12x10-2

= 15600J

22

2. Two slabs of thickness L1 and L2 and thermal conductivities k1 and k2 are in thermal

contact with each other, as shown below. The temperatures of their outer surfaces are Tc

and Th , respectively, and Th > Tc . Determine the temperature at the interface in the

steady-state condition.

Sol. Steady-state condition means that energy transfers through the compound slab is at the

same rate at all points and the temperature at the interface is fixed.

The energy transfer through slab 1 is;

Energy transfer through slab 2 is;

When a steady state is reached, these two rates must be equal; hence,

Solving for T gives;

SECOND LAW OF THERMODYNAMICS

The second law of thermodynamics, establishes which thermodynamic processes do and which

do not occur. The second law can be stated in three equivalent ways;

1. Heat flows spontaneously from a hotter to a colder object but not vice-versa

2. No heat engine that cycles continuously can change all its heat into useful work

3. If a system undergoes spontaneous change, it will change in such a way that its entropy

will increase or, at best remain constant.

Elastic Properties of Solids

23

All objects are deformable when subjected to external forces. That is, it is possible to change the

shape or the size (or both) of an object by applying external forces. As these changes take place,

however, internal forces in the object resist the deformation.

The deformation of solids can be described in terms of the concepts of stress and strain. Stress is

a quantity that is proportional to the force causing a deformation; more specifically, stress is the

external force acting on an object per unit cross-sectional area. The result of a stress is strain,

which is a measure of the degree of deformation. It is found that, for sufficiently small stresses,

strain is proportional to stress; the constant of proportionality depends on the material being

deformed and on the nature of the deformation. The proportionality constant is called the elastic

modulus. The elastic modulus is therefore defined as the ratio of the stress to the resulting strain

Elastic modulus = stress/strain

Because strain is dimensionless, the unit of the elastic modulus is simply the unit of stress (F/A =

N/m2). The elastic modulus in general relates what is done to a solid object (a force is applied) to

how that object responds (it deforms to some extent). Three types of deformation and their

respective elastic modulus are;

1. Young’s modulus, which measures the resistance of a solid to a change in its length

2. Shear modulus, which measures the resistance to motion of the planes within a solid parallel

to each other

3. Bulk modulus, which measures the resistance of solids or liquids to changes in their volume

Young’s Modulus: Elasticity in Length

A long bar clamped at one end is stretched by an amount ΔL under the action of a force F.

The tensile stress is the ratio of the magnitude of the external force F to the cross-sectional area

A. The tensile strain in this case is defined as the ratio of the change in length ΔL to the original

length Li. the Young modulus E is thus defined as;

E = stress/strain =

Young’s modulus is typically used to characterize a rod or wire stressed under either tension or

compression.

For relatively small stresses, the bar will return to its initial length when the force is removed.

The elastic limit of a substance is defined as the maximum stress that can be applied to the

substance before it becomes permanently deformed and does not return to its initial length. It is

possible to exceed the elastic limit of a substance by applying a sufficiently large stress, as seen

in the stress-strain curve.

24

Stress – strain curve for an elastic rod

Initially, a stress-versus-strain curve is a straight line. As the stress increases, however, the curve

is no longer a straight line. When the stress exceeds the elastic limit, the object is permanently

distorted and does not return to its original shape after the stress is removed. As the stress is

increased even further, the material ultimately breaks.

Energy stored in a wire:

The work done when a wire of length l and area A is extended by an amount e due to a force that

increase from zero to F is;

Work done = force x distance = average force x distance = ½Fe

Using the expression for Young modulus E, the energy can be expressed as;

Shear Modulus: Elasticity of Shape

Another type of deformation occurs when an object is subjected to a force parallel to one of its

faces while the opposite face is held fixed by another force as shown in (a) below. The stress in

this case is called a shear stress. If the object is originally a rectangular block, a shear stress

results in a shape whose cross section is a parallelogram.

A book pushed sideways, as shown in (b), is an example of an object subjected to a shear stress.

25

The shear stress (F/A), is the ratio of the tangential force to the area A of the face being sheared.

The shear strain is defined as the ratio .Δx/h, where Δx is the horizontal distance that the sheared

face moves and h is the height of the object. In terms of these quantities, the shear modulus is

Bulk Modulus: Volume Elasticity

Bulk modulus characterizes the response of an object to changes in a force of uniform magnitude

applied perpendicularly over the entire surface of the object, as shown below.

The volume stress is defined as the ratio of the magnitude of the total force F exerted on a

surface to the area A of the surface. The quantity P = F/A is called pressure. If the pressure on an

object changes by an amount ΔP = ΔF/A, then the object will experience a volume change ΔV.

The volume strain is equal to the change in volume ΔV divided by the initial volume Vi.

The Bulk modulus is defined as;

A negative sign is inserted in this defining equation so that B is a positive number. This is

necessary because an increase in pressure (positive ΔP) causes a decrease in volume (negative

ΔV ) and vice versa.

The reciprocal of the bulk modulus is called the compressibility of the material.

26

Exercises:

1. A 200-kg load is hung on a wire having a length of 4.00 m, cross-sectional area

0.200 x 10-4

m2, and Young’s modulus 8.00 x 10

10 N/m

2. What is its increase in length?

2. Assume that Young’s modulus is 1.50 x 1010

N/m2 for bone and that the bone will

fracture if stress greater than 1.50 x108 N/m

2 is imposed on it. (a) What is the maximum

force that can be exerted on the femur bone in the leg if it has a minimum effective

diameter of 2.50 cm? (b) If this much force is applied compressively, by how much does

the 25.0-cm-long bone shorten?

27

FLUIDS MECHANICS

Fluid statics:

A fluid is a collection of molecules that are randomly arranged and held together by weak

cohesive forces and by forces exerted by the walls of a container. Both liquids and gases are

fluids. Fluids do not sustain shearing stresses or tensile stresses; thus, the only stress that can be

exerted on an object submerged in a static fluid (a fluid at rest) is one that tends to compress the

object from all sides. In other words, the force exerted by a static fluid on an object is always

perpendicular to the surfaces of the object.

For a fluid at rest, the average pressure on a surface of area A is defined as the normal force

acting on the area.

Variation of Pressure with Depth

The pressure in a liquid increases with depth. Consider a liquid of density ρ at rest as shown in

the figure below. It is assumed that the liquid is incompressible (i.e. ρ is uniform throughout the

liquid).

Choosing a sample of the liquid contained within an imaginary cylinder of cross-sectional area A

extending from depth d to depth d + h.

The liquid exerts forces at all points on the surface of the sample, perpendicular to the surface.

The pressure exerted by the liquid on the bottom face of the sample is P, and the pressure on the

top face is Po. Therefore, the upward force exerted by the outside fluid on the bottom of the

cylinder has a magnitude PA, and the downward force exerted on the top has a magnitude PoA.

The mass of liquid in the cylinder is M = ρV = ρAh; therefore, the weight of the liquid in the

28

cylinder is Mg = ρAhg. Because the cylinder is in equilibrium, the net force acting on it must be

zero. Choosing upward to be the positive y direction, it follows that

PA - PoA - ρAhg = 0

PA - PoA = ρAhg

or,

P = Po + ρhg

That is, the pressure P at a depth h below a point in the liquid at which the pressure is Po is

greater by an amount ρgh. P is known as the absolute pressure, while the difference P – Po is

known as the gauge pressure.

If the liquid is open to the atmosphere and Po is the pressure at the surface of the liquid, then Po

is atmospheric pressure.

The pressure of the atmosphere, which varies over a small range from the standard value (1.00

atm = 1.013 x 105Pa) is usually measured with a barometer. Other simple pressure measuring

devices include the open-tube manometer.

In view of the fact that the pressure in a fluid depends on depth and on the value of Po, any

increase in pressure at the surface must be transmitted to every other point in the fluid. This

concept was first recognized by the French scientist Blaise Pascal (1623–1662) and is called

Pascal’s law: which states that a change in the pressure applied to a fluid is transmitted

undiminished to every point of the fluid and to the walls of the container.

An important application of Pascal’s law is the hydraulic press illustrated in the figure below.

Diagram of a hydraulic press

A force of magnitude F1 is applied to a small piston of surface area A1. The pressure is

transmitted through an incompressible liquid to a larger piston of surface area A2. Because the

pressure must be the same on both sides,

Therefore, the force F2 is greater than the force F1 by a factor A2/A1. By designing a hydraulic

press with appropriate areas A1 and A2, a large output force can be applied by means of a small

input force. Hydraulic brakes, car lifts, hydraulic jacks, and forklifts all make use of this

principle.

29

Because liquid is neither added nor removed from the system, the volume of liquid pushed down

on the left in Figure 14.4a as the piston moves downward through a displacement 'x1 equals the

volume of liquid pushed up on the right as the right piston moves upward through a displacement

Δx2. That is, A1Δx1 = A2Δx2; thus, A2/A1 = Δx1/Δx2 . Recalling that A2/A1 = F2/F1 . Thus, F2/F1

= Δx1/Δx2 , so F1Δx1 = F2 Δx2. Each side of this equation is the work done by the force. Thus, the

work done by F1 on the input piston equals the work done by F2 on the output piston, as it must

in order to conserve energy.

Examples

1. A 50.0-kg woman balances on one heel of a pair of high-heeled shoes. If the heel is circular

and has a radius of 0.500 cm, what pressure does she exert on the floor?

Sol.

P = F/A

F = mg; m = 50.0kg, g = 9.8m/s2, A = πr

2, r = 0.50cm = 0.5 x 10

-2m

Therefore,

P = 50.0 x 9.8/ π(0.5 x 10-2

)2 = 6.24 x 10

6N/m

2

2. In a car lift used in a service station, compressed air exerts a force on a small piston that has a

circular cross section and a radius of 5.00 cm. This pressure is transmitted by a liquid to a piston

that has a radius of 15.0 cm. What force must the compressed air exert to lift a car weighing 13

300 N? What air pressure produces this force?

Sol.

F1/A1 = F2/A2

r1 = 5.0cm = 0.05m, r2 = 15.0cm = 0.15m, A1 = πr1

2, A2 = πr2

2, F2 = 13300N, F1 = ?

Therefore,

F1 = F2 x A1/A2 = 13300 x π(0.05)2/π(0.15)

2 = 1.48 x 10

3N

3. Calculate the absolute pressure at an ocean depth of 1000 m. Assume the density of seawater

is 1024 kg/m3 and that the air above exerts a pressure of 101.3 kPa.

Sol.

P = Po + ρhg

Po = 101.3kPa = 1.013x105Pa, ρ = 1024 kg/m

3, h = 1000m, g = 9.80m/s

2.

Therefore, P = 1.013x105 + 1024 x 1000 x 9.80

= 1.014x107Pa

4. The four tires of an automobile are inflated to a gauge pressure of 200 kPa. Each tire has an

area of 0.024 0 m2 in contact with the ground. Determine the weight of the automobile.

Sol.

Let the automobile weight = Fm. Each tire therefore supports Fm/4

P = F/A = Fm/4A. therefore, Fm = 4AP = 4 x 0.024 x 200000 = 1.92 x 104N

30

Density and Relative Density

The density of a substance is defined as its mass per unit volume.

The density of a substance is often expressed relative to the density of water. This is called

relative density or specific gravity of the substance

If the weight of a solid object in air is mo, and its weight when totally immersed in water is m1,

then

upthrust = mo – m1 = weight of water displaced.

The relative density of the solid is then given by;

and density of solid

The density or relative density of a liquid can be found by weighing a solid in air (mo), then

weighing it totally immersed in the liquid (m1), and finally weighing it totally immersed in water

(m2). Hence;

mo – m2 = upthrust in water = weight of water displaced

mo – m1 = upthrust in liquid = weight of liquid displaced

or

Buoyant Forces and Archimedes’s Principle

An object immersed in a fluid experiences a resultant upward force due to the pressure of the

fluid on it. The upward force is called a buoyant force,.or upthrust. The upthrust is the difference

between (i) the weight of the object in air when attached to a spring balance, and (ii) the reduced

weight on the spring balance when the object is totally immersed in a liquid. The magnitude of

the buoyant force always equals the weight of the fluid displaced by the object. This statement is

known as Archimedes’s principle.

When an object is totally submerged in a fluid of density ρflu, the magnitude of the upward

buoyant force is B = ρflugV = ρflugVobj, where Vobj is the volume of the object. If the object has a

mass M and density ρobj, its weight is equal to F = Mg = ρobjgVobj, and the net force on it is B - F

= (ρflu - ρobj)gVobj. Hence, if the density of the object is less than the density of the fluid, then the

downward gravitational force is less than the buoyant force, and the unsupported object

31

accelerates upward (Fig. a). If the density of the object is greater than the density of the fluid,

then the upward buoyant force is less than the downward gravitational force, and the

unsupported object sinks (Fig. b).

If the density of the submerged object equals the density of the fluid, the net force on the object

is zero and it remains in equilibrium. Thus, the direction of motion of an object submerged in a

fluid is determined only by the densities of the object and the fluid.

Floating Object

When an object of volume Vobj and density ρobj < ρflu in static equilibrium float on the surface of

a fluid—that is, an object that is only partially submerged, the upward buoyant force is balanced

by the downward gravitational force acting on the object. If Vflu is the volume of the fluid

displaced by the object (this volume is the same as the volume of that part of the object that is

beneath the surface of the fluid), the buoyant force has a magnitude B = ρflugVflu. Since the

weight of the object is F = Mg = ρobjgVobj, and because F = B, it follows that

ρflugVflu = ρobjgVobj, or

This equation implies that the fraction of the volume of a floating object that is below the fluid

surface is equal to the ratio of the density of the object to that of the fluid.

32

Example

A piece of aluminum with mass 1.00 kg and density 2 700 kg/m3 is suspended from a string and

then completely immersed in a container of water. Calculate the tension in the string (a) before

and (b) after the metal is immersed.

Sol.

(a) Before the metal is immersed,

Therefore T1 = Mg = 1.00 x 9.8 = 9.8N

(b) After the metal is immersed,

Therefore T2 = Mg – B = Mg – (ρwV)g, where V = M/ρm = 1/2700

Thus T2 = 1.0 x 9.8 – 1000 x 1/2700 x 9.8

= 9.8 – 3.63 = 6.17N

Fluid dynamics:

When fluid is in motion, its flow can be characterized as being one of two main types. The flow

is said to be steady, or laminar, if each particle of the fluid follows a smooth path, such that the

paths of different particles never cross each other. In steady flow, the velocity of fluid particles

passing any point remains constant in time.

The path taken by a fluid particle under steady flow is called a streamline. The velocity

of the particle is always tangent to the streamline

33

Above a certain critical speed, fluid flow becomes turbulent; turbulent flow is irregular flow

characterized by small whirlpool-like regions.

The term viscosity is commonly used in the description of fluid flow to characterize the degree

of internal friction in the fluid. This internal friction, or viscous force, is associated with the

resistance that two adjacent layers of fluid have to moving relative to each other. Viscosity

causes part of the kinetic energy of a fluid to be converted to internal energy. This mechanism is

similar to the one by which an object sliding on a rough horizontal surface loses kinetic energy.

Because the motion of real fluids is very complex, some simplifying assumptions are made in

order to model ideal fluid flow. The assumptions are;

1. The fluid is non-viscous. In a nonviscous fluid, internal friction is neglected. An object moving

through the fluid experiences no viscous force.

2. The flow is steady. In steady (laminar) flow, the velocity of the fluid at each point remains

constant.

3. The fluid is incompressible. The density of an incompressible fluid is constant.

4. The flow is irrotational. In irrotational flow, the fluid has no angular momentum about any

point. If a small paddle wheel placed anywhere in the fluid does not rotate about the wheel’s

center of mass, then the flow is irrotational.

Consider an ideal fluid flowing through a pipe of non-uniform size, as shown below.

The particles in the fluid move along streamlines in steady flow. In a time interval Δt, the fluid at

the bottom end of the pipe moves a distance Δx1 = v1Δt. If A1 is the cross-sectional area in this

region, then the mass of fluid contained in the left shaded region is m1 = ρA1Δx1 = ρA1v1Δt,

where ρ is the (unchanging) density of the ideal fluid. Similarly, the fluid that moves through the

upper end of the pipe in the time interval Δt has a mass m2 = ρA2v2Δt. However, because the fluid

is incompressible and because the flow is steady, the mass that crosses A1 in a time interval Δt

must equal the mass that crosses A2 in the same time interval. That is, m1 = m2, or

ρA1v1 = ρA2v2; this means that

This expression is called the equation of continuity for fluids. It states that;

―the product of the area and the fluid speed at all points along a pipe is constant for an

incompressible fluid‖.

34

The continuity equation implies that the speed is high where the tube is constricted (small A) and

low where the tube is wide (large A). The product Av (denoted by Q), which has the dimensions

of volume per unit time, is called either the volume flux or the flow rate. The condition Av =

constant is equivalent to the statement that the volume of fluid that enters one end of a tube in a

given time interval equals the volume leaving the other end of the tube in the same time interval

if no leaks are present.

Example:

1. Every second, 5525 m3 of water flows over the 670-m-wide cliff of the Gurara Falls. The

water is approximately 2 m deep as it reaches the cliff. What is its speed at that instant?

Sol.

Q = 5525m3/s, A = 670 x 2 = 1340m

2, v = ?

Q = Av, Therefore, v = Q/A = 5525/1340 = 4.12 m/s

Exercise

1. Water flows through a fire hose of diameter 6.35 cm at the rate of 0.0120 m3/s. The fire hose

ends in a nozzle of inner diameter 2.20 cm. What is the speed with which the water exits the

nozzle?

Bernoulli’s Equation

As a fluid moves through a region where its speed and/or elevation above the Earth’s surface

changes, the pressure in the fluid varies with these changes. The relationship between fluid

speed, pressure, and elevation was first derived in 1738 by the Swiss physicist Daniel Bernoulli.

Consider the flow of a segment of an ideal fluid through a non-uniform pipe in a time interval Δt,

A fluid in laminar flow through a constricted pipe. The volume of the shaded portion on the left

is equal to the volume of the shaded portion on the right

Forces are exerted on this segment by fluid to the left and the right. The force exerted by the luid

on the left end has a magnitude P1A1. The work done by this force on the segment in a time

interval Δt is W1 = F1Δx1 = P1A1Δx1 = P1V, where V is the volume of portion 1. In a similar

manner, the work done by the fluid to the right of the segment in the same time interval Δt is W2

= -P2A2Δx2 = -P2V. (The volume of portion 1 equals the volume of portion 2.) This work is

35

negative because the force on the segment of fluid is to the left and the displacement is to the

right. Thus, the net work done on the segment by these forces in the time interval 't is

W == (P1 - P2)V

Part of this work goes into changing the kinetic energy of the segment of fluid, and part goes into

changing the gravitational potential energy of the segment–Earth system. the change in the

kinetic energy of the segment of fluid of mass m is

The net change in gravitational energy is that the mass of the fluid in portion 1 has effectively

been moved to the location of portion 2. Consequently, the change in gravitational potential

energy is

ΔU = mgy2 - mgy1

The total work done on the system by the fluid outside the segment is equal to the change in

mechanical energy of the system: W = ΔK + ΔU. Substituting for each of these terms gives,

Dividing each term by the portion volume V and recalling that ρ = m/V, this expression reduces

to

Rearranging terms gives,

This is Bernoulli’s equation as applied to an ideal fluid. It is often expressed as

This expression shows that the pressure of a fluid decreases as the speed of the fluid increases. In

addition, the pressure decreases as the elevation increases. This explains why water pressure

from faucets on the upper floors of a tall building is weak unless measures are taken to provide

higher pressure for these upper floors.

When the fluid is at rest, v1 = v2 = 0, and

Example:

A horizontal pipe 10.0 cm in diameter has a smooth reduction to a pipe 5.00 cm in diameter. If

the pressure of the water in the larger pipe is 8.00 x 104 Pa and the pressure in the smaller pipe is

6.00 x 104 Pa, at what rate does water flow through the pipes?

Sol.

36

Venturi flow meter

The flow speed of a fluid through a constricted horizontal pipe (known as Venturi tube) can be

determined if the pressure difference P1 - P2 is known.

Because the pipe is horizontal, y1 = y2 and,

Using the continuity equation, A1v1 = A2v2, then v1 = A2v2/A1. Substituting for v1 gives

So that,

Because A1 > A2, it follows that v2 >v1, and consequently, P1 > P2.

Torricelli’s Law

When an enclosed tank containing a liquid of density ρ has a hole in its side at a distance y1 from

the tank’s bottom, the speed at which the liquid escape can be obtained by applying Bernoulli’s

equation.

Because A2 >> A1, the liquid is approximately at rest at the top of the tank, where the pressure is

P. Noting that at the hole P1 is equal to atmospheric pressure P0, Bernoulli’s equation reduces to,

37

Since y2 – y1 = h, the velocity of escape of the liquid is given by

When P is much greater than P0 (so that the term 2gh can be neglected), the exit speed of the

water is mainly a function of P. If the tank is open to the atmosphere, then P = Po and

. This implies that for an open tank, the speed of liquid coming out through a hole a

distance h below the surface is equal to that acquired by an object falling freely through a

vertical distance h. This phenomenon is known as Torricelli’s law.

REFERRENCE

S. Jewett, ―PHYSICS FOR SCIENTISTS AND ENGINEERS‖, 6TH

EDITION

ISBN:0534408427