# Work & Energy

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```					                          Thermodynamics
I. Temperature and Temperature Scales
A. Definitions
1. Thermodynamics is the study of thermal energy and thermal motion.
2. Temperature is a direct measure of the average KE of atoms or
molecules in a system.
3. Two or more objects are said to be in thermal contact if they have the
ability to exchange heat with one another. This does not necessarily
require physical contact!
4. Two or more systems in thermal contact are said to be in thermal
equilibrium when they achieve the same temperature. Energy
exchange between the two systems is constant so no heat is exchanged.
5. Heat is energy that is exchanged between two systems in thermal
contact due exclusively to a temperature difference between the two
systems. Note that heat and temperature are NOT the same things!!!
B. Zeroth Law of Thermodynamics
1. If object A is in thermal equilibrium with object C, and if object B is in
thermal equilibrium with object C, then object A is in thermal
equilibrium with object B.
2. Objects A, B, and C are all at the same temperature
3. If they are not at the same temperature, then hot objects cool down and
cool objects warm up until an equilibrium temperature between the
extremes is achieved.
C. Temperature Scales
1. Fahrenheit Scale
a. water freezes at 32 °F at atmospheric pressure
b. water boils at 212 °F at atmospheric pressure
c. scale is based upon human body temperature of 100 °F, but the
calibration was too high by 1.4 °F
2. Celsius Scale
a. water freezes at 0 °C at atmospheric pressure
b. water boils at 100 °C at atmospheric pressure
c. scale is based upon properties of water
3. Kelvin Scale
a. water freezes at 273.15 K at atmospheric pressure
b. water boils at 373.15 K at atmospheric pressure
c. absolute zero = 0 K is basis for Kelvin scale
d. same size unit as Celsius degree but offset by 273.15 K
4. Conversions
a. TF  9 TC  32  F
5

b. TK  TC  273 .15 K
c. Celsius and Kelvin have same size unit; Fahrenheit unit is 9
5

the size of the Celsius and Kelvin unit.
d. Temperature only doubles with a doubling of absolute
temperature (e.g., 100 °C ≠ 2 * 50 °C, 100 °F ≠ 2 * 50 °F, but
100 K = 2 * 50 K). Therefore, absolute temperatures MUST be
used in most thermodynamic calculations unless a temperature
difference is specified.
5. Example 19.1
D. Thermometers
1. Thermometers as measuring devices are based upon the principle that
some physical property of a system is a function only of temperature.
a. volume of a liquid
b. length of a solid
c. gas pressure at constant volume
d. gas volume at constant pressure
e. electric resistance of a conductor
f. color of an object
2. mercury and alcohol thermometers are based upon a fluid height at a
given temperature
a. because of differences in expansion properties, temperatures
may be slightly inaccurate far from calibration points
b. only a limited range of temperature can be covered before the
liquid changes state at high or low temperatures
3. gas thermometers
a. good agreement with gases as pressure is reduced, but vacuums
are tough to approach
b. must stay above condensation point or gas turns to liquid
c. when lines on pressure-temperature graphs are extended, they
all intersect at –273.15 °C = 0 K = absolute zero
E. Definition of Absolute Zero
1. Absolute zero is the absence of all motion of individual particles and
the KE of the system is zero
2. zero-point energy must exist (Heisenberg Uncertainty Principle) so 0 K
can never be truly attained. It is a theoretical limit.

II. Heat and Internal Energy
A. Definitions
1. The internal energy of a system is all of the energy associated with its
atoms and molecules when viewed from a reference frame at rest with
respect to the system
2. Heat is the transfer of energy to or from a system and its surroundings
due to a temperature difference between them
3. Units of Heat
a. A calorie is the amount of energy required to raise the
temperature of 1 gram of water by 1 °C. The common Calorie
(capital C) listed in food products is really 1000 calories
(lowercase c) = 1 kilocalorie.
b. A British Thermal Unit (BTU) is the amount of energy
required to raise the temperature of 1 pound of water by 1 °F.
This is a VERY large unit!
c. A Joule is the SI unit of energy. 1 calorie = 4.186 Joules
B. Work, Heat, Energy
1. Work is the change in KE due to external forces acting on the system
as a whole.
2. Heat is the amount of energy transferred to or from a system due to
temperature differences.
3. Internal energy is defined by the total average energy of all atoms and
molecules in the system. Internal energy is directly related to
temperature.
4. work, heat, and internal energy all have the same units, but represent
three fundamentally different, but related physical concepts
C. Mechanical Equivalent of Heat
1. Mechanical energy lost as friction is stored as internal energy in
another system or as heat
2. Joule showed that the loss in mechanical energy is proportional to the
change in the temperature of water, and the constant of proportionality
is 4.186 J/cal.
D. Heat Capacity and Specific Heat
1. When heat is added to a system or work is done, the temperature of the
system usually rises (a phase transition is the only exception)
2. the temperature rise varies with the ability of a substance to absorb or
emit heat
3. the heat capacity of a substance is the energy per unit mass per unit
temperature required to raise 1 gram of a substance by 1 °C
4. the specific heat (capacity) of substance is heat capacity per unit mass
5. Q = mcpT
Q = heat gained or lost from the substance
m = mass of the substance
cp = specific heat capacity at constant pressure
T = change in temperature of the substance
6. specific heats for various substances are given in Table 19.3
7. specific heat at constant volume (cv) differs by a few percent from cp .
These differences will be discussed later.
8. specific heat actually varies as a function of temperature but we will
not be concerned with that at this level
9. water has a very high specific heat leading coastal areas to maintain
more moderate temperatures than inland areas
10. cheese releases more heat than pizza crust at the same temperature so
hot cheese is more likely to burn you than hot pizza crust at the same
temperature
E. Calorimetry
1. measures specific heat
2. sample at Tx > Twater is placed in known mass of water
3. little mechanical work is done and little energy is lost to environment,
so most energy is converted to internal energy of the water
4. Derivation
Qwater = -Qx
At equilibrium temperature Tf
m w c w (T f  Tw )   m x c x (T f  Tx )
m w c w (T f  Tw )
cx 
m x (Tx  T f )
5. Example 19.3 in class
F. Latent Heat
1. Phase changes involve energy transfer but no change in temperature
a. solid  liquid (melting, freezing)
b. liquid gas (vaporizing, condensing)
c. solid  gas (sublimating, condensing)
2. Latent heat is the energy required to change the phase of a substance
a. Lf = latent heat of fusion (melting)
b. Lv = latent heat of vaporization
c. Ls = latent heat of sublimation (rarely used)
d. Table 19.4 latent heats for various substances
3. Q = mL represents heat energy required to change state. Therefore L =
Q/m
4. Example: What is the energy required to convert 1 gram of ice at –30
°C to steam at 120 °C. The specific heats of ice, water, and steam are
respectively ci = 2090 J/kg °C, cw = 4186 J/kg °C, cs = 2010 J/kg °C.
The latent heats of fusion and vaporization are respectively Lf = 3.33
x 105 J/kg and Lv = 2.26 x 106 J/kg.
a. Qtot = Qi + Qf + Qw + Qv + Qs
b. Qi = mciTi
Qi = (1.0 x 10-3 kg)(2090 J/kg °C)(0 °C-(-30 °C))
Qi = 62.7 J (raises ice to melting point)
c. Qf = mLf
Qf = (1.0 x 10-3 kg)(3.33 x 105 J/kg)
Qf = 333 J (melts ice to water)
d. Qw = mcwTw
Qw = (1.0 x 10-3 kg)(4186 J/kg °C)(100 °C - 0 °C)
Qw = 419 J (raises water to boiling point)
e. Qv = mLv
Qv = (1.0 x 10-3 kg)(2.26 x 106 J/kg)
Qv = 2260 J (vaporizes water to steam)
f. Qs = mcsTs
Qs = (1.0 x 10-3 kg)(2010 J/kg °C)(120 °C - 100 °C)
Qs = 40.2 J (raises steam to 120 °C)
g. Qtot = 62.7 J + 333 J + 419 J + 2260 J + 40.2 J
Qtot = 3114.9 J
G. Example 19.4 in class
H. Note that melting/freezing and boiling/condensation points depend upon
pressure and composition
1. higher elevations are at lower pressure, so the boiling point of water is
lower
2. salt + water lowers freezing point so roads are salted in cold weather
III. Work in Thermodynamic Processes
A. Equation of State
1. Pressure
2. Internal energy
3. Volume
4. Temperature
5. Density
B. Equation of state relates two or more of these variables when system is in
thermal equilibrium
C. Work on a System
1. quasi-static (slow changes) are required to maintain thermodynamic
equilibrium
2. dW = F dx = PA dx = P dV
3. if system expands, dV > 0 and dW > 0; if system contracts, dV < 0 and
dW < 0
4. work done BY gas is negative; work done ON gas is positive
5. + work = transfer of energy out of the system; - work = transfer of
energy into the system. This is the opposite of mechanics due to an
unfortunate historical convention.
Vf

6. dW = P dV so W      P(V)dV
Vi

7. Work is area under P-V curve and it is
path dependent!
8. Consider several different paths as shown
below. Case (b) represents the greatest amount of work, case (c)
represents the least amount of work, and case (a) represents an
intermediate situation.

(a)                          (b)                         (c)

9. Work done by/on a system depends upon
a. initial state (i)
b. final state (f)
c. path followed from i to f
10. Energy transfer as heat is also path dependent

IV. First Law of Thermodynamics
A. Two methods for exchanging heat with Surroundings
1. Work is mechanical intervention
2. Heat is energy transfer controlled by temperature differences
B. Path Dependence
1. Work and heat depend upon the path taken from the initial to the final
state
2. Difference between work and heat (Q – W) is independent of path!
3. Eint = Q – W is the First Law of Thermodynamics (conservation of
energy), and it depends only upon the initial and final states of the
system. The path is unimportant!
4. Q > 0 when energy enters the system and W > 0 when system does
work. Typically, if Q > 0, then W > 0 unless outside forces are added.
Note that Q and W are not properties of the system; they are simply
agents that affect the system.
C. Special Cases of the First Law of Thermodynamics
1. dEint = dQ – dW (state function)
2. isolated system
a. no energy transfer occurs
b. Q = W = 0
c. Eint = Q – W
d. the internal energy of an isolated system is constant
3. cyclic processes
a. begin and end process at the same state
b. Eint = Q – W = 0
c. Q = W
d. In a cyclic process, the net work done by the system per cycle
equals the area enclosed by the path on a P-V diagram
4. no work done (see isovolumetric processes)
a. W = 0
b. Eint = Q
c. If energy enters the system, Eint increases; if energy leaves the
system, then Eint decreases
5. Adiabatic processes (no heat is exchanged)
a. Q = 0 because no heat is exchanged
b. can achieve adiabatic behavior if system is either well-insulated
or if change occurs too rapidly for heat transfer
c. Eint = - W
1. If system expands, W > 0 and Eint decreases
2. If system is compressed, W < 0 and Eint increases
d. adiabatic free expansion
1. insulated so Q = 0
2. expansion into vacuum so W = 0
3. Eint = 0 and no temperature change occurs
6. Isobaric processes (pressure is constant)
a. constant pressure
b. neither Q nor W equals zero
c. W = P V = P(Vf – Vi)
7. Isovolumetric processes (volume is constant)
a. constant volume implies W = 0 because V = 0
b. Eint = Q
  c. all changes in the system occur as heat transfer
d. this is a specific case of a “no work” process
e. can of spray paint tossed into a fire
1. energy enters system from fire
2. P and T rise, but V = 0
3. eventually P exceeds yield strength of can and can
explodes
8. Isothermal processes (constant temperature)
a. P vs. V at constant T yields a curve called an isotherm
b. Eint = Eint(T) for ideal gases so Eint = 0 in an isothermal
process
c. Q = W is similar to a cyclic process
D. Example 19.5

V. Energy Transfer mechanisms (3 types)
A. Conduction
1. two systems at different temperatures T1 >
T2 are in direct physical contact
2. energy is transferred when atoms of T1
collide with atoms of T2 increasing the
energy in T2 and decreasing the energy in
T1
3. heat transfer ends when T1 = T2 = Tequilibrium
Q        T       dT
4.      A       A
t       x       dx
Q
P  power 
t
dT
P  kA
dx
5. k = thermal conductivity in Watts/m °C
dT
dx
6. good thermal conductors (metals) have large k
7. good thermal insulators (non-metals) have small k (home insulation)
B. Convection
1. heat energy is transferred by the
movement of a heated substance
2. two systems not necessarily in
direct contact will come to an
equilibrium temperature if the
medium between them is convecting
3. examples of natural convection
a. heated air around a fire
b. mixing of cool and warm air at the beach
c. water evaporated by sunlight
4. examples of forced convection
a. fans cool air
b. heat pumps
c. boiling water on stove
1. energy released as E-M waves
2. does not require direct physical
contact between objects (Sun and
Earth)
3. Stefan’s law: P  AT 4
a.  = 5.6696 x 10-8 W/m2 K4 (Stefan-Boltzmann constant)
b.  = emissivity (0 ≤  ≤ 1) equals the fraction of incoming
radiation that is absorbed
c. T = temperature
4. Isolar = 1340 W/m2 from sun on average
5. At night, Earth re-radiates most (99.98%) of energy it absorbs from the
Sun during the day
6. Pnet   A (T  T0 ) where T is the object’s temperature and T0 is the
4     4

temperature of the surroundings
7.  = 1 is an ideal absorber and emitter (also called a blackbody)
8.  = 0 is an ideal reflector
1. one example is a Thermos bottle
2. all Dewar flasks minimize all three types of heat transfer
E. Examples 19.6 and 19.7

VI. Second Law of Thermodynamics
A. Heat always flows from hot to cold
B. Mechanical energy cannot be completely converted to other useful forms of
energy without some loss of heat
C. Some processes are irreversible which means that the probability of a process
occurring is vastly greater in one direction than the reverse direction
1. oscillating pendulum slows down due to collisions with air molecules;
they never speed up
2. breaking a triangle of pool balls is easy, but striking a ball in such a
way as to restore the triangle after the break is virtually impossible
3. death is irreversible
4. time flows in the forward direction only
D. All closed (isolated) systems proceed from order to disorder. If a part of the
system becomes more ordered, it does so at the expense of greater disorder to
the system as a whole
1. easier to make a mess than to clean it up
2. easier to grow older than younger
3. life on Earth becomes more ordered (humans vs. bacteria) but only
because Earth receives a tremendous energy input from the sun
4. entropy is a measure of disorder
a. entropy is a state function and is therefore independent of
thermodynamic path
b. entropy of the universe increases in all real processes
c. in reversible processes, entropy always increases
d. in reversible processes (extremely rare but not impossible) the
entropy remains zero
e. entropy of a non-closed system can be decreased, but only at
the expense of a greater entropy in the combined system
dQ            Q
f. dS        S 
T             T
E. Examples of Entropy Change
1. Thermal Conduction (T2 > T1)
a. –Qlost = Qgained = Q
Q       Q
b. S 2   lost        (entropy decreases)
T2     T2
Qgained  Q
S2                (entropyincreases)
T1     T1
c. S net  S1  S 2
Q Q             1 1
 S            Q  
T T 
T1 T2             1  2 

d. Define S = 0 before heat transfer begins
 S  S net  0  S net
1 1
S net  Q  
T T 
 1   2 

1  1
Since T2  T1       S net  0
T2 T1
 S  0
2. Free Expansion (T = constant)
f         f
dQ       dQ 1
a. dS                dQ
T     i
T T i
b. If T = constant then dQ = dW = P dV
k
c. Suppose P         (k  0)
V
f          f
1           k dV
dS   PdV  
T i         T i V
k Vf
dS    ln( )
T    Vi
d. Since k > 0, T > 0, and Vf > Vi in expansion, dS > 0
3. Calorimetric Processes (T2 > Tf > T1)
dQ
a. dS 
T
b. Qgained = -Qlost
c. Q = mc T
d. Q gained  m1c1 (T f  T1 )
Qlost  m2 c 2 (T f  T2 )
m1c1 (T f  T1 )   m2 c 2 (T f  T2 )
(m1c1T1  m 2 c 2T2 )
Tf 
(m1c1  m 2 c 2 )
e. Snet = S1 +S2
f. Sinitial = 0 so S = Snet – 0 = Snet
Tf        Tf
dQ           m1c1 dT           Tf
g. S1                        m1c1 ln( )
T1
T         T1
T               T1
Tf       T
dQ
f
m c dT               Tf
S2          2 2         m2 c 2 ln( )
T2
T    T2
T                  T2
Tf             Tf
h. Snet  m1c1 ln( )  m2 c2 ln( )
T1             T2
if T1 = T2 = Tf then Snet = 0
if Tf  T1 then m2c2 0 and Snet > 0
Tf
if Tf T2 then m1c1 0 but ln( )  0 and Snet > 0
T2
4. Examples 21.1 and 21.2

VII. Heat Engines
A. Heat engines convert internal energy to mechanical energy
1. Power plants produce electricity by burning coal, nuclear fuel, etc.,
which heats water to steam. Steam turns a turbine that drives an
electric generator.
2. General operation of a heat engine
a. energy is absorbed from a high-temperature reservoir
b. work is done by the engine
c. energy is expelled by the engine into a low-temperature
reservoir
d. the process is cyclic so Eint = 0
e. Eint = Q – W = 0
W = Qnet
W = Qh – Qc        (Qh > 0, Qc > 0)
3. Net work done in a cyclic process is the
area enclosed by the process on a P-V
diagram
4. Efficiency = e
W
e
Q
Q  Qc
e h
Qh
Qc
e  1
Qh
5. Efficiency of real engines is always < 100%
6. 2nd Law statements
a. cannot get more energy out of a cyclic process than the amount
of energy put into the process
b. cannot break even because Qc = 0 in that case, and the machine
would be a perpetual motion machine. Furthermore the cold
reservoir would not remain at 0 K after that.
B. Refrigerators and Heat Pumps
1. refrigerators and heat pumps are heat engines running in reverse
2. energy is absorbed from Qc and deposited to Qh
3. requires work to be done on the engine because heat flows from hot to
cold according to the 2nd law of thermodynamics
4. since no engine is perfect (100%) efficiency, it takes more work to
remove heat than the amount of heat that is removed.
5. cooling a room with a refrigerator does not work because the
compressor expels more heat than the refrigerator can remove
C. Carnot Engine
1. heat engine that operates through a theoretical, ideal, reversible cycle
(Carnot cycle) is theoretically the most efficient engine possible
2. represents theoretical upper limit on efficiency, because real engines
do not operate through a reversible cycle and also suffer from friction
and other energy losses
Q
3. e  1  c
Qh
Qc = mc(Tc- 0 K) = mcTc (heat of cold reservoir relative to 0 K)
Qh = mc(Th- 0 K) = mcTh (heat of hot reservoir relative to 0 K)
Qc Tc

Qh Th
Tc
e  1
Th
4. Note that e = 100% at Tc = 0 K; however, the addition of energy to a
reservoir initially at 0 K raises its temperature instantaneously and
therefore lowers the efficiency of the engine.
5. Since 0 K is neither attainable nor sustainable, and because Carnot
engines represent the theoretical upper limit of efficiency, no real
engines can ever attain efficiencies of 100%
6. Examples 21.3 and 21.4
D. Coefficients of Performance
energy from high temp Qh
1. COP (heating mode)                                 
work done by pump         W
a. pumping heat into hot reservoir requires external work
b. Qh is generally greater than W so COP > 1
c. higher COP is desirable because it means more heat is
transferred for a given amount of work
d. at 25 °F, COP ~ 4
e. for temperatures below ~ 15 °F COP < 1 and use of heat pumps
for heating houses is inappropriate
Qh     Qh       Th
f. COP (heating mode) =                 
W    Q h Q c Th  Tc
energy removed low temp Qc
2. COP (cooling mode)                                   
work done by pump          W
a.   pumping heat out of cold reservoir requires external work
b.   higher COP is desirable because it means more heat is removed
for a given amount of work
Q        Qc        Tc
c.   COP (cooling mode) = c                  
W      Q h Q c Th  Tc
d.   As Th  Tc  0, COP   , but practically COP < 10. This
arises because it is more difficult to oppose the second law of
thermodynamics with greater temperature differences

VIII. Ideal Gases at the Macroscopic Level
A. Assumptions
1. T must not be too high or too low
2. P must be relatively low
3. molecular volume << volume of container
B. Equation of State
total mass       M
1. # moles = n =                   
molecular mass m
2. PV = nRT = NkbT
3. # molecules = N; NA = Avogadro’s # = 6.02 x 1023 molecules/mol
4. kb = Boltzmann’s constant = 1.38 x 10-23 J/K
5. R = gas constant = 8.315 J/mol K = 0.08214 L atm/mol K
C. Ideal gases always have PV  cnst at all pressures
nT

D. P, V, T are thermodynamic variables
E. Examples 20.1 and 20.2
F. Special Cases of Ideal Gas behavior
1. Isothermal expansion
Vf       Vf
nRT
a. W   PdV            dV
Vi       Vi
V
Vf
b. W  nRT ln(      )
Vi
2. Quasi-static and Reversible Process
a. cv = specific heat capacity at constant volume
b. dQ = dEint + P dV
c. dEint = ncv dT
nRT
d. dQ  ncv dT         dV
V
dQ        dT      dV
e. dS         ncv      nR
T         T        V
Sf       f                f
dT       dV
 dS   ncv T   nR V
Si     i         i

Tf                Vf
S  ncv ln(           )  nR ln(        )
Ti                Vi

IX. Ideal Gases at the Microscopic Level
A. Kinetic Theory of Gases
1. individual molecules are considered to be hard solid spheres
2. spheres move randomly
3. spheres collide elastically with each other and container walls
4. spheres do not deform
5. works for monatomic gases and must be modified for diatomic,
triatomic, and other molecules
B. Microscopic Model
1. # of molecules is large
2. volume of molecules << container volume
3. molecules obey Newton’s laws of motion
4. molecules move in random directions with a distribution of speeds
5. all collisions are elastic
6. forces between molecules are negligible
7. molecules interact with each other only during collisions
8. all molecules are identical
9. molecular rotations and vibrations do not affect the average motions of
particles
C. Derivation of pressure on Container Walls
1. consider the collision of 1 molecule moving
initially in the +x direction with the container
wall
2.  p x  p f  pi  mv x  ( mv x )  2mv x
 p x  F1 t  2mv x         (F1  force on molecule )
3. t = time between collisions with the same wall
2d
t 
vx
 2mv x  2mv x
4. F1            
t      2d 
 v 
    x

 mv x 2
F1 
d
5. F1w   F1 (Newton' s 3 rd Law)
2
mv x
F1w 
d
m 2
6. Fnet   Fiw          (v1x  v 2 x    vix )
2           2

i            d
1 2
7. v x2   (v1x  v2 x    v Nx )
2          2

N
m
8. Fnet  ( Nv x2 )
d
9. v  v x  v y  v z
2    2     2     2

v x2  v y2  v z2 if motion is random
v 2  v x2  v y2  v z2  3v x2
m             Nm v 2
10. Fnet     ( Nv x2 )       ( )
d               d 3
F      F
11. P   2
A d
1 Nm v 2           Nmv 2
P 2           ( )
d d 3                3d 3
Nmv 2
P
3V
2N 1
P          ( mv 2 )
3V 2
12. Pressure is proportional to the number density of molecules and the
average KE of the molecules
13. You can increase the pressure in a container by either increasing the #
of molecules or by increasing their average KE
D. Ideal Gas law
1. Standard Temperature and Pressure (STP) are 0 °C and 1 atm
2 N 1                    1
2. P  ( )( mv 2 )  N ( mv 2 )
3 V 2                    3
1
3. PV  Nk B T  N ( mv 2 )
3
1
k B T  mv 2
3
2 1
T         ( mv 2 )
3k B 2
4. T is therefore a direct measure of average molecular KE
1           3
5.     mv 2  k B T
2           2
1
6. v x2  v 2
3
1           1          1         1
mv x2  k BT  mv y2  mv z2
2           2          2         2
7. Each direction is a single degree of freedom and contains some of the
molecular energy
8. Equipartition of Energy Theorem: Each degree of freedom
contributes 1 k B T to the energy of the system
2
9. Etrans  N ( 1 mv 2 )  3 Nk B T  3 nRT
2          2          2

3k B T     3RT
10. v rms  v 2            
m          M
1
11. v rms 
m
E. Molar Specific Heat of an Ideal Gas
1. Energy required to raise Ti to Tf depends upon the path because Eint =
Q – W is independent of path, but Q and W alone are not
2. Molar Specific Heats
a. Qv  ncv T (constant volume)
b. Q p  nc p T (constant pressure)
c. c p  c v because at constant pressure, both internal energy AND
work are increased; in constant volume processes, no work is
done
3. Monatomic Ideal Gas
a. Eint  3 Nk B T  3 nRT
2           2

b. at constant volume, dV = 0 so dW = PdV = 0
Eint = Qv – W = Qv = ncv T
c. dEint = ncv dT
Eint = ncv T
d. Eint  3 nRT
2

dEint 3
 2 nR  ncv
dT
c v  3 R (monatomic gases only)
2
e. At constant pressure
Eint = Qp – W = Qp = ncp T - PV
Eint = ncv T
PV = nR T
ncv T = ncp T - nR T
cv = cp - nR for all ideal gases
f. monatomic gases
1. cv  3 R2

cv  c p  R
cp  5 R
2

cp
2. Define          (ratio of specific heats)
cv
5R
  2  5  1.667
3R     3
2
3.  = 1.667 agrees very well with experimental results for
monatomic gases, but disagrees with experiment for
more complex gases
g. complex gases
1. add a degree of freedom for the rotation and vibration
of each atom in a molecule
2. each degree of freedom contributes 1 k B T to the total
2

internal energy of the system
3. diatomic gas example
a. a second molecule adds two degrees of freedom,
so Eint  3 k B T  2 ( 1 k B T )  5 k B T
2             2            2

b. cv  5 R
2

cv  c p  R
cp  7 R
2

7R
c.       2  7  1.4
5R       5
2
4. for more than 2 atoms per molecule, different vibration
patterns make the results more complex
h. important note for the future is that energy is quantized in steps
of 1 k B T which forbids non-integer steps in ideal gases
2

i. Brief Aside for Solids
1. if an atom in a solid is slightly displaced, it can rotate
and vibrate in x, y, z, but it cannot translate (it is stuck
in place)
2. rotation and vibration provide 6 additional degrees of
freedom so Eint  0  6( 1 k B T )  3k B T
2

3.    cv  3R is the Dulong-Petit Law for solids
4. at high temperatures, all solids appear to obey the
Dulong-Petit Law and cv  3R  25 J/mol K
4. Examples 20.7 and 20.8
F. Adiabatic Processes for an Ideal Gas
1. adiabatic requires that Q = 0
2. Eint = Q – W = -W and dEint = -dW
3. examples include the gasoline engine and slow expansion of
thermally-insulated gas
4. PV = nRT so d(PV) = PdV + VdP = nRdT
5. Derivation
dEint  dW
ncv dT   PdV
PdV  VdP  nRdT
 PdV
dT 
ncv
 PdV      R
PdV  VdP  nR(          )   PdV
ncv       cv
R  c p  cv
cv  c p
PdV  VdP  (               ) PdV  (1   ) PdV
cv
PdV  VdP  (1   ) PdV
Divide by PV
dV dP                 dV
      (1   )
V      P             V
dV dP
           0
V      P
 ln V  ln P  ln( cnst )
PV   cnst
6. PiVi  Pf V f   (adiabatic)
nRT
7. Substitute P 
V
TiVi 1  T f V f 1
8. Example 20.9
G. Gasoline and Diesel Engines
1. Otto cycle
a. intake stroke (O to A):
piston moves down
drawing in air and fuel
mixture at atmospheric
Pressure

pressure. Volume
increases from V1 to V2.
Energy is carried into the
system by mass transfer
(fuel) and is represented by
Volume

Qh.
b. compression stroke (A to B): system is compressed
adiabatically from V2 to V1; temperature rises from TA to TB;
work done is negative.
c. spark plug fires (B to C): pressure and temperature rise, but
volume remains constant; no work is done
d. power stroke (C to D): gas expands adiabatically from V1 to V2;
positive work is done on the piston
e. exhaust valve opens (D to A): energy is released and pressure
suddenly drops; no work is done
f. exhaust stroke (A to O): piston moves upward while exhaust
valve remains open and volume decreases from V2 to V1.
2. Efficiency of engine if air-fuel mixture is ideal gas
V             1
a. e  1  ( 1 )  1  1 
V2          V1  1
V2

V2
b.     is called the compression ratio
V1
c. ideal gas engines attain e = 56%
d. real engines attain e = 15% - 20%
3. Diesel engines operate on the Otto cycle but do not use a spark plug.
Instead, the compressed gas is itself at a temperature high enough for
self-ignition
H. Entropy on a Microscopic Scale
1. Gas molecules move randomly so there is an infinitesimally tiny
chance that they may all end up in an ordered state on one side of the
container. This is just one possible state.
V
2. # states for a single molecule is wi  i where Vi is the container
Vm
volume and Vm is the volume of a single molecule
3. For N molecules, the number of possible states is W*  w N
V                  Vf
4. W*i  ( i ) N and W* f  ( ) N
Vm                 Vm
W* f      Vf
5.        ( )N
W*i      Vi
W* f          Vf            Vf
ln(      )  N ln( )  nN A ln( )
W*i           Vi            Vi
W* f                      Vf                Vf
k B ln(          )  k B nN A ln(        )  nR ln(        )
W*i               Vi          Vi
6. Entropy for free expansion of a gas
Vf
S  nR ln( )
Vi
W* f
S  k B ln(    )
W*i
7. Entropy is a measure of microscopic disorder.
8. Entropy is described by statistical mechanics. The probability of
finding all molecules in 1 mol of a gas in one corner of a container is
1
P(W* )       1.6  10  23
NA
9. Examples 21.5 and 21.6

X. Maxwell-Boltzmann Distribution
A. Molecules in a gas do not all move at the same speed.
B. Distribution of molecular speeds depends upon the molecular mass and the
temperature of the gas and is called the Maxwell-Boltzmann Distribution
1. If N = # molecules in a gas, then # molecules with speeds between v
and v + dv is dN = Nvdv
mv2

m      3
2. N v  4N (           ) 2 v 2 e 2 k BT
2 k B T
3. v mp  v  v rms where vmp is the most probable speed
3k B T
4. v rms            root mean square speed
m
8k B T
v               average speed
m
2k B T
v mp          most probable speed
m
5. Function produces a bell-shaped curve with most molecules near vmp,
but with outliers moving extremely fast or extremely slow
6. The difference in speeds allows some molecules to escape Earth, fewer
to escape Jupiter, and more to escape the Moon. This accounts in
large part for the differences in atmospheric compositions between
planets of various sizes.
7. Example 20.3
C. Boltzmann Distribution
1. Probability of finding molecules in a particular energy state depends
E

upon e   k BT
where E is the energy of the state
E

2. nV ( E )  n0 e    k BT

3. Examples 20.5 and 20.6
D. Mean Free Path
1. The mean free path of an atom or
molecule is the average distance
between collisions with other gas
particles
2. In time t, a particle of diameter d
sweeps out a cylinder whose length is v t
vt        1
3. The # of collisions in time t is therefore l                       where nv
    2        nv
d v tnv
4
is the number density of gas molecules in the container and  is the
cross-sectional area of the molecule.
1 v
4. The collision frequency is f    nv v
t l
1      1
5. The mean free time is t mf  
f n v v
6. actual calculations including the motion of molecules in the cylinder
shows that more exactly
1                         v
l           and f  2nv v 
2nv                       l
7. Example 20.4
E. Exponential Atmospheres
1. PV  Nk BT (assumes atmosphere is
isothermal)
N
2. nv 
V
3. P  nv k BT
4. Consider air of thickness dy
weight  mgn vV  mgn v Ady
F    y    0  ( P  dP) A  nv A dy mg  PA
dP   nv mg dy
P  nv k B T
dP  k B T dnv
k B T dnv   nv mg dy
dnv    mg
      dy
nv     k BT
mgy                        E
                        
nv ( y )  n0 e           k BT
 n0 e        k BT

mgy

P  P0 e        k BT

where P0 = n0kBT
5. Law of Atmospheres
mgy                  y
                    
a. nv ( y)  n0e                        k BT
 n0e         H

mgy                y
                  
b. P  P0 e                      k BT
e         H

k B T thermal energy
c. H  scale height                                         
mg        weight

XI. Thermal Expansion of Solids and Liquids
A. Thermometers are based upon the principles that:
1. as T increases, V increases
2. as T decreases, V decreases
3. water is a weird exception because as T falls toward 4 °C, water
contracts. Below 4 °C, water expands and density decreases. As a
result, ice floats on water and water freezes from the top down instead
of the bottom up.
B. Engineering Applications
1. Metal and concrete expand when heated, so joints are used on roads,
bridges, and railroad tracks to allow expansion when heated by the
sun.
2. Plastic containers enclosing liquids will expand or contract with
temperature differences, and the material must withstand those
stresses (e.g., water bottle partially filled sits in sun all day, and then
when placed in refrigerator the plastic collapses
C. Linear Expansion of Solids
1. Assume expansion is small relative to original size.
L
linear strain        Li
2.                             cnst if T is small
changein temp        T
3.  L   Li T  L f  Li
4.  has units of °C –1 or K-1
5. Table 19.2 lists coefficients of linear expansion
6. Note that if  is negative, the substance contracts as temperature
increases
D. Volume Expansion of Solids and Liquids
1. linear expansion in 3 dimensions so define  = 3
2. derivation
Vi  V  (l   l )(w   w)(h   h)
Vi  V  (l  l  T )(w  w  T )(h  h  T )
Vi  V  lwh (1    T ) 3
Vi  V  Vi (1    T ) 3
Vi  V  V i [1  3  T  3(  T ) 2  (  T ) 3 ]
V  V i [3  T  3(  T ) 2  (  T ) 3 ]
If   T  1 then high orders vanish rapidly
V  3  T V i    T V i
3. You can also show that A  2  T V i for area expansion
4. Example 19.2
E. Bimetallic strips
1. Different metals have different coefficients
of expansion.
2. Place 2 long strips of different metals
together at room temperature
3. raise temperature and metal with higher 
bends more than metal with lower ; so
bimetallic strip bends into an arc with higher  on the outside.
4. lower temperature cause strip to bend into an arc with lower  on the
outside
5. thermostats make use of this principle

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