Transcript Using the “Clicker” - Boston University: Physics

Constant volume (isochoric) process
No work is done by the gas: W = 0. The P-V diagram is a
vertical line, going up if heat is added, and going down if heat
is removed.
Applying the first law:
Q  Eint
For a monatomic ideal gas:
Eint
3
 nRT
2
Q  Eint
3
 nR T
2
Constant pressure (isobaric) process
In this case the region on the P-V diagram is rectangular, so
its area is easy to find. W  P V
For a monatomic ideal gas:
Q  Eint  W
3
nR T  P V
2
3
 nR T  nR T
2
5
 nR T
2

Heat capacity
For solids and liquids: Q  mc T
For gases: Q  nCT , where C, the heat capacity,
depends on the process.
For a monatomic ideal gas
Constant volume:
3
3 
Q  nCV T  n  R  T , so CV  R
2
2 
5
5 
Constant pressure: Q  nCP T  n  R  T , so CP  R
2
2 
In general:
CP  CV  R
and
Eint  nCV T
Constant temperature (isothermal) process
No change in internal energy: Eint  0
The P-V diagram follows the isotherm.
Applying the first law, and
using a little calculus:
 Vf 
Q  W  nRT ln  
 Vi 
Zero heat (adiabatic) process
Q = 0. The P-V diagram is an interesting line, given by:
PV   constant
CP
 
CV
For a monatomic ideal gas:
CP 5R / 2 5
 


CV 3R / 2 3
Applying the first law:
Eint  W
Worksheet
A thermodynamic system undergoes a three-step process. An
adiabatic expansion takes it from state 1 to state 2; heat is
added at constant pressure to move the system to state 3;
and an isothermal compression returns the system to state 1.
The system consists of a diatomic ideal gas with CV = 5R/2.
The number of moles is chosen so nR = 100 J/K.
The following information is known about states 2 and 3.
Pressure: P2 = P3 = 100 kPa
Volume: V3 = 0.5 m3
What is the temperature
of the system in state 3?
Apply the ideal gas law
P3V3 (100  103 Pa)  0.5 m3
T3 

 500 K
nR
100 J/K
The system does 20000 J of work in the constant pressure
process that takes it from state 2 to state 3. What is the
volume and temperature of the system in state 2?
The temperature in state 2
What is the temperature of the system in state 2?
1.
2.
3.
4.
5.
200 K
300 K
500 K
700 K
None of the above
Finding work
The system does 20000 J of work in the constant pressure
process that takes it from state 2 to state 3. What is the
volume and temperature of the system in state 2?
For constant pressure, we can use: W  P V  nR T
Finding volume: V  W  20000 J  0.2 m3
P 100000 Pa
V  V3  V2 so V2  V3  V  0.5 m3  0.2 m3  0.3 m3
Finding temperature: (use the ideal gas law, or …)
W
20000 J
T 

 200K
nR
100 J/K
T  T3  T2 so T2  T3  T  500K  200K  300K
Complete the table
For the same system, complete the table. The total work done
by the system in the cycle is –19400 J.
Process
Q
ΔEint
W
1 to 2
2 to 3
+20000 J
3 to 1
Entire
cycle
-19400 J
First fill in all the terms that are zero.
Each row satisfies the First Law of Thermodynamics.
Also remember that Eint  nCV T
Complete the table
For the same system, complete the table. The total work done
by the system in the cycle is –19400 J.
Process
Q
1 to 2
0
ΔEint
2 to 3
W
+20000 J
3 to 1
0
Entire
cycle
0
-19400 J
Q is zero for an adiabatic process.
The change in internal energy is zero for an isothermal
process, and is always zero for a complete cycle.
Complete the table
For the same system, complete the table. The total work done
by the system in the cycle is –19400 J.
Process
Q
1 to 2
0
ΔEint
2 to 3
+20000 J
3 to 1
Entire
cycle
W
0
-19400 J
0
-19400 J
Even the last row has to satisfy the first law:
Q  Eint  W
Complete the table
For the same system, complete the table. The total work done
by the system in the cycle is –19400 J.
Process
Q
1 to 2
0
2 to 3
W
+50000 J +20000 J
3 to 1
Entire
cycle
ΔEint
0
-19400 J
0
-19400 J
Find the change in internal energy for the 2  3 process.
Eint
5
5
 nCV T  nR T  (100 J/K)( 200K)  50000 J
2
2
Complete the table
For the same system, complete the table. The total work done
by the system in the cycle is –19400 J.
Process
Q
ΔEint
1 to 2
0
-50000 J
W
2 to 3
+70000 J +50000 J +20000 J
3 to 1
0
Entire
cycle
-19400 J
0
-19400 J
Rows have to obey the first law.
Columns have to sum to the value for the entire cycle.
Complete the table
For the same system, complete the table. The total work done
by the system in the cycle is –19400 J.
Process
Q
1 to 2
0
ΔEint
W
-50000 J +50000 J
2 to 3
+70000 J +50000 J +20000 J
3 to 1
-89400 J
0
-89400 J
Entire
cycle
-19400 J
0
-19400 J
Rows have to obey the first law.
Columns have to sum to the value for the entire cycle.
A heat engine
A heat engine is a device that uses heat to do work. A
gasoline-powered car engine is a good example.
To be useful, the engine must go through cycles, with work
being done every cycle. Two temperatures are required. The
higher temperature causes the system to expand, doing work,
and the lower temperature re-sets the engine so another
cycle can begin. In a full cycle, three things happen:
Heat QH is added at a relatively high temperature TH.
Some of this energy is used to do work W.
The rest is removed as heat QL at a lower temperature TL.
For the cycle: QH = W + QL (all positive quantities)
Efficiency
In general, efficiency is the ratio of the work done divided by
the heat needed to do the work.
e
Q
W QH  QL

 1 L
QH
QH
QH
The net work done in one cycle
is the area enclosed by the cycle
on the P-V diagram.
Carnot’s principle
Sadi Carnot (1796 – 1832), a French engineer, discovered an
interesting result that is a consequence of the Second Law of
Thermodynamics.
Even in an ideal situation, the efficiency of a heat engine is
limited by the temperatures between which the engine
operates. 100% efficiency is not possible, and most engines,
even in ideal cases, achieve much less than 100% efficiency.
Carnot’s principle:
QL TL

QH TH
Ideal (Carnot) efficiency:
TL
eC  1 
TH
A refrigerator
If you had a refrigerator in a closed, well-insulated room and
you left the fridge door open for a long time, what would
happen to the temperature in the room?
1. It would increase
2. It would decrease
3. It would stay the same
Heat engines running backwards
Refrigerators and air conditioners
are heat engines that run backward.
Work is done on the system to pump
some heat QL from a low
temperature region TL. An amount of
heat QH = QL + W is then removed
from the system at a higher
temperature TH.
(a) Represents the cylinder in a car
engine; while (b) represents a
refrigerator.
A heat pump
If you heat your home using electric heat, 1000 J of
electrical energy can be transformed into 1000 J of heat.
An alternate way of heating is to use a heat pump, which
extracts heat from a lower-temperature region (outside the
house) and transfers it to the higher-temperature region
(inside the house). Let's say the work done in the process
is 1000 J, and the temperatures are TH = 27°C = 300 K and
TL = -13 °C = 260 K. What is the maximum amount of heat
that can be transferred into the house?
1. Something less than 1000 J
2. 1000 J
3. Something more than 1000 J
A heat pump
The best we can do is determined by the Carnot relationship.
QL TL
TL


QL 
 QH
QH TH
TH
Using this in the energy equation gives:
TL
QH  QL  W 
 QH  W
TH

THQH  TLQH  THW
For our numerical example this gives:
QH 
THW
(300 K)(1000 J)

 7500 J
TH  TL
300 K  260 K
This is why heat pumps are much better than electric heaters.
Instead of 1000 J of work going to 1000 J of heat we have
1000 J of work producing 7500 J of heat.
Entropy
Entropy is in some sense a measure of disorder.
The symbol for entropy is S, and the units are J/K.
A container of ideal gas has an entropy value, just as it has a
pressure, a volume, and a temperature. Unlike P, V, and T,
which are quite easy to measure, the entropy of a system is
difficult to calculate.
On the other, a change in entropy is easy to determine.
Change in entropy
Entropy changes whenever there is a transfer of heat. The
change in entropy is the heat added divided by the
temperature at which the transfer took place.
If the heat transfer takes place at a single temperature, the
change in entropy is simply:
Q
isothermal process: S 
T
If the heat transfer takes place over a range of temperatures
then, as long as ΔT is small compared to the absolute
temperature T, the change in entropy is approximately:
Q
S 
Tavg
The Second Law of Thermodynamics
The entropy of a closed system is constant for reversible
processes and increases for irreversible processes. Entropy
never decreases (for a closed system).
For a closed system, S  0
Dropping a glass
You drop a glass of milk and the glass smashes into 27 pieces
and milk spills all over the floor. If you videotaped this and
ran the film backwards, it would be obvious to you that
the film was running backwards. Why?
The process violates:
1. The Law of Conservation of Energy
2. The Law of Conservation of Momentum
3. The Second Law of Thermodynamics
4. All of the above
Entropy: time’s arrow
In the process of smashing the glass of milk, both energy and
momentum are conserved. However, the entropy is increased
The direction of time is the direction of increasing entropy.
Reversible and irreversible processes
In an irreversible process, the entropy of a closed system
increases.
In a reversible process, the entropy stays the same.
Reversible or not?
You have two styrofoam containers of water. Each holds 1 kg
of water. In one the water temperature is 17°C, while in the
other it is 37°C. The colder water is then poured into the
warmer water, and the system is allowed to come to
equilibrium.
Is this process reversible or irreversible?
1. Reversible
2. Irreversible
Irreversible!
The container of water will not spontaneously separate into
two parts that differ in temperature by 20°, so this process is
irreversible. Let’s calculate the change in entropy.
Find the heat: Q  mc T  1 kg (4186
J
)(10C)  41860 J
kg C
Use the change in entropy equation, using an average
temperature of 22°C = 295 K for the cooler water and
32°C = 305 K for the warmer water.
Qcool Qwarm 41860 J 41860 J
S 



 4.65 J/K
Tcool Twarm
295 K
305 K
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