IB Physics

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IB Physics

• Topic 10 Thermodynamic Processes
• Mr. Jean

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The plan

• Video clip of the day
• https//www.youtube.com/watch?vnBVL3lp9fGo
• https//www.youtube.com/watch?vs3tjOEtoArg
• Returning Exams
• Thermodynamic processes

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Returning Exams

• Review your exams
• If you have any questions we can go over some of
  the solutions briefly.

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Gravitational Waves

• https//www.youtube.com/watch?vgw-i_VKd6Wo

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Equation of state for an ideal gas
Combining the gas laws gives

PV (a constant)T
Therefore, for n moles of an ideal gas
P Pressure V Volume n number of
moles R Universal gas constant T Temperature
(KELVIN)
PV nRT
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Describe the concept of the absolutezero of
temperature and the Kelvinscale of temperature.
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Work done in compressing a gas
Deduce an expression for the work involved in a
volume change of a gas at constant pressure.

The work done by this force is w Fs
PAs, since FPA
but As is the change in the volume occupied by
the gas, ?V. therefore
W PDV
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1st Law of Thermodynamics
State the first law of thermodynamics.
We can add energy to a gas by heating Q
(temperature gradient) Or by working (mechanical
energy) W
Students should be familiar with the terms system
and surroundings. They should also appreciate
that if a system and its surroundings are at
different temperatures and the system undergoes a
process, the energy transferred by non-mechanical
means to or from the system is referred to as
thermal energy (heat).
Q ?U W
Q Heat energy added to the gas ?U Internal
energy increase of the gas W Work done by the
gas.
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P-V changes
                                                                                  Change of p (and T) at constant volume an isovolumetric change. 2. Change of V (and T) at constant pressure an isobaric change. 3. Change in p and V at constant temperature an isothermal change. 4. Change in p and V in an insulated container (no heating of the gas) an adiabatic change.
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Work done in a thermodynamic process
The product of pressure and volume represents a
quantity of work. This is represented by the area
below a p-V curve. Therefore, the area enclosed
by the four curves represents the net work done
by the engine during one cycle.
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Second Law Entropy
Second Law of Thermodynamics It is impossible to
extract an amount of heat QH from a hot reservoir
and use it all to do work W . Some amount of heat
QC must be exhausted to a cold reservoir.
Entropy is a measure of the disorder (of the
energy) of a system
Every time we change energy from one form to
another, we increase the entropy of the Universe
even though local entropy may decrease.
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Thermodynamic Processes
A system can change its state
A state is a unique set of values for P, V, n, T
(so PV nRT is also called a State Equation)
When you know the state of a system you know U
since U ? NkT ? nRT ? PV, for a monatomic
gas
A process is a means of going from 1 state to
another
There are 4 basic processes with n constant
iso means same
Isobaric, a change at constant pressure
Isochoric or isovolumetric, a change at constant
volume, W 0
Isothermal, a change at constant temperature (?U
0, Q W)
Adiabatic, a process is one in which no heat is
gained or lost by the system.
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Thermodynamic Processes
Isobar
P
(P1,V1) T1
(P2,V2) T2
1
2
Isochore
Adiabat
4
(P4,V4) T4
Q 0
3
T3 T4
(P3,V3) T3
Isotherm
The trip from 1?2?3?4?1 is call a thermodynamic
cycle
V
Each part of the cycle is a process
All state changes can be broken down into the 4
basic processes
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Thermodynamic Processes
Isobar, expansion at constant pressure, work is
done
P
1
2
Isochoric pressure change, W 0
Adiabatic expansion no heat, Q 0
4
3
Isothermal compression W Q, U is constant
The area enclosed by the cycle is the total work
done, W
The work done, W, in a cycle is if you travel
clockwise
V
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Heat Engines and Refrigerators
Engines use a working fluid, often a gas, to
create motion and drive equipment the gas moves
from 1 state (P, V, n, T define a state) to
another in a cycle
The Stirling Cycle 2 isotherms 2 isochores
Stirling designed this engine in the early 18th
century simple and effective
The Stirling Engine
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Isobaric expansion of a piston in a cylinder
The work done W Fd PAd P?V
The work done is the area under the process W
P?V
4 stroke engine
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Isochoric expansion of a piston in a cylinder
The work done W 0 since there is no change in
volume
Thus ?U Q W Q
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Adiabatic expansion of an ideal gas
The work done W 0 here because chamber B is
empty and P 0
Thus ?U Q W 0, that is adiabatic expansion
against no resistance does not change the
internal energy of a system
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EXAMPLE
How much work is done by the system when the
system is taken from (a)  A to B  (900 J)(b)
 B to C  (0 J)(c)  C to A  (-1500 J)
Each rectangle on the graph represents 100 Pa-m³
100 J
(a) From A? B the area is 900 J, isobaric
expansion
(b) From B ? C, 0, isovolumetric change of
pressure
(c) From C? A the area is -1500 J
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EXAMPLE
10 grams of steam at 100 C at constant pressure
rises to 110 C P   4 x 105 Pa             ?T
10 C   ?V 30.0 x 10-6 m3        c 2.01 J/g
What is the change in internal energy?
?U Q W
?U mc?T P?V
?U 189 J
So heating the steam produces a higher internal
energy and expansion
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EXAMPLE
Aluminum cube of side L is heated in a chamber at
atmospheric  pressure. What is the change in
the cube's internal energy if L 10 cm and ?T
5 C?
Q mc?T m  ?V0V0  L3W P?V?V  ?V0?T
?U Q W
?U mc?T P?V
cAl 0.90 J/gC
?Al 72(10-6) C-1
?U ?V0c?T P?V0?T
Patm 101.5 kPa
?U V0?T (?c P?)
?Al 2.7 g/cm³
?U L³?T (?c P?)
?U 0.10³(5)((2700)(900) 101.5(10³)(72(10-6))
?U 12,150 J
NB P? is neglible
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Find the work done for a cycle if P1 1000 kPa,
V1 0.01 m³, V2 0.025 m³, V3 V4 0.04 m³,
T1 400 K, T2 600K, n 2 mol
EXAMPLE
W Area enclosed
? (P2P3)?V23
P1?V12
? (P1P4)?V41
P
1, (P1,V1) T1
2, (P2,V2) T2
Isobar
Isotherm
3, (P3,V3) T3
Isotherm
Isochore
4, (P4,V4) T4
V
1. P2 P1 1000 kPa
W Area enclosed P1?V12 ?
(P2P3)?V23 ? (P1P4)?V41 (15 12.188
18.75)(10³) 8.44 kJ
2. T4 T1 400 K
3. T3 T2 600 K
4. P3 P2V2/V3 625 kPa
5. P4 P1V1/V4 250 kPa
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Find the internal energy for each state if P1
1000 kPa, V1 0.01 m³, V2 0.025 m³, V3 V4
0.04 m³, T1 400 K, T2 600K, n 2 mol
EXAMPLE
1. P2 P1 1000 kPa
2. T4 T1 400 K
P
3. T3 T2 600 K
1, (P1,V1) T1
2, (P2,V2) T2
Isobar
4. P3 P2V2/V3 625 kPa
5. P4 P1V1/V4 250 kPa
Isotherm
3, (P3,V3) T3
Isotherm
Isochore
4, (P4,V4) T4
V
6. U1 ? nRT1 9972 J
7. U4 U1 9972 J
8. U2 ? nRT2 14958 J
9. U3 U2 14958 J
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Find the thermal energy change Q for each state
if P1 1000 kPa, V1 0.01 m³, V2 0.025 m³,
V3 V4 0.04 m³, T1 400 K, T2 600K, n 2
mol
EXAMPLE
1. P2 P1 1000 kPa
Q12
2. T4 T1 400 K
P
3. T3 T2 600 K
1, (P1,V1) T1
2, (P2,V2) T2
Isobar
4. P3 P2V2/V3 625 kPa
5. P4 P1V1/V4 250 kPa
Q34
Q41
Isotherm
3, (P3,V3) T3
Q34
Isotherm
Isochore
4, (P4,V4) T4
V
10. Q12 ?U12 W12 34986 J
6. U1 ? nRT1 9972 J
11. Q23 W23 (?U23 0) W23 ? (P2P3)?V23
12.188 kJ
7. U4 U1 9972 J
12. Q34 ?U34 -4986 J
8. U2 ? nRT2 14958 J
13. Q41 W41 (U41 0) W41 ? (P4P1)?V41 -
18.75 kJ
9. U3 U2 14958 J
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Heat Engines and Refrigerators
The Wankel Rotary engine is a powerful and simple
alternative to the piston engine used by Nissan
and invented by the German, Wankel in the 1920s
The Wankel Cycle 2 adiabats 2 isochores
The Wankel Engine