Add a slide just before efficiency that explains why Qlow exhaust is necessary.
-
Upload
giana-wickwire -
Category
Documents
-
view
219 -
download
3
Transcript of Add a slide just before efficiency that explains why Qlow exhaust is necessary.
Ch15 ThermodynamicsZeroth Law of ThermodynamicsIf two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium with each other.
First Law of ThermodynamicsThe Internal Energy of a closed system will be equal to the energy added to the system by heating minus the work done by the system to its surrounding
Second Law of ThermodynamicsHeat flows out from hot objects to cold; heat does NOT flow from cold to hot
Thermal Energy
• Thermal Energy: The total internal Energy
• Internal Energy: The sum of the kinetic and potential energies of the internal motion of particles that make up an object.
Internal Energy• The sum of all the energy of all the molecules
in an object (thermal energy)
• Internal Energy of an Ideal Gas
3 32 2U NKT nRT 3
2avgKE kT
BrassBrass
Ch15 Thermodynamics
• Heat – Transfer of energy due to ΔT
• Work – Transfer of energy NOT due to ΔT
• Q – Heat
• W – Work– W done on the system is negative (Giancoli)– W done by the system is positive (AP WS)
• ΔU – Change in energy
Ch15 ThermodynamicsFirst Law of Thermodynamics
U Q W
Heat added is +
Heat lost is -
Work on system is –
Work by system is +
Ch15 ThermodynamicsThe distinction between work done on the gas and work done by the gas is one your equation sheet and book define differently
The area under the P-V curve will always be the work done by the gas during the process
First Law of Thermodynamics
2500J of heat is added to a system, and 1800J of work is done on the system. What is the changed in internal energy of the system?
U Q W
(Q) 2500J of heat will increase the Internal Energy
(W) 1800J of work done ON the system will …
Is the work positive or negative? Why?
2500 ( 1800 )U J J 4300U J
Did the temperature increase or decrease?
Ch15 Thermodynamics• Isothermal process: Constant temperature
– The system is in contact with a heat reservoir– Change of phase
32 0U nR T
U Q W Q W
• The work done by the gas in an isothermal process equals the heat added to the gas
Isothermal
0
1
2
3
4
5
0 1 2 3 4 5
Volume
Pre
ssu
re
Isothermal process:
Constant temperature, i.e. PV is constant
Which Isothermal process is at a higher Temperature?
Which Isothermal process does more work?
Isothermal
0
1
2
3
4
5
0 1 2 3 4 5
Volume
Pre
ssu
re
Adiabatic
• Adiabatic Process: No heat in or out of the system– Well insulated (like a thermos)– The process happens very quickly (firing of a car
cylinder)
U Q W U W
Adiabatic Isothermal
0
1
2
3
4
5
6
7
8
9
10
0 1 2 3 4 5 6 7 8 9 10
Volume
Pre
ssu
re
WorkGiven the following two processes: Isothermal and Adiabatic.
Both processes start at 10Pa and end with a volume of 10m3
During which process is more work done?
Estimate the work done in each process.
3 36.5 (10 3 )P V Pa m m
Adiabatic Isothermal
0
1
2
3
4
5
6
7
8
9
10
0 1 2 3 4 5 6 7 8 9 10
Volume
Pre
ssu
re45.5P V J 3 35.5 (10 3 )P V Pa m m
38.5P V J
W P V
Isovolumetric
• Isovolumetric: (Isochoric) No change in volume– Inside a ridged container
W P V (0)W P
0W
Isovolumetric
0
1
2
3
4
5
6
7
8
9
10
0 1 2 3 4 5 6 7 8 9 10
Volume
Pre
ssu
re
Isobaric• Isobaric: No change in pressure
– Movable pistonW P V
Isobaric
0
1
2
3
4
5
6
7
8
9
10
0 1 2 3 4 5 6 7 8 9 10
Volume
Pre
ssu
re
Internal Energy ΔU• 1 mole of an ideal gas is brought from point a to point c
by 3 different process paths. Which path has the highest change in internal energy?
• 1)• 2)• 3)• 4) All the same• 5) Unknown
10 20 30 40 50 60
100
400
300
200
a
b c
dPre
ssur
e (P
a)
Volume (m3)
abcUadcUacU
32U nRT
PV nRT32U PV
332 400 (60 )U Pa m
Work (W)• 1 mole of an ideal gas is brought from point a to point c
by 3 different process paths. During which path did the gas do the most work?
• 1)• 2)• 3)• 4) All the same• 5) Unknown
10 20 30 40 50 60
100
400
300
200
a
b c
dPre
ssur
e (P
a)
Volume (m3)
abcW
adcW
acW
W P V 400(50)abcW
100(50)adcW 250(50)acW
abcW
Heat (Q)• 1 mole of an ideal gas is brought from point a to point c
by 3 different process paths. During which path was the most heat added?
• 1)• 2)• 3)• 4) All the same• 5) Unknown
10 20 30 40 50 60
100
400
300
200
a
b c
dPre
ssur
e (P
a)
Volume (m3)
abcQ
adcQ
acQ
U Q W Q U W
400(50)abcQ U 100(50)adcQ U
250(50)acQ U
abcQ
• One mole of monatomic ideal gas is enclosed under a frictionless piston. A series of processes occur, and eventually the state of the gas returns to its initial state with a P-V diagram as shown below. Answer the following in terms of P0, V0, and R.
• Find the temperature at each vertex.
• Find the change in internal energy for each process.
• Find the work done by the gas for each process.
o o APV nRT o oA
PVT
nR
4 o oB
PVT
nR
32U nR T 3
2
4( )A B
PoVo PoVoU nR
nR nR
0(4 )AB AB o oW P V P V V
2Volume m3
10
• An ideal gas is slowly compress at constant pressure (2.0 ATM) from 10.0L to 2.0L• Heat is then added to the gas holding the volume constant and the pressure and
temperate are allowed to rise until the temperature reaches its original value.
a) Calculate the total work done by the gasb) Calculate the total heat flow into the gas
Pre
ssur
e P
a
In an engine 0.25 moles of an ideal gas in the cylinder expands rapidly against the piston. In this process, the temperature of the gas drops from 1150K to 400K.
a)What type of process is this?b) How much work does the gas do?
U Q W
U W 3
2U nR T
32 (0.25 )(8.314 (400 1150 )J
mol KU moles K K
2300J
2300W J
Is the work done by the gas positive or negative?
Efficiency
• Efficiency (e): the ratio of work W done by the system to the input heat QH
H
WeQ
H LQ W Q
1 1H L L L
H H H H
Q Q Q TWeQ Q Q T
H
WeQ
H
WQ
e
An automobile engine has an efficiency of 20% and produces an average of 23,000J of mechanical work per second. a) How much input heat is required?b) How much heat is discharged as wasted per second?
23,000
0.20H
JQ
51.15 10HQ x J
1L
H
Qe
Q
(1 ) .8(115 )L HQ e Q kJ
92LQ kJ
a) b)
When a gas is taken from a to c along the curved path shown, the work done by the gas is W = -35 J and the heat added to the gas is Q = -62 J. Along path abc, the work done is W = -51J.
a) What is Q for path abc? b) If Pc = 1/2 Pb, what is W for path cda?
c) What is Q for path cda? d) What is Ua - Uc? e) If Ud - Uc = 4 J, what is Q for path da?
When a gas is taken from a to c along the curved path in shown, the work done by the gas is W = -35 J and the heat added to the gas is Q = -62 J. Along path abc, the work done is W = -51J.
a) What is Q for path abc? -78Jb) If Pc = 1/2 Pb, what is W for path cda? 25.5Jc) What is Q for path cda?
52.5Jd) What is Ua - Uc? 27Je) If Ud - Uc = 4 J, what is Q for path da? 23J