Results from kinetic theory, 1

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Results from kinetic theory, 1

• 1. Pressure is associated with collisions of gas
  particles with the walls. Dividing the total
  average force from all the particles by the wall
  area gives the pressure.
• Increasing temperature increases pressure for two
  reasons. There are more collisions, and the
  collisions involve a larger average force.
• For a fixed volume and temperature, adding more
  particles increases pressure because there are
  more collisions.

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Results from kinetic theory, 2

• What is the average velocity of the ideal gas
  particles?

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Results from kinetic theory, 2

• What is the average velocity of the ideal gas
  particles?
• 2. The average velocity is zero, because, on
  average, the velocity of particles going in one
  direction is cancelled by the velocity of
  particles going in the opposite direction. When
  you do the analysis, you find that what really
  matters is the rms-speed (rms stands for root
  mean square).
• Square the speeds, take the average, and take the
  square root of that result.

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Results from kinetic theory, 3

• 3. The really fundamental result of kinetic
  theory is that temperature is a direct measure of
  the average kinetic energy of the particles of
  ideal gas.
• Kinetic theory
  Ideal gas law
• This tells us that the average translational
  kinetic energy of the molecules is
• Here we have a fundamental connection between
  temperature and the average translational kinetic
  energy of the atoms - they are directly
  proportional to one another.

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Light and heavy atoms

• A box of ideal gas consists of light particles
  and heavy particles (the heavy ones have 16 times
  the mass of the light ones). Initially all the
  particles have the same speed. When equilibrium
  is reached, what will be true?
• 1. All the particles will still have the same
  speed
• 2. The average speed of the heavy particles
  equals the average speed of the light particles
• 3. The average speed of the heavy particles is
  larger than that of the light particles
• The average speed of the heavy particles is
  smaller than that of the light particles

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Light and heavy atoms

• Coming to equilibrium means coming to a
  particular temperature, which means the average
  kinetic energy of the particles is a particular
  value. The average kinetic energy of the light
  particles equals the average kinetic energy of
  the heavy particles - this can only happen if the
  average speed of the heavy particles is smaller
  than that of the light particles.
• Smaller by a factor of what?

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Light and heavy atoms

• Simulation
• Coming to equilibrium means coming to a
  particular temperature, which means the average
  kinetic energy of the particles is a particular
  value. The average kinetic energy of the light
  particles equals the average kinetic energy of
  the heavy particles - this can only happen if the
  average speed of the heavy particles is smaller
  than that of the light particles.
• Smaller by a factor of what? 4, so that

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Three cylinders

• Three identical cylinders are sealed with
  identical pistons that are free to slide up and
  down the cylinder without friction. Each cylinder
  contains ideal gas, and the gas occupies the same
  volume in each case, but the temperatures differ.
  In each cylinder the piston is above the gas, and
  the top of each piston is exposed the atmosphere.
  In cylinders 1, 2, and 3 the temperatures are
  0C, 50C, and 100C, respectively. How do the
  pressures compare?
• 1gt2gt3
• 3gt2gt1
• all equal
• not enough information to say

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Free-body diagram of a piston

• Sketch the free-body diagram of one of the
  pistons.

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Free-body diagram of a piston

• Sketch the free-body diagram of one of the
  pistons.

The internal pressure, in this case, is
determined by the free-body diagram. All three
pistons have the same pressure, so the number of
moles of gas must decrease from 1 to 2 to 3.
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Introducing the P-V diagram

• P-V (pressure versus volume) diagrams can be very
  useful.
• What are the units resulting from multiplying
  pressure in kPa by volume in liters?

Rank the four states shown on the diagram based
on their absolute temperature, from greatest to
least.
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Introducing the P-V diagram

• P-V (pressure versus volume) diagrams can be very
  useful.
• What are the units resulting from multiplying
  pressure in kPa by volume in liters?

Rank the four states shown on the diagram based
on their absolute temperature, from greatest to
least. Temperature is proportional to PV, so rank
by PV 2 gt 13 gt 4.
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Isotherms

• Isotherms are lines of constant temperature.
• On a P-V diagram, isotherms satisfy the equation
• PV constant

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Thermodynamics

• Thermodynamics is the study of systems involving
  energy in the form of heat and work.
• Consider a cylinder of ideal gas, at room
  temperature.
• When the cylinder is placed in a
• container of hot water, heat is
• transferred into the cylinder.
• Where does that energy go?
• The piston is free to move up
• or down without friction.

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Thermodynamics
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The First Law of Thermodynamics

• Some of the added energy goes into raising the
  temperature of the gas (we call this raising the
  internal energy). The rest of it does work,
  raising the piston. Conserving energy
• (the first law of thermodynamics)
• Q is heat added to a system (or removed if it is
  negative)
• is the internal energy of the system (the
  energy associated with the motion of the atoms
  and/or molecules), so is the change in
  the internal energy, which is proportional to the
  change in temperature.
• W is the work done by the system.
• The First Law is often written as

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Work

• We defined work previously as
•     (true if the force is constant)
• F PA, so
• At constant pressure the work done by the system
  is the pressure multiplied by the change in
  volume.
• If there is no change in volume, no work is done.
  
• In general, the work done by the system is the
  area under the P-V graph. This is why P-V
  diagrams are so useful.

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Work the area under the curve

• The net work done by the gas is positive in this
  case, because the change in volume is positive,
  and equal to the area under the curve.

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A P-V diagram question
An ideal gas initially in state 1 progresses to a
final state by one of three different processes
(a, b, or c). Each of the possible final states
has the same temperature. For which process is
the change in internal energy larger? 1. a
2. b 3. c 4. Equal for all three 5. We cant
determine it
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A P-V diagram question

• Because the change in temperature is the same,
  the change in internal energy is the same for all
  three processes.

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Another P-V diagram question
An ideal gas initially in state 1 progresses to a
final state by one of three different processes
(a, b, or c). Each of the possible final states
has the same temperature. For which process is
more heat transferred into the ideal gas? 1.
a 2. b 3. c 4. Equal for all three 5. We
cant determine it
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Another P-V diagram question

• The heat is the sum of the change in internal
  energy (which is the same for all three) and the
  work (the area under the curve), so whichever
  process involves more work requires more heat.

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Another P-V diagram question

• The heat is the sum of the change in internal
  energy (which is the same for all three) and the
  work (the area under the curve), so whichever
  process involves more work requires more heat.
• Process c involves more
• work, and thus requires
• more heat.

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Constant volume vs. constant pressure
We have two identical cylinders of ideal gas.
Piston 1 is free to move. Piston 2 is fixed so
cylinder 2 has a constant volume. We put both
systems into a reservoir of hot water and let
them come to equilibrium. Which statement is
true? 1. Both the heat Q and the change in
internal energy will be the same for the two
cylinders 2. The heat is the same for the two
cylinders but cylinder 1 has a larger change in
internal energy. 3. The heat is the same for the
two cylinders but cylinder 2 has a larger change
in internal energy. 4. The changes in internal
energy are the same for the two cylinders but
cylinder 1 has more heat. 5. The changes in
internal energy are the same for the two
cylinders but cylinder 2 has more heat.
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Constant volume vs. constant pressure

• Each cylinder comes to the same temperature as
  the reservoir. How do the changes in internal
  energy compare?
• Which cylinder does more work?

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Constant volume vs. constant pressure

• Each cylinder comes to the same temperature as
  the reservoir. How do the changes in internal
  energy compare?
• The same number of moles of the same gas
  experience the same temperature change, so the
  change in internal energy is the same.
• Which cylinder does more work?
• Cylinder 2 does no work, so cylinder 1 does more
  work.
• By the first law, cylinder 1 requires more heat
  to produce the same change in temperature as
  cylinder 2. The heat required depends on the
  process.

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Solving thermodynamics problems

• A typical thermodynamics problem involves some
  process that moves an ideal gas system from one
  state to another.
• Draw a P-V diagram to get some idea what the
  work is.
• Apply the First Law of Thermodynamics (this is a
  statement of conservation of energy).
• Apply the Ideal Gas Law.
• the internal energy is determined by the
  temperature
• the change in internal energy is determined by
  the change in temperature
• the work done depends on how the system moves
  from one state to another (the change in internal
  energy does not)

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Constant volume (isochoric) process

• No work is done by the gas W 0. The P-V
  diagram is a vertical line, going up if heat is
  added, and going down if heat is removed.
• Applying the first law
• For a monatomic ideal gas

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Constant pressure (isobaric) process

• In this case the region on the P-V diagram is
  rectangular, so its area is easy to find.
• For a monatomic ideal gas

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Heat capacity

• For solids and liquids
• For gases , where C, the heat
  capacity, depends on the process.
• For a monatomic ideal gas
• Constant volume
• Constant pressure
• In general

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Constant temperature (isothermal) process

• No change in internal energy
• The P-V diagram follows the isotherm.
• Applying the first law, and
• using a little calculus

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Zero heat (adiabatic) process

• Q 0. The P-V diagram is an interesting line,
  given by
• For a monatomic ideal gas
• Applying the first law

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Worksheet

• You have some monatomic ideal gas in a cylinder.
  The cylinder is sealed at the top by a piston
  that can move up or down, or can be fixed in
  place to keep the volume constant. Blocks can be
  added to, or removed from, the top of the
  cylinder to adjust the pressure, as necessary.
• Starting with the same initial conditions each
  time, you do three experiments. Each experiment
  involves adding the same amount of heat, Q.
• A Add the heat at constant pressure.
• B Add the heat at constant temperature.
• C Add the heat at constant volume.

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Worksheet

• A Add heat Q to the system at constant
  pressure.
• B Add heat Q to the system at constant
  temperature.
• C Add heat Q to the system at constant volume.

Sketch these processes on the P-V diagram. The
circle with the i beside it represents the
initial state of the system. One of the processes
is drawn already. Identify which one, and draw
the other two.
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Rank by final temperature

• Rank the processes based on the final
  temperature.
• 1. A gt B gt C
• 2. A gt C gt B
• 3. B gt A gt C
• 4. B gt C gt A
• C gt A gt B
• C gt B gt A
• Equal for all three

• Add heat Q to the system at
• A - constant pressure.
• B - constant temperature.
• C - constant volume.

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Rank by final temperature

• In process B, the temperature is constant.
• In process A, some of the heat added goes to
  increasing the temperature.
• In process C, all the heat added goes to
  increasing the temperature.
• C gt A gt B

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Rank by work

• Rank the processes based on the work done by the
  gas.
• 1. A gt B gt C
• 2. A gt C gt B
• 3. B gt A gt C
• 4. B gt C gt A
• C gt A gt B
• C gt B gt A
• Equal for all three

• Add heat Q to the system at
• A - constant pressure.
• B - constant temperature.
• C - constant volume.

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Rank by work

• In process C, no work is done.
• In process A, some of the heat added goes to
  doing work.
• In process B, all the heat added goes to doing
  work.
• B gt A gt C

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Rank by final pressure

• Rank the processes based on the final pressure.
• 1. A gt B gt C
• 2. A gt C gt B
• 3. B gt A gt C
• 4. B gt C gt A
• C gt A gt B
• C gt B gt A
• Equal for all three

• Add heat Q to the system at
• A - constant pressure.
• B - constant temperature.
• C - constant volume.

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Rank by pressure

• In process A, the pressure stays constant.
• In process B, the pressure decreases.
• In process C, the pressure increases.
• C gt A gt B

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Example problem

• A container of monatomic ideal gas contains just
  the right number of moles so that nR 20 J/K.
  The gas is in state 1 such that
• P1 20 kPa V1 100 x 10-3 m3
• (a) What is the temperature T1 of the gas?

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Example problem

• A container of monatomic ideal gas contains just
  the right number of moles so that nR 20 J/K.
  The gas is in state 1 such that
• P1 20 kPa V1 100 x 10-3 m3
• (a) What is the temperature T1 of the gas?
• Use the ideal gas law.
• PV nRT, so
• The factor of 1000 in the kPa cancels the 10-3 in
  the volume.

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Example problem

• A container of monatomic ideal gas contains just
  the right number of moles so that nR 20 J/K.
  The gas is in state 1 such that P1 20 kPa
  V1 100 x 10-3 m3
• (b) If Q 2500 J of heat is added to the gas,
  and the gas expands at constant pressure, the gas
  will reach a new equilibrium state 2. What is the
  final temperature T2?

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Example problem

• A container of monatomic ideal gas contains just
  the right number of moles so that nR 20 J/K.
  The gas is in state 1 such that P1 20 kPa
  V1 100 x 10-3 m3
• (b) If Q 2500 J of heat is added to the gas,
  and the gas expands at constant pressure, the gas
  will reach a new equilibrium state 2. What is the
  final temperature T2?
• At constant pressure for a monatomic ideal gas
• Therefore

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Example problem

• A container of monatomic ideal gas contains just
  the right number of moles so that nR 20 J/K.
  The gas is in state 1 such that P1 20 kPa
  V1 100 x 10-3 m3
• (c) Q 2500 J of heat is added to the gas, and
  the gas expands at constant pressure. How much
  work is done by the gas?

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Example problem

• A container of monatomic ideal gas contains just
  the right number of moles so that nR 20 J/K.
  The gas is in state 1 such that P1 20 kPa
  V1 100 x 10-3 m3
• (c) Q 2500 J of heat is added to the gas, and
  the gas expands at constant pressure. How much
  work is done by the gas?
• At constant pressure, we have
• We can do this only for a constant pressure
  process.

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Example problem

• A container of monatomic ideal gas contains just
  the right number of moles so that nR 20 J/K.
  The gas is in state 1 such that P1 20 kPa
  V1 100 x 10-3 m3
• (d) Q 2500 J of heat is added to the gas, and
  the gas expands at constant pressure. What is the
  final volume V2?

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Example problem

• A container of monatomic ideal gas contains just
  the right number of moles so that nR 20 J/K.
  The gas is in state 1 such that P1 20 kPa
  V1 100 x 10-3 m3
• (d) Q 2500 J of heat is added to the gas, and
  the gas expands at constant pressure. What is the
  final volume V2? One approach is to bring in
  the ideal gas law again

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