# org key - thermal physics ib2 08 by nuhman10

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```									                                    Thermal Physics                                            IB 12

Internal Energy: total potential energy and random kinetic                         Symbol: U         Units: J
energy of the molecules of a substance

Internal Kinetic Energy: arises from random translational,
vibrational, and rotational motion

Internal Potential Energy: arises from forces between the molecules

Temperature (Definition #1): a measure of the average random                                         Symbol: T
kinetic energy of all the particles of a system
Units: oC, K

Thermal Energy (Heat): the transfer of energy between two substances by non-       Symbol: Q         Units: J
mechanical means – conduction, convection and radiation

Temperature (Definition #2): a property that determines the direction of thermal energy transfer between two
objects

Thermal Equilibrium: at same temperature – no thermal energy transfer – independent of mass, etc.

Thermal Capacity: amount of energy required to raise the temperature of a
substance by 1 K

Formula: C = Q/ΔT            Q = CΔT                                      Symbol: C        Units: J/K

Specific Heat Capacity: amount of energy required per unit mass to raise
the temperature of a substance by 1 K

Formula: c = Q/mΔT           Q = mcΔT                                     Symbol: c        Units: J/(kg K)
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IB 12
1. Compare the thermal capacities and specific heat capacities of these samples.

Why do different amounts of the same substances have different
thermal capacities? more molecules to store internal potential and
kinetic energy
A

lower C                higher C
same c                 same c

B

higher C                 lower C
higher c                 lower c

Why do the same amounts of different substances have different specific heat capacities? substances contain
different numbers of molecules with different molecular masses

2. The thermal capacity of a sample of lead is 3.2 x 103 J K-1. How much thermal energy will be released if it cools from 61 0 C to 250 C?

Q = CΔT
Q = 1.2 x 105 J

3. How much thermal energy is needed to raise the                                                                           Slope
temperature of 2.50 g of water from its freezing point                               mercury
to its boiling point?
T   1
water                   slope        
Q = mcΔT                                                                                                                Q mc
Q = (2.50 x 10-3) (4.186 x 103) (100 – 0)
Q = 1.05 x 103J

raise the temperature of liquid mercury the same amount.                  more Q needed for water since higher c

2
4. A hole is drilled in an 800g iron block and an electric heater is placed inside. The heater provides               IB 12
thermal energy at a constant rate of 600 W.

a) Assuming no thermal energy is lost to the surrounding environment, calculate how long it will take the iron block to
increase its temperature by 15 0 C. 9.0 s

b) The temperature of the iron block is recorded as it varies with time and is
shown at right. Comment on reasons for the shape of the graph.

begins at room temp
increases linearly as Q = cmΔT
as gets hotter, more energy lost to environment
levels out when heat gained by heater = heat lost to room

c) Calculate the initial rate of increase in temperature. 1.7 0C/s

5. An active solar heater is used to heat 50 kg of water initially at 12 0 C. If the average rate that the thermal energy is absorbed in
a one hour period is 920 J min-1, determine the equilibrium temperature after one hour. 120 C

3
Calorimetry                                                 IB 12

Calorimetry: determining the specific heat capacity (or latent heat capacity) of a
substance

Conservation of Energy                                 Qc  Qh
mc cc Tc  mh ch Th

Assumption: no thermal energy lost to environment, container,
thermometer

1. A 0.10 kg sample of an unknown metal is heated to 100 0 C by placing it in boiling water for a few            Method of Mixtures
minutes. Then it is quickly transferred to a calorimeter containing 0.40 kg of water at 10 0 C.
After thermal equilibrium is reached, the temperature of the water is 150 C.

a) What is the specific heat capacity of the metal sample?           983 J/(kg 0C)

b) What is the thermal capacity of the metal sample?            98.3 J/ 0C

2. A 3.0 kg block of copper at 90 0 C is transferred to a calorimeter containing 2.00 kg of water at 20 0 C. The mass of the
calorimeter cup, also made of copper, is 0.210 kg. Determine the final temperature of the water. 28.30 C

4
Phases of Matter                                              IB 12
Kinetic theory says that:
1. All matter is made up of atoms, and
2. the atoms are in continuous random motion at a variety of speeds.
3. Whether a substance is a solid, liquid, or gas basically depends on how close together its molecules are and how strong the
bonds are that hold them together.

Solid                        Liquid                              Gas
Macroscopic                 Definite volume                  Definite volume                  Variable volume
description                 Definite shape                   Variable shape                   Variable shape
Molecules are closely            Molecules are widely spaced
Molecules are held in fixed
packed with strong bonds         apart without bonds, moving
positions relative to each
Microscopic                                                  but are not held as rigidly in   in random motion, and
other by strong bonds and
description                                                  place and can move relative      intermolecular forces are
vibrate about a fixed point in
to each other as bonds break     negligible except during
the lattice
and reform                       collisions
Comparative                 High                             High                             Low
density
Vibrational                      Mostly translational
Kinetic energy              Vibrational                      Rotational                       Higher rotational
Some translational               Higher vibrational
Potential energy            High                             Higher                           Highest
separation
Molecules per m3                          1028                           1028                              1025
Volume of
molecules/volume                           1                               1                               10-3
of substance

Phase Changes

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1. Describe and explain the process of phase changes in terms of molecular behavior.                              IB 12

When thermal energy is added to a solid, the molecules gain kinetic energy as they vibrate at an increased rate. This is seen
macroscopically as an increase in temperature. At the melting point, a temperature is reached at which the kinetic energy of
the molecules is so great that they begin to break the permanent bonds that hold them fixed in place and begin to move
about relative to each other. As the solid continues to melt, more and more molecules gain sufficient energy to overcome
the intermolecular forces and move about so that in time the entire solid becomes a liquid. As heating continues, the
temperature of the liquid increases due to an increase in the vibrational, translational and rotational kinetic energy of the
molecules. At the boiling point, a temperature is reached at which the molecules gain sufficient energy to overcome the
intermolecular forces that hold them together and escape from the liquid as a gas. Continued heating provides enough
energy for all the molecules to break their bonds and the liquid turns entirely into a gas. Further heating increases the
translational kinetic energy of the gas and thus its temperature increases.

2. Explain in terms of molecular behavior why temperature does not change during a phase change.

The making or breaking of intermolecular bonds involves energy. When bonds are broken (melting and vaporizing), the
potential energy of the molecules is increased and this requires input energy. When bonds are formed (freezing and
condensing), the potential energy of the molecules is decreased as energy is released. The forming or breaking of bonds
happens independently of the kinetic energy of the molecules. During a phase change, all energy added or removed from
the substance is used to make or break bonds rather than used to increase or decrease the kinetic energy of the molecules.
Thus, the temperature of the substance remains constant during a phase change.

3. Explain in terms of molecular behavior the process of evaporation.

Evaporation is a process by which molecules leave the surface of a liquid, resulting in the cooling of the
liquid. Molecules with high enough kinetic energy break the intermolecular bonds that hold them in the
liquid and leave the surface of the substance. The molecules that are left behind thus have a lower
average kinetic energy and the substance therefore has a lower temperature.

Factors affecting the rate of evaporation:

a) surface area            b) drafts               c) temperature           d) pressure       e) latent heat of vaporization

4. Distinguish between evaporation and boiling.

Evaporation – process whereby liquid turns to gas, as explained above

- occurs at any temperature below the boiling temperature

- occurs only at surface of liquid as molecules escape

- causes cooling of liquid

Boiling –      process whereby liquid turns to gas when the vapor pressure of the liquid equals the atmospheric
pressure of its surroundings

-   occurs at one fixed temperature, dependent on substance and pressure

-   occurs throughout liquid as bubbles form, rise to surface and are released

-   temperature of substance remains constant throughout process

6
Specific Latent Heat                                           IB 12
Specific Latent Heat: amount of energy per unit mass required to change                                           Symbol: L
phase of a substance at constant temperature and pressure
Units: J/kg

Formula: L = Q/m               Q = mL

Specific latent heat of fusion:              Lf
melting and freezing

Specific latent heat of vaporization: Lv
boiling and condensing

1. How much energy is needed to change 500 grams of ice into water?

a) Assume the ice is already at its melting point.

b) Assume the ice is at -150 C.

2. Thermal energy is supplied to a pan containing 0.30 kg of water at 20 0 C at a rate of 400 W for 10 minutes. Estimate the mass of
water turned into steam as a result of this heating process.                                            0.060 kg

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The Kinetic Model of an Ideal Gas                                            IB 12

Kinetic theory views all matter as consisting of individual particles in continuous motion in an attempt to
relate the macroscopic behaviors of the substance to the behavior of its microscopic particles.

Certain microscopic assumptions need to be made in order to deduce the behavior of an ideal gas, that is, to
build the Kinetic Model of an Ideal Gas.

Assumptions:

1. A gas consists of an extremely large number of very tiny particles (atoms or molecules) that are in continuous random
motion with a variety of speeds.

2. The volume of the particles is negligible compared to the volume occupied by the entire gas.

3. The size of the particles is negligible compared to the distance between them.

4. Collisions between particles and collisions between particles and the walls of the container are assumed to be perfectly
elastic and take a negligible amount of time.

5. No forces act between the particles except when they collide (no intermolecular forces). As a consequence, the internal
energy of an ideal gas consists solely of random kinetic energy – no potential energy.

6. In between collisions, the particles obey Newton’s laws of motion and travel in straight lines at a constant speed.

Explaining Macroscopic Behavior in terms of the Kinetic Model

Pressure

Macroscopic definition: force per unit area acting on a surface

Formula:       P = F/A                               Units: N/m2 = Pa (Pascals)

Atmospheric Pressure                       1. A cylinder with diameter 3.00 cm is open to the air.
What is the pressure on the gas in this open cylinder?

Weight per unit area of all air above

2. What is the pressure on the gas after a 500. gram
piston and a 5.00 kg block are placed on top?

Atmospheric pressure at sea level
1.01x 105 N/m2 = 1.01 x 105 Pa
= 101 kPa
= 760 mm Hg

8
Pressure                                                                                                           IB 12

Microscopic definition: total force per unit area from the collisions of gas particles with walls of
container

Explanation:

1) A particle collides with the wall of container and changes momentum. By Newton’s second law, a
change in momentum means there must have been a force by the wall on the particle.
p
2) By Newton’s third law, there must have been an equal and opposite force by the particle on the wall.           F
3) In a short interval of time, there will be a certain number of collisions so the average result of all these
t
collisions is a constant force on the container wall.                                                            F
P
4) The value of this constant force per unit area is the pressure that the gas exerts on the container walls.         A

1. Macroscopic behavior: Ideal gases increase in pressure when more gas is added to the container.

Microscopic explanation: More gas means more gas particles in the container so there will be an increase in the number of
collisions with the walls in a given interval of time. The force from each particle remains the same but an increased number
of collisions in a given time means the pressure increases.

2. Macroscopic behavior: Ideal gases increase in temperature when their volume is decreased.
animation: Serway: chap 12: TDM06AN1
Microscopic explanation: As the volume is reduced, the walls of the container move inward. Since the particles are now
colliding with a moving wall, the wall transfers momentum (and kinetic energy) to the particles, making them rebound faster from
the moving wall. Thus the kinetic energy of the particles increases and this means an increase in the temperature of the gas.

3. Macroscopic behavior: At a constant temperature, ideal gases increase in pressure when their volume decreases.

Microscopic explanation: The decrease in volume means the particles hit a given area of the wall more often. The force
from each particle remains the same but an increased number of collisions in a given time means the pressure increases.

1
Relationship: pressure is inversely related to volume (Boyle’s Law)                                   P
V
4. Macroscopic behavior: At a constant volume, ideal gases increase in pressure when their temperature increases.

Microscopic explanation: The increased temperature means the particles have, on average, more kinetic energy and are
thus moving faster. This means that the particles hit the walls more often and, when they do, they exert a greater force on
the walls during the collision. For both these reasons, the total force on the wall in a given time increases which means that
the pressure increases.
Relationship: pressure is directly related to temperature (Pressure Law – Admonton Law)
PT
5. Macroscopic behavior: At a constant pressure, ideal gases increase in volume when their temperature increases.

Microscopic explanation: A higher temperature means faster moving particles that collide with the walls more often and
with greater force. However, if the volume of the gas is allowed to increase, the rate at which these particles hit the walls
will decrease and thus the average force exerted on the walls by the particles, that is, the pressure can remain the same.

Relationship: volume is directly related to temperature (Charles Law – Gay-Lussac Law)
VT

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Ideal Gas Laws                                                    IB 12
Squeeze a balloon                                  Hot air balloon                                Heat can of soup

volume

pressure
pressure

volume                                         temperature                                         temperature

Control = temperature                         Control = Pressure                            Control = volume

P α 1/V           P = k/V                     VαT        V = kT                             PαT               P = kT
PV = k
pressure

K = C + 273

temperature (0 C)                                                                        temperature (K)

Absolute Zero: temperature at which gas would exert no pressure

Kelvin scale of Temperature: an absolute scale of temperature in which 0 K is the absolute zero of temperature

Mole: an amount of a substance that contains as many particles as there are atoms in 12 grams of carbon-12.

Avogadro’s constant: the number of atoms in 12 g of carbon 12.

NA = 6.02 x 1023 particles/mole                            NA = 6.02 x 1023 mol-1

Molar mass: the mass of one mole of a substance.

As a general rule, the molar mass in grams of a substance is numerically equal to its mass number.

a) 1 mole of       7           has a mass of 7 g
3   Li

b) 2 moles of 27               has a mass of 54 g
13        Al
8 g/ 4 = 2 moles
c) How atoms are in 8 grams of helium (mass number = 4)?
2 moles x NA = 1.20 x 1024 atoms

10
Ideal Gas Equation of State                                            IB 12
Derivation:
P α 1/V         PαT                             PV α nT

VαT             Vαn                             PV = nRT

Equation of State: PV = nRT

The “state” of a fixed amount of a gas is described by the values of its pressure, volume, and temperature.

Gas constant: R = 8.31 J/(mol K)

Ideal Gas: a gas that follows the ideal gas equation of state PV = nRT for all values of p, V, and T

Compare real gases to an ideal gas:

a) real gases only behave like ideal gases only at low pressures and high temperatures

b) ideal gases cannot be liquefied but real gases can

Combined Gas Law derivation:

State 1: P1V1 = nRT1                         P1V1/T1 = nR                           P1V1/T1 = P2V2/T2

State 2: P2V2 = nRT2                         P2V2/T2 = nR
T must be in K

1. What is the volume occupied by 16 g of oxygen                 2. A weather balloon with a volume of 1.0 m3 contains helium (mass
(mass number = 8) at room temperature and                        number = 4) at atmospheric pressure and a temperature of 35 0 C.
atmospheric pressure?         200 C 0.049 m3                     What is the mass of the helium in the balloon?    0.160 kg

11
Thermodynamics                                              IB 12

Thermodynamics is the branch of physics that deals with the way in which a system interacts with its surroundings.

Thermodynamic System substance - usually an ideal gas

Surroundings everything else – walls of container, outside environment

State of the system for a gas, a particular set of values of P, V, n, and T

Internal energy: total potential energy and random kinetic energy of the molecules of a substance             Symbol: U

Units: J

Work: product of force and displacement in the direction of the force                                         Symbol: W

Units: J

Thermal Energy (Heat): the transfer of energy between two substances by non-mechanical                        Symbol: Q
means – conduction, convection and radiation
Units: J

The internal energy of a system can change by . . .

Heating                            Cooling                               Expansion                      Compression
Q = +400 J                        Q = -400 J                             W = +100 J                      W = -100 J

Qin = +400 J                      Qin = -400 J                            Wby = +100 J                    Wby = -100 J

Qout = -400 J                     Qout = +400 J                           Won = -100 J                    Won = +100 J

Definitions:   Q = thermal energy added to system

W = work done by the system

ΔU = change in internal energy of the system
12
IB 12
1. A sample of gas is heated with a Bunsen burner                2. A sample of gas is warmed by placing it in a bath
and allowed to expand. If 400 J of thermal energy                of hot water, adding 400 J of thermal energy. At the
are transferred to the gas during heating and the gas            same time, 100 J of work is done compressing the
does 100 J of work by expanding, what is the                     gas manually. What is the resulting change in the
resulting change in the internal energy of the gas?              internal energy of the gas?

ΔU = 400 J -100 J                                                         ΔU = 400 J – (-100 J)

ΔU = 300 J                                                                ΔU = 500 J

ΔU = Q - W                                                                ΔU = Q - W

Formula:                             ΔU = Q - W                            Q = ΔU + W

First Law of Thermodynamics: The thermal energy transferred to a system from its surroundings is equal to the work
done by the system plus the change in internal energy of the system.

NOTE: The First Law is a statement of . . . the principle of conservation of energy.

In each case, determine the change in the internal energy of the gas.
a) A gas gains 1500 J of heat from its surroundings, and                b) A gas gains 1500 J of heat at the same time as an external
expands, doing 2200 J of work on the surroundings.                      force compresses it, doing 2200 J of work on it.

Heating a hot air balloon                        Q = ΔU + W
Q = ΔU + W              and let canvas expand                                                   Heating and
1500 J= ΔU + (-2200 J)      stirring water
1500 J= ΔU + (+2200 J)
ΔU = +3700 J
ΔU = -700 J

Internal energy of many substances depends on . . .        temperature and intermolecular bonds

Internal energy of an ideal gas depends on . . .

1. only on temperature since there are no intermolecular bonds

U α T so ΔU α ΔT                              U increases if T increases, if +ΔT then +ΔU

2. the change in internal energy of ideal gas is path independent

13
Four Common Thermal Processes                                      IB 12

1. An isobaric process is one that occurs at constant pressure.                                                            ΔP = 0

2. An isochoric (isovolumetric) process is one that occurs at constant volume.                                             ΔV = 0

3. An isothermal process is one that occurs at constant temperature.                                                       ΔT = 0

4. An adiabatic process is one that occurs without the transfer of thermal energy.                                         ΔQ = 0
Isobaric Process                      Isochoric                        Isothermal Process                     Adiabatic
(Isovolumetric)                                                             Process
Process

The gas in the cylinder is expanding    The gas in the cylinder is being     The gas in the cylinder is being
isobarically because the pressure is    heated isochorically since the      allowed to expand isothermally      The gas in the cylinder is
held constant by the external      volume of the cylinder is held          since it is in contact with a         being compressed
atmosphere and the weight of the          fixed by the rigid walls.        water bath (heat reservoir) that      adiabatically since the
piston and the block. Heat can enter    Heat can enter or leave through      keeps the temperature constant.    cylinder is surrounded by an
or leave through the non-insulating       the non-insulating walls.         Heat can enter or leave through         insulating material.
walls.                                                           the non-insulating walls.

Isobaric Process
Work Involved in a Volume Change at Constant Pressure

How is pressure held constant? weight of brick, piston, and atmosphere constant

How is work done by the gas?

Molecules strike piston and transfer momentum and KE to it causing it to
move upward/outward - as KE decreases, so does internal energy and T

How much work is done by the gas if it expands at constant pressure?
W = F s cos θ

W = p A s cos 00                                                         animation: Serway: chap 12: TDM06AN1

W = p ΔV

14
IB 12
What does an isobaric process look like on a diagram of pressure vs. volume (P-V diagram)?

Expansion of gas                                              Compression of gas

How can the amount of work done by a gas during a process be determined from a P-V diagram?

work done by the gas = area underneath curve
arrow to right = positive work done by the gas = expansion
arrow to left = negative work done by the gas = compression

Isobaric Processes and the First Law of Thermodynamics
Expansion at
constant pressure                             Gas laws:                    1st law:
If gas is ideal, U increases when T increases
UαT
PV/ T = PV/ T                                 So + ΔT means + ΔU

P1 = P2
ΔQ = ΔU + ΔW
So V1/T1 = V2/T2
(+) = (+) + (+)
If V increases, so does T        More heat is added than work done if isobaric

Example: A gas is allowed to expand
isobarically by adding 1000 J of thermal
energy, causing the gas to increase its internal               800 J
energy by 200 J. How much work is done by
the gas in expanding?

Compression at                           Gas laws:                                1st law:
constant pressure
If gas is ideal, U decreases when T decreases
PV/ T = PV/ T
UαT
P1 = P2                                      So - ΔT means - ΔU

So V1/T1 = V2/T2                                 ΔQ = ΔU + ΔW

If V decreases, so does T                         (-) = (-) + (-)

More heat is removed than work done if
isobaric - Heat leaves system   15
IB 12
Isochoric (Isovolumetric) Process

Work:                                           Gas law:

PV/T = PV/T
Means ΔV = 0
V=V
If ΔV = 0                                      So P/T = P/T
then W = 0
If T increases, P
No area = no work                              increases

1st law:                  Q = ΔU + W                            If Q+, then ΔU+

ΔU = Q                              If ideal gas, ΔU α ΔT
(since W = 0)                          so ΔT+

1. One mole of an ideal gas is heated at a constant volume of 2.0 x 10 -3 m3 from an initial pressure of 1.0 x 105 Pa to a final
pressure of 5.0 x 105 Pa.

a) Determine the initial and final temperatures of the gas.

b) Does the internal energy of the gas increase or decrease? Justify your answer.

c) Determine the work done by the gas during this process.

c) If the change in internal energy of the gas is 1200J, determine the amount of thermal energy added to the gas.

16
2. In each case shown below, an ideal gas at 5.0 x 105 Pa and 1.0 x 10-3 m3 expands to 4.0 x 10-3 m3 at a      IB 12
pressure of 1.0 x 105 Pa by a different process or series of processes.
III.
I.                                          II.

a) Compare the change in internal energy of the gas as a result of each process. Justify your answer.

b) Compare the work done by (or on) the case during each process. Justify your answer.

c) Compare the thermal energy added to or removed from the gas during each process. Justify your answer.

d) If the change in internal energy in each case is 500 J, calculate the work done and thermal energy exchanged in each case.

Conclusions:

1) Change in internal energy does not depend on the path taken – only on the change in temperature – path
independent.

2) Work done and thermal energy transferred depend on the path taken between the initial and final states.

17
Isothermal Process                                          IB 12

Heat reservoir: hot or cold water bath that maintains constant temperature of gas by supplying or removing thermal energy

Gas Law:                                         1st Law:

PV/T = PV/T                                           Q = ΔU + W

T=T                                         If ideal gas, ΔU α ΔT
so ΔT= 0 means ΔU = 0
PV = PV
so       Q= W
If V increases, P decreases

Expansion: thermal energy flows in at same rate as work is done by gas

Compression: thermal energy flows out at same rate as work done on the gas

Ideal Gas Equation of State
P1V1/T1 = P2V2/T2
PV = nRT

P = nRT/V - hyperbola for fixed T                P1V1 = P2V2 on one isotherm

Isotherm: hyperbola of constant temperature

Conclusions:

1) all states on one isotherm have same U since have same T – ΔT and ΔU = 0 moving along same isotherm

2) isotherms further from origin – higher T so higher U

3) ΔU between two isotherms is path independent – same ΔU since same ΔT and ΔU α ΔT

Expansion                                                 Compression
arrow to left
arrow to right                                              work done on -
work done by +                                              Q removed -

18
animation: Serway: chap 10: TDA06AN3

Adiabatic walls: insulating walls so no thermal energy can enter or leave system

NOTE: Rapid expansion or compression of gas is approximately adiabatic

Q = ΔU + W
1st Law:                    Q=0
so ΔU = - W

If ideal gas, ΔU α ΔT

so W α -ΔT

Expansion: work done by gas cools gas down as it loses internal energy

PV/T = PV/T                               W α -ΔT so +W means
temperature goes down = jumps
P decreases and V                         to lower isotherm = gas cools
increases and T                           down
decreases

Compression: work done on gas heats gas up as it gains internal energy

PV/T = PV/T                               W α -ΔT so -W means
temperature goes up = jumps to
P increases and V                         higher isotherm = gas gets hotter
decreases and T
increases

1. If 410 J of heat energy are added to an ideal gas              2. If an ideal gas is allowed to expand adiabatically, the internal energy
causing it expand at constant temperature,                         of the gas changes by2500 J.

a) what is the change in internal energy of the gas?              a) Does the internal energy of the gas increase or decrease? Justify your

b) how much work is done by the gas?

Determine:

b) the thermal energy added or removed from the gas.

c) how much work is done on the gas?
c) the work done by the gas.

19
IB 12
Cycles

Cycle: a series of processes that returns a gas to its initial state

The cycle shown below represents processes performed on an ideal gas initially at P 0 = 1.0 x 105 Pa and V0 = 2.0 x 10-3 m3.

Q                  ΔU                   W

A→B

B→C

C→D

D→A

Cycle

1. Compare the temperatures at each state A, B, C, and D.

2. During process A→B, 600 J of thermal energy were added to the gas. Complete the chart.

20
IB 12
Properties of the individual thermal processes

isochoric, temperature increase, U increase, W = 0,
Q in
A→B          Q + = Q in

isobaric expansion, +W, temperature increase, U
B→C          increase, Q in (more than W by)
Q in                                  Q out
isochoric, temperature decrease, U decrease, W = 0,
C→D          Q - = Q out

isobaric compression, - W, temperature decrease, U
Q out
decrease, Q out (more than Won)
D→A

Net Work for a Cycle

+                            =

Properties of the entire cycle

1) gas returns to same P, V, and T

2) ΔT = 0 so ΔU = 0 (for all ideal gases)

3) ΔU = 0 so net Q = netW

4) net W = area enclosed by figure so positive area enclosed means positive net work = work done by gas = net work out

5) net Q = W so Q+ so more heat added than removed during cycle = net heat in
21
IB 12
An ideal gas is confined in a cylinder with a movable piston. The gas starts at
300 K in state A and proceeds through the cycle shown in the diagram.

a) Find the temperatures at B and at C.

900 K isothermal

b) State whether ΔU, W and Q are +, - , or 0 for each of the three processes and for the entire cycle.

Q                  ΔU                    W                    A to B: W = 0, Q +, ΔU+

B to C: ΔU = 0, W+ and Q+
A→B
C to A: Q -, W -, ΔU –
B→C
Cycle: ΔU = 0, Q+, W+

C→A

Cycle

c) The internal energy of the gas changes by 1520 J during process A to B. 1700 J of heat are added to the gas
during process B to C. Find ΔU, W, and Q for each process and for the entire cycle.

Q                  ΔU                   W
A to B: Q = ΔU = 1520 J, W = 0
A→B
B to C: ΔU = 0, Q = +1700J, W = +1700J

B→C                                                                          C to A: ΔU = -1520 J, W = - 1000 J, Q = -2520 J

Cycle: ΔU = 0, Q = 700 J, W = 700 J

C→A

Cycle

22
The Second Law of Thermodynamics and Entropy                                                  IB 12

The Second Law of Thermodynamics implies that . . . thermal energy cannot spontaneously transfer from a
region of low temperature to a region of high temperature.

Entropy: a system property that expresses the degree of disorder in the system.

Second Law of Thermodynamics:

1) the overall entropy of the universe is increasing

2) all natural processes increase the entropy of the universe

Although local entropy can decrease, any process will increase the total entropy of a system and its surroundings (the universe).

1. Discuss this statement for the case of a puddle of water freezing into a block of ice.

2. A block of ice is placed in a thermally insulated room          3. An operating refrigerator with its door open is placed in a
initially at room temperature. Discuss any changes in the          thermally insulated room. The refrigerator operates for a long
total energy, total entropy, and temperature of the room.          period of time. Discuss any changes in the total energy, total
entropy, and temperature of the room.

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