Transcript Chapter 12: Laws of Thermodynamics Work in Thermodynamics Processes

Chapter 12: Laws of Thermodynamics
Suggested homework assignment: 4,21,33,39,48
Work in Thermodynamics Processes
 Work
done on a gas
• Energy can be transferred to a system by
heat and by work done on the system.
• In this chapter, all the systems we are
concerned with are volumes of gas and
they are in thermodynamic equilibrium:
every part of the gas is at the same
temperature and pressure.
• Consider a gas contained by a cylinder
with a movable piston and in equilibrium.
The gas occupies a volume V and exerts
a uniform pressure P on the cylinder wall.
Work in Thermodynamics Processes
 Work
done on a gas (cont’d)
• The gas is compressed slowly enough so
that the system remains essentially in
thermodynamic equilibrium.
• As the piston is pushed downward by an
external force F through a distance Dy, the
work done on gas is:
W   FDy   PADy
DV  ADy
W  PDV
under constant pressure
• If the gas is compressed, DV is negative and the work done on the
gas is positive. If the gas expands, the work done on the gas is
negative.
Work in Thermodynamics Processes
 PV
diagram
• PV diagram
W   PDV   P(V f  Vi )
 (area under the graph)
• In general, the area under the graph
in PV diagram is equal in magnitude
to the work done on the gas, whether
or not the process proceeds at
constant pressure. Note that Pa x m3
= J.
isobaric process
First Law of Thermodynamics
 First
law of thermodynamics
• The first law of thermodynamics is another energy conservation law
that relates changes in internal energy – the energy associated with
the position and movement of all the molecules of a system – to
energy transfers due to heat and work.
• Two mechanisms to transfer energy between a system and its
environment : a macroscopic displacement of an object by a force
and by heat which occurs through random molecular collisions.
Both mechanisms result in a change in internal energy DU.
First law of thermodynamics
DU  U f  U i  Q  W
Ui : initial internal energy
Uf : final internal energy
Q : energy transferred to the system by heat
W : work done on the system
First Law of Thermodynamics
 First
law of thermodynamics (cont’d)
• The internal energy of any isolated system must remain constant.
 Molar
specific heat at constant volume
• Molar specific heat at constant volume Cv is the amount of
heat Q needed to change the temperature by DT per mole
at constant volume and is defined as:
 Q 

  nCV n : number of mole
 DT V
 Internal
energy of a monatomic ideal gas
• We learned that the internal energy of a monatomic ideal gas is:
U
3
nRT
2
• Therefore a change in U is proportional to a change in T:
DU 
3
nRDT
2
First Law of Thermodynamics
 Molar
specific heat
• From the first law of thermodynamics :
DU  U f  U i  Q  W
• At constant volume, W=0. Therefore the first law becomes:
DU  Q  nCV DT
• However, for an ideal gas:
3
DU  nRDT
2
Therefore, whether its volume is constant or not,
3
CV  R
2
First Law of Thermodynamics
 Molar
specific heat (cont’d)
• A gas with a larger specific heat requires more energy to realize
a given temperature change. The size of molar specific heat
depends on the structure of the gas molecule and how many ways
it can store energy.
 A monatomic gas (He etc.) can store energy as motion in three
different directions (3-dimension). Degrees of freedom = 3
 A diatomic gas (H2 etc.) can store energy as motion in three
different directions (3-dimension), and can tumble and rotate
in two different directions.
Degrees of freedom = 5
 Other molecules can store energy as vibrations, which add more
degrees of freedom.
Each degree of freedom contribute to the molar specific heat
by (1/2)R.
So, for example, the molar specific heat of a diatomic gas such
as oxygen O2 is : 5 x (1/2)R = (5/2)R.
First Law of Thermodynamics
 Molar
specific heat (cont’d)
• See Table of molar specific heats
 Isobaric
process
• An isobaric process is a process
where the pressure remains
constant.
W  PDV
• Isobaric process and first law:
DU  Q  W
DU  Q  PDV
Q  DU  PDV
First Law of Thermodynamics
 Isobaric
process (cont’d)
• Ideal gas in isobaric process
Q  DU  PDV
For a monatomic ideal gas :
DU 
3
nRDT
2
For an isobaric process :
PV  nRT  PDV  nRDT
3
5
nRDT  nRDT  nRDT
2
2
 Q 
Define molar specific heat at constant pressure as:  DT   nC p

p
For an ideal gas in an isobaric process:
Q
Q  nC p DT 
5
5
nRDT  C p  R
2
2
C p  CV  R
This is valid for all ideal gasses
First Law of Thermodynamics
 Adiabatic
process
• An adiabatic process is a process where no energy enters or leaves
the system by heat – the system is thermally isolated.
For an adiabatic process,
Q0
First law of thermodynamics
DU  W
• It can also be shown that for an ideal gas in an adiabatic process :

PV  constant w here  
Cp
CV
First Law of Thermodynamics
 Isovolumetric
process
• An isovolumetric process, also called an isochoric process, is a
process where the volume is constant.
• Since the volume does not change, there is no work done.
DU  Q
For an ideal gas :
DU  Q  nCV DT
First Law of Thermodynamics
 Isothermal
process
• An isothermal process is a process where the temperature is constant.
• For an ideal gas, since the internal energy depends only on the
temperature:
DU  0
First law of thermodynamics
W  Q
• It can be shown that the work
done on an ideal gas is given by:
Vf
W  nRT ln 
 Vi



First Law of Thermodynamics
 Examples
• Example 12.4 : Expanding gas
Suppose a system of monatomic ideal gas at 2.00x105 Pa and an
initial temperature of 293 K slowly expands at constant pressure from
a volume of 1.00 L to 2.50 L.
(a) Find the work done on the environment.
Wenv  PDV  (2.00 105 Pa)(2.50 10-3 m 3  1.00 10 3 m 3 )  3.00 10 2 J
(b) Find the change in the internal energy of the gas.
2.50 10-3 m3

 T f  Ti
 (239 K)
 733 K
3
3
PiVi
Ti
Vi
1.00 10 m
Pf V f
Tf
Vf
PiVi (2.00 105 Pa)(1.00 10-3 m3 )
n

 8.2110 2 mol
RTi
(8.31 J/(K mol)(293 K)
3
DU  nCV DT  nRDT  4.50 10 2 J
2
First Law of Thermodynamics
 Examples
• Example 12.4 : Expanding gas (cont’d)
(c) Use the first law to obtain the energy transferred by heat.
DU  Q  W  Q  DU  W
W  Wenv  3.00 10 2 J
Q  4.50 102 J  (3.00 102 J)  7.50 102 J
(d) Use the molar heat capacity at constant pressure to obtain Q.
5
Q  nC p DT  nRDT  7.50 10 2 J
2
(e) How would the answers change for a diatomic gas?
3 
DU  nCV DT    1nRDT  7.50 10 2 J
2 
5 
Q  nC p DT    1nRDT  1.05 103 J
2 
First Law of Thermodynamics
 Examples
• Example 12.5 : Work and engine cylinder
In a car engine operating at 1.80x103 rev/min, the expansion of hot,
high-pressure gas against a piston occurs in about 10 ms. because
energy transfer by heat typically takes a time on the order of minutes
or hours, it is safe to assume that little energy leaves the hot gas
during expansion. Estimate the work done by the gas on the piston
during this adiabatic expansion by assuming the engine cylinder
contains 0.100 moles of an ideal monatomic gas which goes from
1.20x103 K to 4.00x102 K typical engine temperatures, during the
expansion.
W  DU  Q  DU  0  DU
DU  U f  U i 
3
nR(T f  Ti )  9.97 10 2 J
2
Wpiston  W  DU  9.97 102 J
First Law of Thermodynamics
 Examples
• Example 12.6 : An adiabatic expansion
A monatomic ideal gas at a pressure
1.01x105 Pa expands adiabatically
from an initial volume of 1.50 m3,
doubling its volume.
(a) Find the new pressure.

Cp
CV

(5 / 2) R 5

(3 / 2) R 3
C  P1V1  1.99 105 Pa m5
C  P2V2  P2  3.19 104 Pa
(b) Estimate the work done on the
gas.
First Law of Thermodynamics
 Examples
• Example 12.7 : An isovolumetric process
How much thermal energy must be added to 5.00 moles of monatomic
ideal gas at 3.00x102 K and with a constant volume of 1.50 L in order
to raise the temperature of the gas by 3.80x102 K?
3
Q  DU  nCV DT  nRDT  4.99 103 J
2
• Example 12.8 : An isothermal expansion
A balloon contains 5.00 moles of monatomic ideal gas. As energy is
added to the system by heat, the volume increases by 25% at a
constant temperature of 27.0oC. Find the work Wenv done by the gas
in expanding the balloon, the thermal energy Q transferred to the gas,
and the work W done on the gas.
Wenv
 Vf
 nRT ln 
 Vi

  2.78 103 J  Q  W (note :V f  1.25Vi )

Second Law of Thermodynamics
 Heat
engines
• A heat engine takes in energy by heat and partially converts it to
other forms, such as electrical and mechanical energy.
• A heat engine, in general, carries some working substance through
a cyclic process during which :
(1) energy is transferred by heat from a source at a high temperature
(2) work is done by the engine
(3) energy is expelled by the engine by heat to a source at lower
temperature.
• A steam engine, the working substance is water. The water in the
engine is carried through a cycle in which it first evaporates into
steam in a boiler and then expands against a piston. After the steam
is condensed with cooling water, it returns to the boiler, and the
process is repeated.
Second Law of Thermodynamics
 Heat
engines (cont’d)
• The engine absorbs energy Qh from
the hot reservoir, does work Weng,
then gives up energy Qc to the cold
reservoir. Because the working substance goes through a cycle, always
returning to its initial thermodynamic
state – the initial and final internal
energy is the same, so DU=0.
DU  0  Q  W
Qnet  W  Weng
 Qh  Qc
Weng  Qh  Qc
The work Wenv done by a heat engine
equals the net energy absorbed by
the engine.
Second Law of Thermodynamics
 Heat
engines (cont’d)
• If the working substance is a gas,
the work done by the engine for a
cyclic process is the area enclosed
by the curve representing the process
on a PV diagram.
• The thermal efficiency of a heat
engine is defined by :
e
Weng
Qh

Qh  Qc
Qh
 1
Qc
Qh
Second Law of Thermodynamics
 Examples
• Example 12.10 : Efficiency of an engine
During one cycle, an engine extracts 2.00x103 J of energy from a
hot reservoir and transfers 1.50x103 J to a cold reservoir.
(a) Find the thermal efficiency of the engine.
e  1
Qc
Qh
 0.250 or 25.0%
(b) How much work does this engine do in one cycle?
Weng  Qh  Qc  5.00 102 J
(b) How much power does the engine generate if it goes through four
cycles in 2.50 s?
W 4.00  (5.00 10 2 J)
P

 8.00 10 2 W
Dt
2.50 s
Second Law of Thermodynamics
 Examples
• Example 12.11 : Analyzing an engine cycle
A heat engine contains an ideal gas
confined to a cylinder by a movable
piston. The gas starts at A where
T=3.00x102 K and B->C is an isothermal process.
(a) Find the number n of moles of
the gas and the temperature at B.
PAVA
n
 0.203 mol
RT A
TB 
PBVB
 9.00 10 2 K
nR
Second Law of Thermodynamics
 Examples
• Example 12.11 : Analyzing an engine cycle (cont’d)
(b) Find DU, Q, and W for isovolumetric process A->B.
3 
DU AB  nCV DT  n R DT  1.52 103 J
2 
DV  0  WAB  0
QAB  DU AB  1.52 103 J
Second Law of Thermodynamics
 Examples
• Example 12.11 : Analyzing an engine cycle (cont’d)
(c) Find DU, Q, and W for isothermal process B->C.
DU AB  nCV DT  0
 VC 
WBC  nRT ln    1.67 103 Pa
 VB 
DU BC  0  QBC  WBC
 QBC  WBC  1.67 103 J
Second Law of Thermodynamics
 Examples
• Example 12.11 : Analyzing an engine cycle (cont’d)
(d) Find DU, Q, and W for isobaric process C->A.
WCA   PDV  1.01103 J
(1 atm  1.01105 Pa)
DU CA
3
 nRT  1.52  103 J
2
QCA  DU CA  WCA  2.53 103 J
Second Law of Thermodynamics
 Examples
• Example 12.11 : Analyzing an engine cycle (cont’d)
(e) Find the net change in internal
energy DUnet
DU net  DU AB  DU BC  DU CA  0
(f) Find the energy input, Qh; the energy rejected, Qc; the thermal
efficiency; and the net work performed by the engine.
Qh  QAB  QBC  3.19 103 J
Qc  2.53 103 J
Qc
e  1
 0.207
Qh
Weng  (WAB  WBC  WCA )  6.60 102 J
Second Law of Thermodynamics
 Refrigeration
and heat pump
• A reversed heat engine is called refrigeration!
Energy is injected into the engine called
heat pump and that results in extraction
of energy from the cold reservoir to the
hot reservoir.
Examples are refrigerator and air
conditioner.
• Coefficient of performance
For a heat pump in the cooling mode
COP (cooling mode) 
Qc
W
For a heat pump in the heating mode
COP (heating mode) 
Qh
W
Second Law of Thermodynamics
 Example
12.12 : Cooling the leftovers
• 2.00 L of leftover soup at T=323 K is placed in a refrigerator. Assume
the specific heat of soup is the same as that of water and the density
1.25x103 kg/m3. The refrigerator cools the soup to 283 K.
(a) If the COP of the refrigerator is 5.00, find the energy needed, in the
form of work, to cool the soup.
m  V  2.50 kg
Qc  Q  mcDT  4.19 105 J
COP 
Qc
W
 5.00  W  8.38 10 4 J
(b) If the compressor has a power rating 0.250 hp find the time
needed to cool the food.
P  (0.250 hp )(746 W/1 hp)  187 W
W
Dt 
 448 s
P
Second Law of Thermodynamics
 Second
law of thermodynamics
• There are limits to the efficiency of heat engines.
• An ideal engine which would convert all the input energy into work
does not exist.
• The Kelvin-Planck formulation of the second law of thermodynamics:
No heat engine operating in a cycle can absorb energy from a
reservoir and use it entirely for the performance of an equal
amount of work.
• This means that the efficiency e=Weng/|Qh| of engines must always
be less than one. Some energy Qc must always be lost to the
environment.
• It is theoretically impossible to construct a heat engine with an
efficiency 100%.
We cannot get a greater amount of energy out of a cyclic process
that we put in.
Second Law of Thermodynamics
 Reversible
and irreversible processes
• No engine can operate with 100% efficiency, but different designs
yield different efficiencies.
• One design called Carnot cycle (engine) delivers the maximum
possible efficiency.
• A reversible process is a process in which every state along the same
path is an equilibrium state. In a reversible process, the system can
return to its initial condition (state) by going along the same path in
reverse direction.
• An irreversible process is a process which does not satisfy the
condition for a reversible process.
• Most natural processes are irreversible, but some of them are almost
reversible. If a real process occurs so slowly that the system is virtually
always in equilibrium, the process can be considered reversible.
Second Law of Thermodynamics
 Carnot
engine
• Consider a heat engine operating in an ideal, reversible cycle called
a Carnot cycle.
Second Law of Thermodynamics
 Carnot
engine (cont’d)
• Consider a heat engine operating in an ideal, reversible cycle called
a Carnot cycle.
A->B : Isothermal expansion at Th.
Qh from hot reservoir.
WAB done by gas.
B->C : Adiabatic expansion Th->Tc.
No heat goes out or comes in.
WBC done by gas.
C->D : Isothermal compression at Tc.
Qc to cold reservoir.
WCD done on gas.
D->A : Adiabatic compression Tc->Th.
No heat goes out or comes in.
WDA done on gas.
Second Law of Thermodynamics
 Carnot
engine (cont’d)
• Ratio of heat input to output vs. ratio of temperatures
Qc
Tc

Qh Th
• Thermal efficiency of a Carnot enegine
Qc
Tc
eC  1 
 1
Qh
Th
Th , Tc in K
All Carnot engines operating reversibly between the same two
temperatures have the same efficiency.
• All real engines operate irreversibly, due to friction and brevity of their
cycles, and are therefore less efficient that the Carnot engine.
Second Law of Thermodynamics
 Example
12.13 : Steam engine
• A steam engine has a boiler that operates at 5.00x102 K. The energy
from the boiler changes water to steam which drives the piston. The
temperature of the exhaust is that of the outside air, 300 K.
(a) What is the engine’s efficiency if it is an ideal engine?
Tc
eC  1   0.400 (40%)
Th
(b) If the 3.50x103 J of energy is supplied from the boiler, find the
work done by the engine on its environment.
Qc
Tc
Tc
  Qc  Qh
 2.10 103 J
Qh Th
Th
Weng  Qh  Qc  1.40 103 J
Entropy
 Definition
of entropy
• Let Qr be the energy absorbed or expelled during a reversible,
constant temperature process between two equilibrium states.
Then the change in entropy during any constant temperature
process connecting the two equilibrium states id defined by:
DS 
Qr
T
SI unit: joules/kelvin (J/K)
• A similar formula holds even when the temperature is not constant.
• Although calculation of DS during a transition between two equilibrium
states requires finding a reversible path that connects the states, the
entropy change calculated on that reversible path is taken to be DS
for the actual path. This is valid logic as the change in entropy DS
depends only on the initial and final states and not on the path taken.
• The entropy of the Universe increases in all natural processes.
Entropy
 Meaning
of entropy
• In nature a disorderly arrangement is much more probable than an
orderly one if the laws of nature are allowed act without interference.
• Using statistical mechanics it can be concluded that isolated systems
tend toward great disorder, and entropy is a measure of that disorder.
S  kB ln W
kB : Boltzman constant
W : a number proportional to the probability that
system has a particular configuration.
• The second law of thermodynamics is really a statement of what is
most probable rather than of what must be.
• The entropy of the Universe always increases
Entropy
 Examples
• Example 12.14 : Melting a piece of lead
(a) Find the change in entropy of 300 g of lead when it melts at 327oC.
Lead has a latent heat of fusion of 2.45x104 J/kg.
Q  mLf  7.35 10 J
3
T  TC  273  6.00 10 2 K
Q
DS   12.3 J/K
T
(b) Suppose the same amount of energy is used to melt part of a piece
of silver, which is already at its melting point of 961oC. Find the
change in the entropy of the sliver.
T  TC  273  1.23 10 2 K
Q
DS   5.96 J/K
T
Entropy
 Examples
• Example 12.15 : Ice, steam, and the entropy of the Universe
A block of ice at 273 K is put in thermal contact with a container
at 373 K, converting 25.0 g of ice to water at 273 K while condensing
some of the steam to water at 373 K.
(a) Find the change in entropy of the ice.
Q  mLf  8.33 103 J
DS ice
Qice

 30.5 J/K
Tice
(b) Find the change in entropy of the steam.
Thermal energy lost by the
Qsteam  Qice
DS steam 

 22.3 J/K steam is equal to the thermal
Tsteam Tsteam
energy gained by the ice.
(c) Find the change in entropy of the Universe.
DSUni  DSice  DS steam  8.2 J/K
Entropy
 Examples
• Example 12.16 : A falling boulder
A chunk of rock of mass 1.00x103 kg at 293 K falls from a cliff of
height 125 m into a large lake, also at 293 K. Find the change in
entropy of the lake, assuming that all of the rock’s kinetic energy
upon entering the lake converts to thermal energy absorbed by
the lake.
PE  mgh  1.23 106 J
The rock’s kinetic energy at the time
of entrance to the lake.
Q PE
DS  
 4.20 103 J/K
T
T
Human Metabolism
 Application
of thermodynamics to living organisms
• Animals do work and give off energy by heat, and this lead us to
believe the first law of thermodynamics can be applied to living
organisms.
• Let’s apply the first law in terms of the time rates of change of DU,
Q, and W.
On average, energy Q flows out of the body, and
DU Q W
work is done by the body on its surroundings:


DU/Dt is negative
Dt Dt Dt
Q/Dt and W/Dt are negative.
• Without supply of energy, the internal energy and the body
temperature would decrease. But in reality, all animals acquire
internal energy (chemical potential energy) by eating and breathing.
• Overall the energy from oxidation of food ultimately supplies the work
done by the body and energy lost from the body by heat. From this
point of view, DU/Dt is the rate at which internal energy is added to
our bodies by food, which balances the rate of energy loss by heat
and work.
Human Metabolism
 Measuring
the metabolic rate
• The metabolic rate DU/Dt is the rate at which chemical potential
energy in food and oxygen are transformed into internal energy to
balance the body losses of internal energy by work and heat.
• The metabolic rate DU/Dt is directly proportional to the rate of oxygen
consumption by volume.
For an average diet, the consumption of one
DVO2
DU
liter of oxygen releases 4.8 kcal or 20 kJ of
 4.8
Dt
Dt
energy.
L/s
Human Metabolism
 Metabolic
rate, activity, and weight gain
• Table below summarizes the measured rate of oxygen consumption
in mL/(min kg) and the calculated metabolic rate for 65-kg male
engaged in various activities.
Human Metabolism
 Physical
fitness and efficiency of the human body as a
machine
• One measure of a person’s physical fitness is his or her maximum
capacity to use or consume oxygen. Table below gives some idea
how well a person fit.
• The body’s efficiency e is defined as the
ratio of the mechanical power supplied by
a human to the metabolic rate:
W
Dt
e
DU
Dt
Human Metabolism
 Example
12.17 : Fighting fat
• In the course of 24 hours, a65-kg person spends 8 h at a desk
puttering around the house, 1h jogging 5 miles, 5 h in moderate
activity, and 8 h sleeping. What is the change in her internal energy
during this period?
DU   Pi Dti  (200 kcal/h)(10 h)  (5 mi/h)(120 kcal/mi)(1 h)
 (400 kcal/h)(5 h)  (70 kcal/h)(8 h)
 5000 kcal