the thermodynamic cycle. heat engines and refrigerators operate on thermodynamic cycles where a gas...
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The Thermodynamic Cycle
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Heat engines and refrigerators operate on thermodynamic cycles where a gas is carried from an initial state through a number of intermediate states and ultimately returned to its initial state.
During a thermodynamic cycle:
Heat Engines convert a portion of the heat input to work
Refrigerators convert input work to the transfer of heat from low temperature to high temperature
Examples:Steam engineInternal gas combustion engineDiesel engine
Examples:RefrigeratorHeat Pump
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4.5 moles of O2 are carried through a thermodynamic cycle composed of four steps: (i → 1→ 2→ 3→ i)
The molar specific heats for oxygen are:
CP =29.18 Jmole⋅K
CV =20.83 Jmole⋅K
i → 1: (Isochoric Vi =0.06m3) increase in
2.5pressure from ×105Pa 3.5to ×105Pa
2 → 3 : Isothermal expansion returning
2.5the gas to its initial pressure of×105Pa
1→ 2 : 0.10Isobaric expansion to m3
3 → i : Isobaric compression returning
the gas to its initial state
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3.5×105
2.5 ×105
0.06 0.10 V3
i
1 2
3
isochoric
isobaric
isothermal
isobaric
Pressure, Pa
Volume ,m3
(4.5,0.06, 3.5×105 ,T1 )
(4.5,0.06, 2.5×105 ,Ti )
(4.5,0.10, 3.5×105 ,T2 )
(4.5,V3 , 2.5×105 ,T3 )
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What is the gas temperature at point "i"?
PiVi =nRTi
Ti =PiVinR
(4 .5 , 0.06 , 2 .5 ×10 5 ,T i )
Ti =2.5 ×105 Pa( ) .06m3( )
4.5moles( ) 8.31 Jmole• K( )
Ti =401K
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What is the gas temperature at point "1"?
(4 .5 , 0.06 , 3 .5 ×10 5 ,T 1)
P1V1 =nRT1
T1 =P1V1nR
T1 =3.5×105 Pa( ) .06m3( )
4.5moles( ) 8.31 Jmole• K( )
T1 =562K
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What is the gas temperature at point "2"?
(4 .5 , 0 .10 , 3 .5 ×10 5 ,T 2 )
P2V2 =nRT2
T2 =P2V2nR
T2 =3.5×105 Pa( ) .10m3( )
4.5moles( ) 8.31 Jmole• K( )
T2 =936K
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What is the gas volume at point "3"?
(4 .5 , V3 , 2 .5 ×10 5 ,T 3 )
Since the step 2 → 3 ,is isothermal 3 the temperature at point is the same
2. as the temperature at point T3 =T2 =936KP3V3 =nRT3
V3 =nRT3
P3
V3 =(4.5moles)(8.31 J
mole⋅K )(936K )
2.5×105Pa
V3 =.14m3
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3.5×105
2.5 ×105i
1 2
3
isochoric
isobaric
isothermal
isobaric
0.06 0.10 V3
Pressure, Pa
Volume ,m3
0.06 0.10 0.14
(4.5,0.06, 3.5×105 , 562) (4.5,0.10,3.5×105 , 936)
(4.5, 0.14, 2.5×105 ,936 )5(4.5,0.06, 2.5×10 ,401 )
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What is the magnitude and direction
of the heat flow in the step "i → 1 "?
Qi→ 1 =nCVΔTi→ 1 =nCV T1 −Ti( )
Qi→ 1 = 4.5moles( ) 20.83 Jmole• K( ) 562K −401K( )
Qi→ 1 =1.509 ×104 J K into the gas
What is the magnitude and direction
of the work done in the step "i → 1 "?
ΔVi→1 = 0∴ Wi→1 = 0
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What is the magnitude and direction
of the change in thermal energy in the
step "i → 1 "?ΔUi→1 = Q i→1 + Wi→1
ΔUi→1 = 1.509 × 104 J + 0
ΔUi→1 = 1.509 × 104 JK increase
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What is the magnitude and direction
of the heat flow in the step "1 → 2 "?
Q1→ 2 =nCPΔT1→ 2 =nCP T2 −T1( )
Q1→ 2 = 4.5moles( ) 29.18 Jmole• K( ) 936K −562K( )
Q1→ 2 =4.911×104 J K into the gas
What is the magnitude and direction
of the work done in the step "1→ 2 "?W1→ 2 =−P1ΔV1→ 2 =−P1 V2 −V1( )W1→ 2 =−3.5×105 Pa .1m3 −.06m3( )
W1→ 2 =−1.4 ×104 J K by the gas
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What is the magnitude and direction
of the change in thermal energy
in the step "1→ 2 "?
ΔU1→2 = Q1→2 + W1→2
ΔU1→2 = 4.911 × 104 J + −1.4 × 104 J( )
ΔU1→2 = 3.511 × 104 JK increase
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What is the magnitude and direction
of the work done in the step "2 → 3 "?
W2→ 3 =−nRT2 lnV3V2
⎛ ⎝ ⎜
⎞ ⎠ ⎟
W2→ 3 =−4.5moles( ) 8.31 Jmole• K( ) 936K( ) ln .14m3
.10m3 ⎛ ⎝ ⎜
⎞ ⎠ ⎟
W2→ 3 =−1.178 ×104 J K by the gas
What is the magnitude and direction
of the heat flow in the step "2 → 3 "?
Q2→ 3 =−W2→ 3 =1.178 ×104 J K into the gas
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What is the magnitude and direction
of the change in thermal energy in the
step "2 → 3 "?
ΔU2→3 = 0
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What is the magnitude and direction
of the heat flow in the step "3 → "i ?Q3→ i =nCPΔT3→ i =nCP Ti −T3( )
Q3→ i = 4.5moles( ) 29.18 Jmole• K( ) 401K −936K( )
Q3→ i =−7.025 ×104 J K out of the gas
What is the magnitude and direction
of the work done in the step "3 → "i ?W3→ i =−P3ΔV3→ i =−P3 Vi −V3( )
W3→ i =−2.5×105 Pa .06m3 −.14m3( )
W3→ i =2.0 ×104 J K on the gas
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What is the magnitude and direction
of the change in thermal energy in the
step "3 → "i ?ΔU3→i = Q3→i + W3→i
ΔU3→i = −7.025 × 104 J + 2.0 × 104 J
ΔU3→i = −5.025 × 104 JK decrease
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3.5×105
2.5 ×105i
1 2
3
isochoric
isobaric
isothermal
isobaric
Pressure, Pa
Volume ,m30.06 0.10 0.14
Qi→ 1 =1.509×104 J
Q1→ 2 =4.911×104 J
W1→ 2 =−1.4×104 J
Q2→ 3 =1.178×104 JW2→ 3 =−1.178×104 J
Q3→ i =−7.025×104 J W3→ i =2.0×104 J
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Summary
Net for Cycle
Step Process Work-W J
Heat Flow-Q J
Change in Thermal
Energy-Δ UJ
i→1 Isochoric
1→2 Isobaric
2→3 Isothermal
3→ i Isobaric
0 1.509x104J 1.509x104J≈0
-1.400x104J 4.911x104J 3.511x104J
-1.178x104J 1.178x104J 0
2.000x104J -7.025x104J -5.025x104J
-5.780x103J 5.730x103J -.05x103J ≈0
When an ideal gas is carried through a complete cycle returning it to its initial state the net change in thermal energy is zero.
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What is the efficiency of this heat engine?
net work out = 5.78 ×103 J
total heat in = 1.509×104 J +4.911×104 J +1.178×104 Jtotal heat in = Q i→ 1 +Q1→ 2 +Q2→ 3
total heat in = 7.598 ×104 J
efficiency =η= net work out total heat in×100%
η= 5.78 × 103J7.598 × 104 J
× 100%
η=7.6%
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Why is the molar specific heat at constant pressure, CP, always greater than the molar specific heat at constant volume, CV?
Recall: The specific heat of a substance is the amount of heat required to change the temperature of 1kg of the substance by 1C°,(or 1K). The equation relating heat flow and change in temperature is: Q=mcΔT, or Q=nCΔT. From the second equation an expression for molar specific heat can be obtained:
C = QnΔT
CP = QnΔTP
CV = QnΔTV
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Suppose we take two containers each holding 1 mole of the same gas.
We will take the first contained gas through an isobaric expansion process where heat Q=1000J flows into the gas and the gas does 200J of work.
We will take the second contained gas through an isochoric expansion process where heat Q=1000J flows into the gas and the gas does no work.
For an ideal gas the thermal energy, U only depends on the temperature, T.
Let’s say that for every 100J change in thermal energy there is a corresponding 1C° change in the temperature.
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Isobaric
Qin =1000J
Wout =200J
ΔU = 800J
ΔT = 8K
CP = QnΔTP
CP = 1000J1mole⋅8K
CP =125 Jmole⋅K
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Isochoric
Qin =1000J ΔU = 1000J
ΔT = 10K
CV = QnΔTV
CV = 1000J1mole⋅10K
CV =100 Jmole⋅K
CP >CV
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0
25
50
75
100
125
150
175
200
0 25 50 75 100 125 150 175 200
Cp, J/mol · K
Cv, J/mol · KCp versus Cv
Cv =Cp −8.31
Cv =Cp −R
Gas Cp, J/mol · K Cv, J/mol · K
Air 29.11 20.80Argon 20.78 12.47Butane 99.76 91.45Carbon Dioxide 37.23 28.91Carbon Monoxide 29.13 20.84Ethane 53.11 44.80Ethylene 43.43 35.12Helium 20.79 12.47Hydrogen 28.84 20.53Methane 36.16 27.84Neon 20.79 12.47Nitrogen 29.11 20.81Octane 195.48 187.17Oxygen 29.38 21.06Propane 74.06 65.74Steam 33.73 25.42