thermodynamics wee
DESCRIPTION
PHYSICSEXPERIMENTS PHYSICSEXPERIMENTSTRANSCRIPT
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3b Thermodynamics
Specific heat capacity, cLatent heat capacity, LChange of phase, evaporationFirst law of thermodynamics, -
Real Life Application of specific heat capacity, c
Thinking Question
Cheese is still hot a long time, after being taken out of the oven. Why?
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Heat capacity, C
note: C = m.c
m = Mass of the substance,
c = Specific heat capacityThe heat capacity, C, of a body is defined as
the amount of heat Q required to raise its
temperature by 1 Kelvin without going
through a change in state.
The S.I. unit is J K1
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Why is C worth learning?
Demo video on Heat capacity, CWater Balloon Heat Capacity.MPG -
Nice explanation
http://www.asknlearn.com/contentpackaging/14297/RS14/Container.html -
Specific Heat Capacity
The specific heat capacity, c, of a body is defined as
the amount of heat required to raise the temperature
of 1 kg of the body by 1 Kelvin.
The S.I. unit is J kg1 K1
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Specific Heat Capacity, c
of some substances at 25 oC and atmospheric pressure
SubstanceJ/ kg oCAluminum900Bismuth123Copper386Brass380Gold126Lead128Silver233Tungsten134Zinc387Mercury140Alcohol (ethyl)2400Water4186Ice (-10 0C)2050 -
What does larger specific heat capacity mean?
Specific Heat with Rods and Wax.MPGHint:For the same mass, m, cAl > c steel> c lead meansQAl > Q steel> Q lead for the same -
To solve problems
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Example 1
How much energy is required when a piece of copper of mass 0.275 kg is heated from 14.0 C to 100.0 C? (Specific heat capacity of copper = 380 J kg1 K1)m = 0.275 kg
c = 380 J/(kg.K)
f - i = (100 14)
Q
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Example 2
An electric heater of 5 kW is used to heat up a piece of copper of mass 0.5 kg from 10 C to 90 C. Calculate the time taken. (ccopper = 380 J kg1 K1)m = 0.5 kg
c = 380 J/(kg.K)
f - i = (90 10)
P= 5x103 W
heater
- Q = m c (f - i )(5x103)(t)= (0.5)(380)(90-10)t = 3.0 s
m = 0.5 kg
c = 380 J/(kg.K)
f - i = (90 10)
P= Q/t
= 5x103 W
heater
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Example 3
An ethnic restaurant serves coffee in copper mugs. A waiter fills a cup having a mass of 0.10 kg, initially at 20 C, with 0.20 kg of coffee initially at 70 C. What is the final temperature after the coffee and the cup has attained thermal equilibrium? Assume that there is no heat loss to the surroundings.(ccoffee = 4200 J kg1 K1; ccopper = 380 J kg1 K1)H2O
m = 0.2 kg
c = 4200 J/(kg.K)
f - i = ( 70)
Q= ?
cup
m = 0.1 kg
c = 380 J/(kg.K)
f - i = ( 20)
- Q = 0mcup.ccup ( 20) + mH20.cH20 ( 70) = 0 (0.1)(380)( 20) +
(0.2)(4200)( 70)=0Solve for = 67.8 0C = 68 0C
H2O
m = 0.2 kg
c = 4200 J/(kg.K)
f - i = ( 70)
Q= ?
cup
m = 0.1 kg
c = 380 J/(kg.K)
f - i = ( 20)
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Flash Simulation
heat_metal.swfThinking and reflectionThe random fluctuation of the thermometer instrument, relate back to chapter 1 called uncertainty or random errorthe assumption no heat loss to surrounding is shown in simulation but not easily explain with clarity with wordsThe concept of thermal equilibrium is vividly show in simulation when calculations alone fail to show the process of thermal equilibrium -
Electrical method to determine c
A
V
Heater & material under test
Connect to power supply
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Remember PWS6 electrical setup for heater power?Term 1 practical YJC.
- Example 4Hand write
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Phase Changes
Q= mcsolid
Q= mcliquid
Q= mcgas
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Change of phase-melting
Latent heat of fusion, LfHeat is supplied, QTemperature remains constant. T=constantQ= mLf
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Change of phase-boiling
Latent heat of vaporisation, LvHeat is supplied, QEnergy (Work done by system) is needed to expand against atmosphere WbyTemperature remains constant, T =constantQ= mLv
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Change of phase
Specific Latent heat of vaporisation >
Specific latent heat of fusion. Lv > Lf
Energy is needed to work against atmosphere, Wby
Difference in PE between molecules in liquid to gaseous > in solid to liquidQ= mLv
Q= mLf
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Select 2
http://www.asknlearn.com/contentpackaging/14297/RS14/Container.html -
cooling effect during evaporation
Molecules with higher KE and are near to surface, are able to overcome intermolecular force and escape from liquid surface.
Molecules with lower KE are left behind in the liquid.
As a result, the average molecular KE decreases and hence the temperature drops.
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Evaporation
Occurs at all temperature (including boiling)Average KE of liquid decreasesCooling processhttp://www.colorado.edu/physics/2000/applets/bec.html
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Video on Real Life Application of Evaporation & Condensation
Drinking Bird.MPGClever synthesis of many Physics ideas You can read up about it by Goggling Dippy Bird http://science.howstuffworks.com/question608.htmWhen water evaporates from the fuzz on the Dippy Bird's head, the head is cooled.
The temperature decrease in the head condenses the methylene chloride vapor, decreasing the vapor pressure in the head relative to the vapor pressure in the abdomen.
The greater vapor pressure in the abdomen forces fluid up through the neck and into the head.
As fluid enters the head, it makes the Dippy Bird top-heavy.
The bird tips. Liquid travels to the head. The bottom of the tube is no longer submerged in liquid.
Vapor bubbles travel through the tube and into the head. Liquid drains from the head, displaced by the bubbles.
Fluid drains back into the abdomen, making the bird bottom-heavy.
The bird tips back up.
If the bird dips into a cup of water, the fuzzy material absorbs water again and the cycle starts over.
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Now lets test yourselves
Qn 1
For a given liquid at atmospheric pressure,
which process can occur at any
temperature?
Boiling
Evaporation
Melting
Solidification
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Now lets test yourselves
Qn 2
Latent heat of vaporisation is the energy required to
Separate the molecules of the liquid
Force back the atmosphere to make space for the vapour
Increase the average molecular speed in the liquid phase to that in the gas phase
Separate the molecules and to force back the atmosphere
Separate the molecules and to increase their average molecular speed to that in the gas phase.
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Latent heat, L
The S.I. unit for l is J kg1.
No change in temperature
Specific latent heat of fusion, L fSpecific latent heat of vaporisation, L v Q = m L f Q = m L v - Example 5
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summary of Specific heat capacity, Specific latent heat
Total heat gained = Total heat lost / supplied
Or
(heat) = 0
Specific heat capacity, c : Q=mc
Specific latent heat of fusion, Lf : Q=m Lf
Specific latent heat of vaporisation, Lv: Q= m Lv
In a closed system,
3 most important equations:
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Phase Change of water
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Electrical method to determine L
A
V
Heater & material under test
Connect to power supply
- Example 6Example 7Suggested to hand write, for ease of following the chain of thought
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Example 6
An electric kettle contains 1.5 kg of water at 100 C and is powered by a 2.0 kW electric element. If the thermostat of the kettle fails to operate, calculate the time taken for the kettle to boil dry.
(lv of water = 2000 kJ kg1)
t = 1500 s
heat gained by water (boil) = heat supplied by kettle
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Example 7 (a) better to use h/t
Therefore, lv = 1.98 x 106 J Kg-1
1st set up:
Mass = 0.1 kg
Power = 240 W
Time taken = 15 min
2nd set up:
Mass = 0.15 kg
Power = 350 W
Time taken = 15 min
a) Calculate the specific latent heat of vaporisation.
Heat supplied = heat gained by liquid + heat loss
P t = m lv + h
Since time taken is the same, h will also be the same.
(240)(15 x 60) (0.1) lv = (350)(15 x 60) (0.15) lv
h = P t m lv
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Example 7 (b)
= 8.3 %
b) Find out what percentage of the heat supplied was lost in 1st attempt.
Heat loss, h = (240)(15 x 60) (0.1)(1.98 x 106)
Using h = P t m lv for 1st set up:
= 1.80 x 10 4 J
Percentage of heat loss =
=
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First law of thermodynamics
Conservation of energyTo change the internal energy of systemForces to do work
By means of heat transfer
Combination of both
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First law of Thermodynamics
The increase in internal energy of a closed system is the sum of the heat supplied to the system subtract the work done by the system.
Increase in internal energy of system
Heat supplied to system
Work done by system
(+)
(+)
(+)
Volume will increase.
Temperature will increase.
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First law of Thermodynamics
The increase in internal energy of a closed system is the sum of the heat supplied to the system subtract the work done by the system.
to
In a closed system, the mass is fixed. Therefore no mass can be added into or removed from the system.
Increase in internal energy of system
Work done by system
Heat supplied to system
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Recap: Internal energy
Internal energy, U of a system is the total sum of the kinetic energy, Ek and the potential energy, Ep of the molecules in the system.
Kinetic energy, Ek is associated with the temperature.
Potential energy, Ep depends on the separation of molecules.
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Mechanical Equivalent of Heat.MPG
Video showing Q is a form of energy. -
Nice explanation
http://www.asknlearn.com/contentpackaging/14297/RS15/Container.htmlRemember one form
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First law of Thermodynamics
Sign convention:
DescriptorsMeaning SignsHeat enters systemHeat supplied to system+ QHeat leaves systemHeat given out by system Q -
First law of Thermodynamics
Sign convention:
DescriptorsMeaning SignsVolume of system decreasesWork done on system- WbyVolume of system increasesWork done by system+ Wby -
First law of Thermodynamics
Sign convention:
DescriptorsMeaning SignsTemperature increasesIncrease in internal energy+ UTemperature decreasesDecrease in internal energy U -
Sign conventions
conventionsOpposite conventionUIncrease in internal energyDecrease in internal energyQHeat supplied to the systemHeat loss by the systemWbyWork done by the systemWork done on the system -
Internal energy of ideal gas
Recall internal energy does not possess internal potential energyPE=0 by definition for gasis the sum of the molecular KE of particlesKE temperature -
Internal energy of ideal gas
Average translational KE per molecule =
Total KE for N molecules in the gas =
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Work done by system
V
A
B
p
Work done by gas in compression is a negative number
p
V
A
B
Work done by gas in expansion is a positive number
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Work done by system (gas)
V
p
Wby = area under the p-v graph =
Wby= p V = p (Vf Vi) = negative number
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Work done-cycle
p
V
By gas
On gas
A
B
C
D
Process
A B : by gas
B C : zero
C D : on gas
D A : zero
Overall : work done by gas
WD in a cycle = area of enclosed region
WD ABCDA = WD by gas = area of enclosed region
WD ADCBA = - WD by gas = - area of enclosed region
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Video on Fire Syringe
After watching the video, think about how this video is related to the tableProcessWhat it meansIso-baricConstant pressureIso-choric /Iso-volumetricConstant volumeIso-thermalConstant temperatureAdiabaticNo heat enters or leaves the system -
Notice isothermal lines.
http://www.lon-capa.org/~mmp/applist/pvt/pvt.htm - http://mysite.verizon.net/pmrenault/thermo.html
T=200K
T=150K
- http://mysite.verizon.net/pmrenault/thermo.html
P=constant
- http://mysite.verizon.net/pmrenault/thermo.html
V=constant
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Thermodynamic processes
There are 4 special processes to consider:
Isothermal processes
ProcessWhat it meansIso-baricConstant pressureIso-choric /Iso-volumetricConstant volumeIso-thermalConstant temperatureAdiabaticNo heat enters or leaves the systemp
V
T2
T1
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Thermodynamic Processes
There are 4 special thermodynamic processes:
ProcessWhat it meansImplicationsIso-baricConstant pressureP = constantIso-choric/ Iso-volumetricConstant volumeV = constantV = 0 m3W = P V = 0 JIso-thermalConstant temperatureT = constantU = 0 JAdiabaticNo heat enters or leaves systemQ = 0 JU = W -
Thermodynamic Processes
There are 4 special thermodynamic processes:
ProcessWhat it meansImplicationsIso-baricConstant pressureP = constantIso-choric/ Iso-volumetricConstant volumeV = constantV = 0 m3W = P V = 0 JIso-thermalConstant temperatureT = constantU = 0 JAdiabaticNo heat enters or leaves systemQ = 0 JU = W -
Thermodynamic Processes
There are 4 special thermodynamic processes:
ProcessWhat it meansImplicationsIso-baricConstant pressureP = constantIso-choric/ Iso-volumetricConstant volumeV = constantV = 0 m3Wby = P V = 0 JIso-thermalConstant temperatureT = constantU = 0 JAdiabaticNo heat enters or leaves systemQ = 0 JU = -Wby -
Thermodynamic Processes
There are 4 special thermodynamic processes:
ProcessWhat it meansImplicationsIso-baricConstant pressureP = constantIso-choric/ Iso-volumetricConstant volumeV = constantV = 0 m3W = P V = 0 JIso-thermalConstant temperatureT = constantU = 0 JAdiabaticNo heat enters or leaves systemQ = 0 JU = -Wby -
Cool video on Q=0
Adiabatic Expansion.MPG -
pV diagram
Indicator diagram or pV diagram is simply a Pressure vs Volume graph
p / Pa
V / m3
V = volume of container
p = Pressure exerted on the walls by gas molecules
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pV diagram
Indicator diagram or pV diagram is simply a Pressure vs Volume graph
p / Pa
V / m3
U = Q - Wby
Wby = P V
= area under pV diagram
Heat flow in
U= (3/2).nRT
So
U = (3/2)nRT
pV diagram
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Iso-volumetric process
Constant volume
p / Pa
V / m3
U = Q - Wby
pV diagram
Area = 0 J
Wby = 0 J
U = Q
V remains unchanged when P changes.
60
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Iso-baric process
Constant pressure
p / Pa
V / m3
U = Q - Wby
pV diagram
Area = 100 x 50 = 5000 J
Wby = + 5000 J or 5000 J?
V increases
Work done BY gas
+ Wby
Wby = + 5000 J
P remains unchanged when V changes.
100
V = 50
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Iso-baric process
Constant pressure
p / Pa
V / m3
P remains unchanged when V changes.
U = Q -Wby
pV diagram
100
V = 50
Area = 100 x 50 = 5000 J
Wby = + 5000 J or 5000 J?
V decreases
Work done ON gas
Wby negative number
Wby = - 5000 J
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Iso-thermal process
Constant Temperature
p / Pa
V / m3
U = Q - Wby
pV diagram
Area = W
No need to calculate W!
However, calculation of W in Iso-thermal process is not in A level syllabus!
T = 80 oc
T remains unchanged along the curve.
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Iso-thermal process
Constant Temperature
p / Pa
V / m3
T remains unchanged along the curve.
U = Q - Wby
pV diagram
T remains constant
0= Q Wby
U = 0 J
T = 80 oc
U also remains constant
[ Wby = negative number if process is decrease in volume
Wby = postive number if process is increase in volume]
Higher T?
T = 90 oc
50
60
70
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Finding Wby from pV diagram
Important points:
1. Find the area under graph = magnitude of W.
2. If volume decreases (compression), work is done ON gas Wby is negative number
3. If volume increases (expansion), work is done BY gas Wby is positive number
4. Work done ON gas = Work done BY gas
p / Pa
V / m3
- Example 8Example 9Example 10Example 11Example 12Example 13Example 14
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Example 8
C
200
500
Find the work done by the gas in going from the situation represented by point A, through point B, to point C.
300
500
800
V/cm3
p/KPa
A
B
Area = (200+500)(200/2) + (500)(300)
= 2.20 x 105 J
Magnitude of Wby:
Wrong answer! Can you spot the mistake?
Draw this!
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Example 8
C
200
500
Find the work done on the gas in going from the situation represented by point A, through point B, to point C.
Wby gas = + 220 J
= 220 J
300
500
800
V/cm3
p/KPa
A
B
Since volume decreases (compression) Wby is a positive number
Area = (2 x 105 + 5 x 105)(200 x 10-6)/2 + (5 x 105)(300 x 10-6)
Magnitude of W:
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Example 12 (a)
A thermodynamic system is taken from initial state A to state B and back again to A via state C.
V/cm3
p/KPa
A
B
C
Initial state
3
1
20
40
A B
B C
C A
2
+0.04
0
-0.06
a) Fill up the table with +, or 0.
+
+
+
+
pV diagram
Iso-thermal curve
1st law of thermodynamics
U = Q - Wby
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Example 12 (b)
A thermodynamic system is taken from initial state A to state B and back again to A via state C.
V/cm3
p/KPa
A
B
C
Initial state
3
1
20
40
2
b) Calculate net work done by the system for one complete cycle.
WDA B = + (20 x 103)(2 x 10-6) = 0.04 J
WDB C = 0 J
WDC A = [(20 + 40) x 103)](2 x 10-6)/2 = 0.06 J
Therefore net work done by system = 0.04 + 0 + ( 0.06) = 0.02 J
(area enclosed by loop)
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Example 13
An ideal gas undergoes a cycle of changes A B C D as shown in the figure below. Complete the table.
V/10-3 m3
p/KPa
A
B
D
C
Initial state
0.3
0.1
100
200
A B
B C
C D
D A
50
25
140
+ 20
0
40
0
1. Draw table
2. Known values
3. Find W from pV diagram
U = Q - Wby
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Example 13
An ideal gas undergoes a cycle of changes A B C D as shown in the figure below. Complete the table.
V/10-3 m3
p/KPa
A
B
D
C
Initial state
0.3
0.1
100
200
U = Q - Wby
A B
B C
C D
D A
50
70
25
140
20
+ 25
100
0
+40
0
75
75
4. Find U
3. Find W from pV diagram
5. Finally, find Q
Q may be found using mc
Summary
For cycle, total U = 0 J
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Example 13: Won explanation
An ideal gas undergoes a cycle of changes A B C D as shown in the figure below. Complete the table.
V/10-3 m3
p/KPa
A
B
D
C
Initial state
0.3
0.1
100
200
U = Q - Wby
A B
B C
C D
D A
50
70
25
140
20
+ 25
100
0
+40
0
75
75
Summary
+ 20
0
40
0
Won
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Example 14 (a)
Mass of gas = 1 gInitial pressure = 1.01 x 105 Pa
Molar mass = 18Initial temperature = 27 oc
V/m3
p/ x 105 Pa
A
B
C
V1
1.01
a) Illustrates these changes on a pV diagram.
A: the gas is heated at constant pressure to 127 oc
B: compressed isothermally to its original volume
C: the gas is cooled at constant volume to its original temperature
P2
Initial state?
V2
27 oc
127 oc
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Example 14 (b)
Mass of gas = 1 gInitial pressure = 1.01 x 105 Pa
Molar mass = 18Initial temperature = 27 oc
V/m3
p/ x 105 Pa
A
B
C
V1
1.01
b) How much work does the gas do in pushing the piston during stage A?
P2
Find V1:
V1 = 1.37 x 10-3 m3
V2
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Example 14 (b)
Mass of gas = 1 gInitial pressure = 1.01 x 105 Pa
Molar mass = 18Initial temperature = 27 oc
b) How much work does the gas do in pushing the piston during stage A?
Find V2:
V/m3
p/ x 105 Pa
A
B
C
V1
1.01
P2
V1 = 1.37 x 10-3 m3
V2 = 1.83 x 10-3 m3
V2
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Example 14 (b)
Mass of gas = 1 gInitial pressure = 1.01 x 105 Pa
Molar mass = 18Initial temperature = 27 oc
b) How much work does the gas do in pushing the piston during stage A?
Therefore work done by gas
= + (area under graph)
V/m3
p/ x 105 Pa
A
B
C
V1
1.01
P2
V1 = 1.37 x 10-3 m3
= (1.01 x 105)[(1.83 1.37) x 10-3]
= 46.5 J
V2 = 1.83 x 10-3 m3
V2
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Example 14 (c)
Mass of gas = 1 gInitial pressure = 1.01 x 105 Pa
Molar mass = 18Initial temperature = 27 oc
c) What is the change in the internal energy in stage A?
V/m3
p/ x 105 Pa
A
B
C
V1
1.01
P2
= + 69.3 J
Assuming negligible potential energy, U = KE = (3/2) NK T = (3/2) nR T
V2
27 oc
127 oc
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Example 14 (d)
Mass of gas = 1 gInitial pressure = 1.01 x 105 Pa
Molar mass = 18Initial temperature = 27 oc
d) What is the heat input to the cylinder in stage A?
V/m3
p/ x 105 Pa
A
B
C
V1
1.01
P2
U = Q Wby
69.3 = Q 46.5
115.8 J = Q
Note: Wby gas = + 46.5 J Won gas = 46.5 J
V2
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Example 14 (e)
Mass of gas = 1 gInitial pressure = 1.01 x 105 Pa
Molar mass = 18Initial temperature = 27 oc
e) What is the change in internal energy of the gas in stage B?
V/m3
p/ x 105 Pa
A
B
C
V1
1.01
P2
Since T = 0,
therefore U = 0 J,
since U = (3/2) nR T
Is there any change in temperature in stage B?
V2
-
Example 14 (f)
Mass of gas = 1 gInitial pressure = 1.01 x 105 Pa
Molar mass = 18Initial temperature = 27 oc
f) How much heat must be extracted from the gas in stage C?
V/m3
p/ x 105 Pa
A
B
C
V1
1.01
P2
Therefore U for stage C = 69.3 J
For a cycle, total U = 0 J
U = + 69.3 J
U = 0 J
U = ? J
U = Q - Wby
69.3 = Q - 0
69.3 J =Q = Qin
Q to be released = +69.3 J
V2
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