INTERNAL ENERGY Every object of matter, whether solid, liquid, or gas, consists of atoms or molecules in rapid motion. The kinetic energies of these particles constitute the internal energy of the body of matter.

The temperature of the body is a measure of the average kinetic energy of its particles.  Heat may be thought of as internal energy in transit. When heat is added to a body, its internal energy increases and its temperature rises; when heat is removed from a body, its internal energy decreases and its temperature falls.

TEMPERATURE Temperature is familiar as the property of a body of matter responsible for sensations of hot or cold to the touch. Temperature indicates the direction of internal energy flow: When two objects are in contact, internal energy goes from the one at higher temperature to the one at lower temperature.

A thermometer is a device for measuring temperature. Two things are necessary in constructing a thermometer:

First we must have a property X that varies with temperature T. The thermometric property should be one that is easily measured like the expansion of a liquid, the pressure in a gas, or the resistance of an electric circuit. Other quantities which vary with temperature are radiated energy, the color of emitted light, vapor pressure etc.

The second requirement in constructing a thermometer is the establishment of standard temperatures. This is usually done by selecting lower and upper fixed points: - The lower fixed point (ice point) is the temperature at which water and ice coexist in thermal equilibrium under a pressure of 1 atm.- The upper fixed point (steam point) is the temperature at which water and steam coexist in equilibrium under a pressure of 1 atm.

The Celsius (or centrigrade) temperature scale assigns 0°C to the freezing point of water and 100°C to its boiling point.

TEMPERATURE SCALES

An absolute temperature scale has its zero point in the absolute zero of temperature. Lord Kelvin devised this scale. The Kelvin (K) has been adopted by the SI as the base unit for temperature measurement.

The Kelvin and the Celsius scales are related as follows: 

TK = TC + 273

13.1 A mercury-in-glass thermometer may not be used at temperatures below -40C. This is because mercury freezes at this temperature.a. What is the freezing point of mercury in the Kelvin scale?b. What is the difference between this temperature and the freezing point of water?

a. TK = TC + 273 = -40 + 273 = 233 K

b. 273 - 233 = 40 KΔT = Twater - THg = (0 - (-40)) = 40C

LINEAR EXPANSION A change in any one dimension of a solid due to a change in temperature is called linear expansion.

ΔL = α Lo ΔT Units: m

where is the coefficient of linear expansion (ºC-1)

13.2 An iron pipe is 300 m long at room temperature (20C). If the pipe is to be used as a steam pipe, how much allowance must be made for expansion, and what will the new length of the pipe be?

Lo = 300 m

to = 20CtSTEAM = 100CαFe = 1.2x10-5/C

ΔL = α Lo ΔT

= 1.2x10-5 (300) (100 - 20) = 0.288 m

L = Lo + ΔL

= 300 + 0.288 = 300.29 m

AREA EXPANSION The following formulas apply for area expansion:

where γ is the coefficient of area expansion, γ = 2 α

ΔA = γ Ao ΔT Units: m2

13.3 A brass disk has a hole 80 mm in diameter punched in its center at 20C. If the disk is placed in boiling water, what will be the new area of the hole?

r = 80/2 = 40 mmto = 20C, tf = 100CαBRASS = 1.8x10-5/Cγ = 2α = 3.6x10-5/C

A = πr2 = π (40)2

= 5026.5 mm2

ΔA = γ Ao ΔT

= 3.6x10-5 (5026.5) (100 - 20) = 14.5 mm2

A = Ao + ΔA

= 5026.5 + 14.5 = 5041 mm2

GAS LAWS AND KINETIC THEORY Four measurable quantities of a sample: pressure, volume, temperature and mass determine the state of a given sample. In this unit we will study the thermal behavior of gases. 

IDEAL GASES  In a gas the individual molecules are so far apart that the cohesive forces between them are usually very small. Also, when a large quantity of gas is confined in a rather small volume, the volume occupied by the molecules is still a tiny fraction of the total volume.

An ideal gas is a gas whose behavior is completely unaffected by cohesive forces or molar volumes. Of course no real gas is ideal, but under ordinary conditions of temperature and pressure, the behavior of any gas conforms very closely to the behavior of an ideal gas.

Boyle's Law"If the temperature of a gas is held constant the volume occupied by an enclosed gas is inversely proportional to the pressure applied to it."

13.4 What volume of hydrogen gas at atmospheric pressure is required to fill a 5.6x10-2 m3 tank under an absolute pressure of 1.7 x 106 Pa?

P1 = 101.3x103 Pa

P2 = 1.7x106 Pa

V2 = 5.6x10-2 m3

P1V1 = P2V2

VPV

P12 2

1

17 10 56 10

1013 10

6 2

3

. ( . )

.

x x

x= 0.939 m3

Charles' Law"The volume of a given amount of gas is directly proportional to the absolute temperature when the pressure is kept constant."

A graph of the temperature of a gas and its volume is shown below. If the line is extrapolated to the left it will intersect the x-axis at a particular value. This point is - 273°C or 0°K. As matter cannot contract beyond a zero volume or exert less than a zero pressure, it follows that -273°C has become known as absolute zero - the limit beyond which temperature cannot be lowered.

13.5 A large balloon filled with air has a volume of 200 liters at 0C. Calculate its volume at 57C if the pressure is unchanged.

V1 = 200 L

T1 = 273 K

T2 = 57 + 273 = 330 K

V

T

V

T1

1

2

2

VV T

T21 2

1

200(330)

273 = 241.75 L

Gay-Lussac's Law"At constant volume, the pressure of a gas is directly proportional to the absolute temperature."

General Gas Law: PV

T

PV

T1 1

1

2 2

2

13.6 A tank with internal volume of 20 liters is filled with oxygen under an absolute pressure of 6x106 Pa at 20C. The oxygen is to be used in a high-flying aircraft, where the absolute pressure is 7x104 Pa and the temperature is -20C. What volume of oxygen can be supplied by the tank under these conditions?

V1 = 20 L

P1 = 6x106 Pa

T1 = 20 + 273 = 293 K

P2 = 7x104 Pa

T2 = - 20 + 273 = 253 K

PV

T

PV

T1 1

1

2 2

2

VPV T

P T21 1 2

2 1

6 10 20 253

7 10 293

6

4

x

x

( )( )

( ) = 1480 L

If we consider the effect of a change in mass, the equation becomes: PV

m T

PV

m T1 1

1 1

2 2

2 2

The mass is changed by injecting molecules at the left. The density remains constant for constant pressure and temperature and it increases as the volume is held fixed by the piston and the temperature is fixed

13.7 The pressure on a helium storage tank reads 1.4x106Pa when the temperature is 27C. The container develops a leak overnight, and the pressure the next morning is found to be 1x106Pa at a temperature of 17C. What percentage of the original mass of helium remains inside of the container?

P1 = 1.4x106 Pa T1 = 27 + 273 = 300 K

P2 = 1x106 Pa T2 = 17 + 273 = 290 K

V1 = V2

P

m T

P

m T1

1 1

2

2 2

fraction remaining = m

m2

1

m

m

PT

PT2

1

2 1

1 2

1 10 300

14 10 290

6

6

x

x

( )

. ( )= 0.738

= 73.8% He remains

THE IDEAL GAS LAW PV

n T

PV

n T1 1

1 1

2 2

2 2

PV = nRT where R is the universal gas constantR = 8.31 J/mol K = 0.0821 L atm/mol K = 1.99 cal/mol K

STP conditions: At a temperature of 273 K and a pressure of 1 atm, 1 mol of any gas occupies a volume of 22.4 liters.

13.8 How many grams of oxygen will occupy a volume of 1600 liters at a pressure of 2 atm and a temperature of 190C?

V = 1600 LP = 2 atmT = 190 + 273 = 463 KMoxygen = 2(16) = 32 g/molR = 0.0821 L atm/mol K

PV = nRT

nPV

RT

( )( )

( . )( )

2 1600

0 0821 463= 84.18 mol

m = nM = 84.18 mol (32 g/mol) = 2694 g

AVOGADRO'S NUMBER AND THE IDEAL GAS Avogadro’s hypothesis formulated in 1811 states that “equal volumes of gas at the same pressure and temperature contain equal numbers of molecules.”

The number of molecules per mole is known as Avogadro’s number (NA).

NA = 6.02x1023 molecules.

The total number of molecules in a gas (N) equals the product of the number of moles of gas (n) and the number of molecules per mole (NA); thus N = n NA and the ideal gas equation can be written as follows:

P V = n R T = (N/NA) R T  Where: (N/NA) R = k and k = 1.38 x l0-23 J/K.

k is known as Boltzmann’s constant.

POSTULATES OF THE KINETIC THEORY OF GASES - A gas consists of a large number of molecules moving in random directions with a variety of speeds.- The average distance between any two molecules in a gas is large compared to the size of an individual molecule.- The molecules obey the laws of classical mechanics and are presumed to interact with one another only when they collide.- Collisions between molecules or between a molecule and the walls of the container are perfectly elastic.

The pressure exerted by an enclosed gas may be rewritten in light of these postulates and is given by the equation:

PV Nmv1

3

2

where N is the total number of molecules in the gas, m is the mass of an individual molecule, and v2 is the average value of the squares of the velocities of the molecules of the gas.

ROOT-MEAN SQUARE VELOCITY (vrms)

 The root-mean-square velocity is often confused with average velocity. The root-mean-square velocity refers to the square root of the average of the squares of the magnitudes of the velocities of the molecules in a gas.

v v v v v Nrms 21 2

12

22

32

1 2ej c h/ /... /

The average or the mean speed is equal to the sum of the speeds of the molecules divided by the number of molecules.

v v v v N 1 2 3 ... /b g

KINETIC ENERGY AND ABSOLUTE TEMPERATURE From the kinetic theory, PV Nmv

1

3

2

and from the ideal gas law, P V = n R T

The average kinetic energy of a molecule in a gas is directly proportional to the absolute temperature of the gas:

KE kT3

2k = 1.38 x l0-23 J/K

13.8 A 5.00 L vessel contains 0.010 kg of an ideal gas at 100C and a pressure of 2 atm.a. How many molecules are in the vessel?

V = 5 LT = 100 + 273 = 373 KP = 2 atmR = 0.0821 L atm/mol K

PV = nRT

nPV

RT

( )( )

( . )( )

2 5

0 0821 373= 0.33 mol

N = 0.33 mol (6.02x1023 molecules/mol)

= 2x1023 molecules

b. Determine the root-mean-square speed of the molecules in the gas.

m = 0.01 kgk = 1.38 x l0-23 J/K

m = mass/molecule =0 010

2 1023. kg

x molecules= 5x10-26 kg/molecule

KE kT3

2

1

2

3

22 2mv kT

vx

x

3 138 10 373

5 10

23

26

( . )( )

vkT

m3

= 555.7m/s

13.9 What is the root-mean-square speed of a nitrogen molecule at a temperature of 300 K? The mass of a nitrogen molecule is 4.65 x10-26 kg.

T = 300 Km = 4.65x10-26 kg

vkT

m3

vx

x

3 138 10 300

4 65 10

23

26

( . )( )

.= 517 m/s