Weebly
| |Solid |Liquid |Gas |
|Macroscopic description |Definite volume |Definite volume |Variable volume |
| |Definite shape |Variable shape |Variable shape |
|Microscopic description |Molecules are held in fixed positions|Molecules are closely packed with |Molecules are widely spaced apart |
| |relative to each other by strong |strong bonds but are not held as |without bonds, moving in random motion,|
| |bonds and vibrate about a fixed point|rigidly in place and can move |and intermolecular forces are |
| |in the lattice |relative to each other as bonds break|negligible except during collisions |
| | |and reform | |
|Comparative density |High |High |Low |
|Kinetic energy |Vibrational |Vibrational |Mostly translational |
| | |Rotational |Higher rotational |
| | |Some translational |Higher vibrational |
|Potential energy |High |Higher |Highest |
|Average molecular separation |Atomic radius (10-10 m) |Atomic radius (10-10 m) |10 x atomic radius (10-9) |
|Molecules per m3 |1028 |1028 |1025 |
|Volume of molecules/volume of |1 |1 |10-3 |
|substance | | | |
| |Q |ΔU |W |
|A → B | | | |
|B → C | | | |
|C → D | | | |
|D → A | | | |
|Cycle | | | |
|A → B |isochoric, temperature increase, U increase, W = 0, Q + = Q in |
|B → C |isobaric expansion, +W, temperature increase, U increase, Q in (more than W |
| |by) |
|C → D |isochoric, temperature decrease, U decrease, W = 0, Q - = Q out |
|D → A |isobaric compression, - W, temperature decrease, U decrease, Q out (more than |
| |Won) |
| |Q |ΔU |W |
|A → B | | | |
|B → C | | | |
|C → A | | | |
|Cycle | | | |
| |Q |ΔU |W |
|A → B | | | |
|B → C | | | |
|C → A | | | |
|Cycle | | | |
-----------------------
Symbol: Q Units: J
Thermal Energy (Heat): the transfer of energy between two substances by non-mechanical means – conduction, convection and radiation
Symbol: T
Units: oC, K
Temperature (Definition #1): a measure of the average random kinetic energy of all the particles of a system
Symbol: U Units: J
Internal Energy: total potential energy and random kinetic energy of the molecules of a substance
Thermal Physics
Temperature (Definition #2): a property that determines the direction of thermal energy transfer between two objects
Internal Kinetic Energy: arises from random translational, vibrational, and rotational motion
Internal Potential Energy: arises from forces between the molecules
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
Specific Heat Capacity: amount of energy required per unit mass to raise the temperature of a substance by 1 K
3. How much thermal energy is needed to raise the temperature of 2.50 g of water from its freezing point to its boiling point?
1. Compare the thermal capacities and specific heat capacities of these samples.
Symbol: C Units: J/K
Symbol: c Units: J/(kg K)
Formula: c = Q/m”T Q = mc”T
Q/mΔT Q = mcΔT
Formula: C = Q/ΔT Q = CΔT
A
B
lower C
same c
higher C
same c
higher C
higher c
lower 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
Why do different amounts of the same substances have different thermal capacities? more molecules to store internal potential and kinetic energy
Q = mcΔT
Q = (2.50 x 10-3) (4.186 x 103) (100 – 0)
Q = 1.05 x 103J
[pic]
Slope
water
mercury
Compare your answer to the amount of thermal energy needed to raise the temperature of liquid mercury the same amount.
more Q needed for water since higher c
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 610 C to 250 C?
Q = CΔT
Q = 1.2 x 105 J
Method of Mixtures
Calorimetry: determining the specific heat capacity (or latent heat capacity) of a substance
Assumption: no thermal energy lost to environment, container, thermometer
Conservation of Energy
[pic]
1. A 0.10 kg sample of an unknown metal is heated to 1000 C by placing it in boiling water for a few minutes. Then it is quickly transferred to a calorimeter containing 0.40 kg of water at 100 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 900 C is transferred to a calorimeter containing 2.00 kg of water at 200 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. A hole is drilled in an 800g iron block and an electric heater is placed inside. The heater provides thermal energy at a constant rate of 600 W.
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.
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 150 C. 9.0 s
c) Calculate the initial rate of increase in temperature. 1.7 0C/s
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
5. An active solar heater is used to heat 50 kg of water initially at 120 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
Calorimetry
Phases of Matter
Phase Changes
1. Describe and explain the process of phase changes in terms of molecular behavior.
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.
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.
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.
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
Factors affecting the rate of evaporation:
a) surface area b) drafts c) temperature d) pressure e) latent heat of vaporization
Specific Latent Heat: amount of energy per unit mass required to change phase of a substance at constant temperature and pressure
Specific Latent Heat
Symbol: L
Units: J/kg
Formula: L = Q/m Q = mL
Specific latent heat of vaporization: Lv
boiling and condensing
Specific latent heat of fusion: Lf
melting and freezing
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 200 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
The Kinetic Model of an Ideal Gas
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.
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 at sea level
1.01x 105 N/m2 = 1.01 x 105 Pa
= 101 kPa
= 760 mm Hg
Weight per unit area of all air above
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?
2. What is the pressure on the gas after a 500. gram piston and a 5.00 kg block are placed on top?
Microscopic definition: total force per unit area from the collisions of gas particles with walls of container
Pressure
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.
2) By Newton’s third law, there must have been an equal and opposite force by the particle on the wall.
3) In a short interval of time, there will be a certain number of collisions so the average result of all these collisions is a constant force on the container wall.
4) The value of this constant force per unit area is the pressure that the gas exerts on the container walls.
[pic]
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.
[pic]
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.
[pic]
Relationship: pressure is inversely related to volume (Boyle’s Law)
4. Macroscopic behavior: At a constant volume, ideal gases increase in pressure when their temperature increases.
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.
Relationship: pressure is directly related to temperature (Pressure Law – Admonton Law)
3. Macroscopic behavior: At a constant temperature, ideal gases increase in pressure when their volume decreases.
2. Macroscopic behavior: Ideal gases increase in temperature when their volume is decreased.
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.
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)
[pic]
Ideal Gas Laws
Heat can of soup
Hot air balloon
Squeeze a balloon
temperature
pressure
temperature
volume
volume
pressure
Control = temperature
P α 1/V P = k/V
PV = k
Control = Pressure
V α T V = kT
Control = volume
P α T P = kT
temperature (0 C)
temperature (K)
Absolute Zero: temperature at which gas would exert no pressure
pressure
Kelvin scale of Temperature: an absolute scale of temperature in which 0 K is the absolute zero of temperature
K = C + 273
The “state” of a fixed amount of a gas is described by the values of its pressure, volume, and temperature.
Molar mass: the mass of one mole of a substance.
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
Equation of State: PV = nRT
Mole: an amount of a substance that contains as many particles as there are atoms in 12 grams of carbon-12.
Derivation:
Ideal Gas: a gas that follows the ideal gas equation of state PV = nRT for all values of p, V, and T
P α T
V α n
Gas constant: R = 8.31 J/(mol K)
PV α nT
PV = nRT
P α 1/V
V α T
Ideal Gas Equation of State
Combined Gas Law derivation:
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
T must be in K
P1V1/T1 = P2V2/T2
P1V1/T1 = nR
P2V2/T2 = nR
State 1: P1V1 = nRT1
State 2: P2V2 = nRT2
[pic]
As a general rule, the molar mass in grams of a substance is numerically equal to its mass number.
b) 2 moles of
[pic]
a) 1 mole of
has a mass of 54 g
has a mass of 7 g
c) How atoms are in 8 grams of helium (mass number = 4)?
8 g/ 4 = 2 moles
2 moles x NA = 1.20 x 1024 atoms
1. What is the volume occupied by 16 g of oxygen (mass number = 8) at room temperature and atmospheric pressure? 200 C 0.049 m3
2. A weather balloon with a volume of 1.0 m3 contains helium (mass number = 4) at atmospheric pressure and a temperature of 350 C. What is the mass of the helium in the balloon? 0.160 kg
Thermodynamics
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
Work: product of force and displacement in the direction of the force
Thermal Energy (Heat): the transfer of energy between two substances by non-mechanical means – conduction, convection and radiation
Internal energy: total potential energy and random kinetic energy of the molecules of a substance
Symbol: U
Units: J
Symbol: W
Units: J
Symbol: Q
Units: J
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.
The internal energy of a system can change by . . .
W = -100 J
W = +100 J
Expansion
Compression
Q = -400 J
Q = +400 J
Qin = +400 J
Qout = -400 J
Cooling
Heating
Qin = -400 J
Qout = +400 J
Wby = +100 J
Won = -100 J
Wby = -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
1. A sample of gas is heated with a Bunsen burner and allowed to expand. If 400 J of thermal energy are transferred to the gas during heating and the gas does 100 J of work by expanding, what is the resulting change in the internal energy of the gas?
2. A sample of gas is warmed by placing it in a bath of hot water, adding 400 J of thermal energy. At the same time, 100 J of work is done compressing the gas manually. What is the resulting change in the internal energy of the gas?
ΔU = 400 J -100 J
ΔU = 300 J
ΔU = Q - W
ΔU = 400 J – (-100 J)
ΔU = 500 J
ΔU = Q - W
Formula: ΔU = Q - W Q = ΔU + W
a) A gas gains 1500 J of heat from its surroundings, and expands, doing 2200 J of work on the surroundings.
Heating a hot air balloon and let canvas expand
Q = ΔU + W
1500 J= ΔU + (+2200 J)
ΔU = -700 J
b) A gas gains 1500 J of heat at the same time as an external force compresses it, doing 2200 J of work on it.
Q = ΔU + W
1500 J= ΔU + (-2200 J)
ΔU = +3700 J
Heating and stirring water
In each case, determine the change in the internal energy of the gas.
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
What does an isobaric process look like on a diagram of pressure vs. volume (P-V diagram)?
W = F s cos θ
W = p A s cos 00
W = p ΔV
How is pressure held constant? weight of brick, piston, and atmosphere constant
How much work is done by the gas if it expands at constant pressure?
animation: Serway: chap 12: TDM06AN1
Work Involved in a Volume Change at Constant Pressure
Compression of gas
animation: Serway: chap 12: TDM06AN1
Expansion of gas
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
Expansion at constant pressure
Four Common Thermal Processes
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
The gas in the cylinder is expanding isobarically because the pressure is held constant by the external atmosphere and the weight of the piston and the block. Heat can enter or leave through the non-insulating walls.
Isobaric Process
The gas in the cylinder is being heated isochorically since the volume of the cylinder is held fixed by the rigid walls.
Heat can enter or leave through the non-insulating walls.
Isochoric (Isovolumetric) Process
The gas in the cylinder is being allowed to expand isothermally since it is in contact with a water bath (heat reservoir) that keeps the temperature constant.
Heat can enter or leave through the non-insulating walls.
Isothermal Process
The gas in the cylinder is being compressed adiabatically since the cylinder is surrounded by an insulating material.
Adiabatic Process
Isobaric Process
Compression at constant pressure
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
Gas laws:
PV/ T = PV/ T
P1 = P2
So V1/T1 = V2/T2
If V increases, so does T
1st law:
If gas is ideal, U increases when T increases
U α T
So + ΔT means + ΔU
ΔQ = ΔU + ΔW
(+) = (+) + (+)
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 energy by 200 J. How much work is done by the gas in expanding?
Gas laws:
PV/ T = PV/ T
P1 = P2
So V1/T1 = V2/T2
If V decreases, so does T
1st law:
If gas is ideal, U decreases when T decreases
U α T
So - ΔT means - ΔU
ΔQ = ΔU + ΔW
(-) = (-) + (-)
More heat is removed than work done if isobaric - Heat leaves system
800 J
Isochoric (Isovolumetric) Process
Gas law:
Work:
PV/T = PV/T
V = V
So P/T = P/T
If T increases, P increases
Means ΔV = 0
If ΔV = 0
then W = 0
No area = no work
1st law:
Q = ΔU + W
ΔU = Q
(since W = 0)
If Q+, then ΔU+
If ideal gas, ΔU α ΔT
so ΔT+
d) If the change in internal energy in each case is 500 J, calculate the work done and thermal energy exchanged in each case.
c) Compare the thermal energy added to or removed from the gas during each process. Justify your answer.
b) Compare the work done by (or on) the case during each process. Justify your answer.
a) Compare the change in internal energy of the gas as a result of each process. Justify your answer.
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 pressure of 1.0 x 105 Pa by a different process or series of processes.
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.
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.
I.
II.
III.
Isothermal Process
Heat reservoir: hot or cold water bath that maintains constant temperature of gas by supplying or removing thermal energy
1st Law:
Gas Law:
PV/T = PV/T
T = T
PV = PV
If V increases, P decreases
Q = ΔU + W
If ideal gas, ΔU α ΔT
so ΔT= 0 means ΔU = 0
so Q = W
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
Isotherm: hyperbola of constant temperature
PV = nRT
P = nRT/V - hyperbola for fixed T
P1V1/T1 = P2V2/T2
P1V1 = P2V2 on one isotherm
Expansion
Compression
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
arrow to right
work done by +
Q added +
arrow to left
work done on -
Q removed -
animation: Serway: chap 10: TDA06AN3
Adiabatic Process
Adiabatic walls: insulating walls so no thermal energy can enter or leave system
NOTE: Rapid expansion or compression of gas is approximately adiabatic
1st Law:
PV/T = PV/T
P decreases and V increases and T decreases
Q = ΔU + W
Q = 0
so ΔU = - W
If ideal gas, ΔU α ΔT
so W α -ΔT
Compression: work done on gas heats gas up as it gains internal energy
W α -ΔT so +W means temperature goes down = jumps to lower isotherm = gas cools down
W α -ΔT so -W means temperature goes up = jumps to higher isotherm = gas gets hotter
Expansion: work done by gas cools gas down as it loses internal energy
PV/T = PV/T
P increases and V decreases and T increases
2. If an ideal gas is allowed to expand adiabatically, the internal energy of the gas changes by2500 J.
a) Does the internal energy of the gas increase or decrease? Justify your answer.
Determine:
b) the thermal energy added or removed from the gas.
c) the work done by the gas.
1. If 410 J of heat energy are added to an ideal gas causing it expand at constant temperature,
a) what is the change in internal energy of the gas?
b) how much work is done by the gas?
c) how much work is done on the gas?
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 P0 = 1.0 x 105 Pa and V0 = 2.0 x 10-3 m3.
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.
Q out
Q out
Q in
Q in
Properties of the individual thermal processes
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
Net Work for a Cycle
+
=
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
A to B: W = 0, Q +, ΔU+
B to C: ΔU = 0, W+ and Q+
C to A: Q -, W -, ΔU –
Cycle: ΔU = 0, Q+, W+
b) State whether ΔU, W and Q are +, - , or 0 for each of the three processes and for the entire cycle.
A to B: Q = ΔU = 1520 J, W = 0
B to C: ΔU = 0, Q = +1700J, W = +1700J
C to A: ΔU = -1520 J, W = - 1000 J, Q = -2520 J
Cycle: ΔU = 0, Q = 700 J, W = 700 J
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.
The Second Law of Thermodynamics and Entropy
Second Law of Thermodynamics:
1) the overall entropy of the universe is increasing
2) all natural processes increase the entropy of the universe
Entropy: a system property that expresses the degree of disorder in the system.
The Second Law of Thermodynamics implies that . . . thermal energy cannot spontaneously transfer from a region of low temperature to a region of high temperature.
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.
3. An operating refrigerator with its door open is placed in a thermally insulated room. The refrigerator operates for a long period of time. Discuss any changes in the total energy, total entropy, and temperature of the room.
2. A block of ice is placed in a thermally insulated room initially at room temperature. Discuss any changes in the total energy, total entropy, and temperature of the room.
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.