Download IB Option B.2 Thermodynamics Feb 21 Agenda

Physics 2 – Feb 21, 2017
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
P3 Challenge – a) What is the average kinetic energy of a
molecule of oxygen gas at 298 K? b) What is the root mean
square velocity of an oxygen molecule if its atomic mass is
32 u? (kB = 1.38 x 10-23 J/K; 1 u = 1.66 x 10-27 kg)
Today’s Objective:
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
B.2.1-2 Thermodynamics: Boltzmann
Assignment:
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Ch 3.2, p141, #30-32, B.2 p33 #18-23
Agenda
Boltzmann Distribution
Average K.E.
Internal Energy
State functions
Types of Processes
PV work
PV diagrams
Boltzmann Distribution

Boltzmann distribution is a result of statistical mechanics
that describes how the random particle speeds of an ideal
gas are distributed.

Note: Statistical mechanics is the application of statistics to
the gas particle behavior.

Unsymmetrical distribution. Always a finite probability at
any high speed, but 0 at 0 speed.

The ranking of temperature for the three graphs are
Blue<Red<Green.

Notice the peak location increases and lowers and the
speeds spread out.
Average Particle Speed

Three types of “average speed” values for the
particles of an ideal gas

most probable speed (mode) red peak
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average speed (mean) blue

c = “root mean square” speed corresponds to
average kinetic energy

Best indication of temperature is c because
temperature is average kinetic energy
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IB assumes c represents all three.

Unfortunate that IB uses c for both vrms and the speed
of light. Beware: This is NOT c = 3 x 108 m/s
Average Kinetic Energy
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The average K.E. is related to the average speed, c,
where ma is the atomic mass of a single gas particle.


Note on units: 1 u = 1.66 x 10-27 kg
From the ideal gas law and statistical mechanics:
PM  dRT

1 2
P  dc
3
And other basic relationships:
M  N A ma
m
N
n

M NA
1
2
E K  ma c
2
3
E K  k BT
2
R
23 J
kB 
 1.38 10
K
NA
Internal Energy of a Gas

If the average kinetic energy of one particle of gas is 3/2
kT, and you have a sample containing N particles, the
internal energy of the sample is N times the average
kinetic energy.

From this, and recalling the definitions of kB and moles
you can derive the expression for the internal energy of a
gas, U.

Problem solving is either of the plug and chug variety, or
is algebra derivation of formulas types. So know these two
fundamental relationships. (Only the first is in the IB
packet.)

Notice that both average EK and U only depend on T in K.
3
E K  k BT
2
N
R
n
kB 
NA
NA
3
3
U  nRT  PV
2
2
States
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In thermodynamics, the state of a system is defined by
specifying values for a set of measurable properties sufficient to
determine all other properties. For gases, these properties are P,
V and T.

E.g. Your health state could be said to be set by your vital signs.
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Simple gas law problems always involve two states.

Ideal gas law applies to a single state.

Processes involve at least two states.
State Functions
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A state function is any variable that is only dependent on the state of
the system.

A state function is not dependent on the path used to obtain the
state.

Many variables are state functions: mass, Energy, Entropy, Pressure,
Temperature, Volume, chemical composition, density, number of
moles
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Important variables that are NOT state functions: heat and work

Both heat and work are path dependent
Types of Processes
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A thermodynamic process is one that takes a system from one state to
another.

There are several types of processes dependent on the conditions for the
change. (We’ve already seen most with the simple gas laws.)
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Isothermal process – Temperature is constant
e.g. Boyle’s law
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Isobaric process – Pressure is constant
e.g. Charles’ law
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Isovolumetric process – Volume is constant
e.g. Gay-Lussac’s Law
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Adiabatic process – No heat transfer
PV Diagrams

Pressure vs Volume graphs are the most common
way to describe the state of an ideal gas and the
processes it undergoes.

Isobaric and isovolumetric processes are horizontal
and vertical lines respectively.

Isothermal processes are described by a strict
inverse relationship.
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Recall Boyles law PV = constant when T is constant.

The closer a line is to the origin, the lower the
temperature.
Adiabatic Processes
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In an adiabatic process, Q = 0. From this, the
relationship between P and V can be determined.

The resulting equation means

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Its graph on a PV diagram will be similar to an
isothermal line, but steeper.

Problems similar to Bolye’s law problems can be
solved using this equation if an adiabatic process is
described.
Ex: An ideal gas initially at 105 kPa expands
adiabatically from 3.5 x 10-4 m3 to 8.9 x 10-4 m3. What
is the new pressure?
PV
5
3
 constant
Adiabatic Processes

Notice that for an adiabatic process, P, V and T change.

PV
5
3
The initial and final states of an adiabatic process are located on
different isothermal lines.

Using the ideal gas law to find the VT relationship, solve PV = nRT
for P.

Substituting this into the adiabatic equation and collecting all
constant terms. (n, R and constant)

If a gas expands adiabatically, the temperature drops.

If a gas compresses adiabatically, the temperature rises.
TV
 constant
nRT
P
V
2
3
 constant
PV Work Derived Part 1 – Piston h signs
W  F s

Work is defined as a force applied through a distance.

Recall that work done on a system is positive and work done by a
system is negative.

Therefore, the work done by an external force for a compression will be
positive and the work done by an external force for an expansion will
h=h
h=0
be negative.

Consider a piston on a canister holding a gas. The piston creates an
eternal force. Pick a height = 0 point as the initial piston location.

For an expansion, the h of the gas sample is positive (h – 0)
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For a compression, the h of the gas sample is negative (-h – 0)
h=-h
PV Work Derived – Part 2 Work external
W  F s cos 

Workext for expansion is negative
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Expansion, h positive Compression, h negative
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Recall that Work requires the displacement to be in the same direction as
the force.

Here, our external force is 180 from the direction of h.

Cos 180 = – 1 . Therefore Wext = – F h
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Wext, exp = – F h would be negative, Wext,comp = – F h would be positive.
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Results as expected above. Verify this.
Workext for compression is positive
Wext   F h
PV Work Derived – Part 3 Gas Pressure

To transform our Wext expression into PV work, we first recognize
that the force exerted by the gas pressure is an equal and
opposite reaction force to the external applied force.

W done by the gas is the opposite of the Wext
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To convert force into P and h into V, multiply by A/A as a
form of 1.

Sort the factors and replace by the definitions of P and V.
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Ta Da. That’s why the work done by a gas is called PV work.
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W for an expansion is positive and W for a compression is
negative
Wext   F h
W  F h
A
W  F h 
A
F
W  h  A
A
W  PV
PV work on a PV diagram
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P is the vertical axis and V is the horizontal, so PV
corresponds to the area under the curve for a process.
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If the process is not isobaric, the PV work is still the area
under the curve.
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Sometimes this is easy to calculate, but for isothermals and
adiabatic processes, it requires calculus.
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Area sweeps to the right are positive work (V is positive)
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Area sweeps to the left are negative work (V is negative)
Work done for a PV diagram cycle
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Many processes that are shown on a PV
diagram are cycles that return the state of
the system to the initial conditions.
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The energy change for a cycle is zero
because energy is a state function.
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Work is not a state function.
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Work is the area within the loop created by
the processes that make up the cycle.
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CW loop is pos. W.
CCW loop is neg. W.
Exit Slip - Assignment
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Exit Slip- A gas expands isobarically from a volume of 3.80 x 10-3 m3 to
5.20 x 10-3 m3 at 103.5 kPa. How much PV work is done by the gas?
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What’s Due on Feb 16? (Pending assignments to complete.)
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Ch 3.2, p141, #30-32, B.2 p33 #18-23
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DOWNLOAD Engineering Physics Text B!!!! Read B.2.1-B.2.4, p19-23
What’s Next? (How to prepare for the next day)
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Read B p24-32 about Thermodynamics