Transcript Ch1 - Pat Arnott Web Site

Thermal Physics: 425/625
• Acknowledgement:
I acknowledge the excellent efforts by
Rutgers Physics Professors Misha
Gershenson and Weida Wu in preparing
powerpoint course material for this
textbook. The powerpoint presentations in
this course are rooted in their work, with
permission from Dr Wu, with modifications
by me.
Physics 425/625
Thermal Physics
Fall 2011
Prerequisites: Physics 182, 301.
Professor: Pat Arnott
Office:
Leifson Physics RM 213
e-mail:
[email protected]
phone:
775-784-6834
office hour: Wed 1:00-3:00 PM and by appointment
Class:
Tuesday and Thursday 11:00 am to 12:15 pm
Physics 425/625
Thermal Physics
Fall 2011
Textbook:
An Introduction to Thermal Physics,
D.V. Schroeder,
Addison-Wesley-Longman, 2000
Web Site Access:
http://www.patarnott.com/phys625
Homework assignments, announcements and
notes will be posted here. The link must be
visited regularly.
Thermal Physics = Thermodynamics + Statistical Mechanics
- conceptually, the most important, enjoyable subject of the
undergraduate physics program.
Thermodynamics provides a framework of relating the
macroscopic properties of a system to one another. It is
concerned only with macroscopic quantities and ignores the
microscopic variables that characterize individual molecules
(both strength and weakness).
Statistical Mechanics is the bridge between the microscopic
and macroscopic worlds: it links the laws of thermodynamics to
the statistical behavior of molecules.
Macroscopic Description is Qualitatively Different!
Why do we need to consider macroscopic bodies as a special
class of physical objects?
For a single particle: all equations of classical mechanics,
electromagnetism, and quantum mechanics are time-reversal
invariant (Newton’s second law, F = dp/dt, looks the same if
the time t is replaced by –t and the momentum p by –p).
For macroscopic objects: the processes are often irreversible (a
time-reversed version of such a process never seems to occur).
Examples: (a) living things grow old and die, but never get
younger, (b) if we drop a basketball onto a floor, it will bounce
several times and eventually come to rest - the arrow of
time does exist.
“More is different”, Phil Anderson, Science, 177, 393 (1972)
Ancients struggled with heat: What is it?
“It is astonishing to realize that many modern
conceptions (or “laws”) in the science of heat—
thermodynamics— arose during the nineteenth
century, a period of utter confusion about the
fundamental nature of heat. How could it have
been otherwise, given that the very existence of
atoms was still in question! “
http://www.infinite-energy.com/iemagazine/issue37/mysteries.html
The Main Idea of the Course
Statistical description
of a large system
of identical (mostly,
non-interacting) particles
Equation of state
for macrosystems
(how macroparameters of the
system and the temperature
are interrelated)
all microstates of an isolated
system occur with the same
probability, the concepts of
multiplicity (configuration
space), Entropy
Irreversibility of
macro processes,
the 2nd Law of Thermodynamics
Thermodynamic Systems, Macroscopic Parameters
Open systems can exchange both
matter and energy with the environment.
Closed systems exchange energy but
not matter with the environment.
Isolated systems
can exchange
neither energy nor matter with the
environment.
Internal
and
external
macroscopic
parameters:
temperature, volume, pressure, energy, electromagnetic fields,
etc. (average values, fluctuations are ignored).
No matter what is the initial state of an isolated system,
eventually it will reach the state of thermodynamic
equilibrium (no macroscopic processes, only microscopic
motion of molecules).
A very important macro-parameter:
Temperature
Temperature is a property associated with random motion of
many particles.
Introduction of the concept of temperature in thermodynamics is
based on the the zeroth law of thermodynamics:
A well-defined quantity called temperature exists such
that two systems will be in thermal equilibrium if both
have the same temperature.
Temperature Measurement
Properties of a thermoscope (any device that quantifies temperature):
1. It should be based on an easily measured macroscopic quantity a
(volume, resistance, etc.) of a common macroscopic system.
2. The function that relates the chosen parameter with temperature,
T = f(a), should be monotonic.
3. The quantity should be measurable over as wide a range of T as
possible.
The simplest case – linear dependence T = Aa (e.g., for the ideal gas
thermometer, T = PV/NkB).
the ideal gas thermometer, T = PV/NkB
T
the resistance thermometer with

a semi- conductor sensor, R  exp T
 
a
Thermometer  a thermoscope
calibrated to a standard temp. scale
The Absolute (Kelvin) Temperature Scale
The
absolute
(Kelvin)
temperature scale is based
on fixing T of the triple point
for water (a specific T =
273.16 K and P = 611.73 Pa
where water can coexist in
the solid, liquid, and gas
phases in equilibrium).
T,K
273.16
0
absolute zero
 P 

T  273.16 K 
 PTP 
PTP
P
- for an ideal gas
constant-volume
thermoscope
PTP – the pressure of the gas in a
constant-volume gas thermoscope
at T = 273.16 K
Our first model of a many-particle system: the Ideal Gas
Models of matter:
gas models
(random motion of particles)
lattice models (positions of particles are fixed)
Air at normal conditions:
~ 2.71019 molecules in 1 cm3 of air (Pr. 1.10)
Size of the molecules ~ (2-3)10-10 m, distance
between the molecules ~ 310-9 m
The average speed - 500 m/s
The mean free path - 10-7 m (0.1 micron)
The number of collisions in 1 second - 5 109
The ideal gas model - works well at low densities (diluted gases)
• all the molecules are identical, N is huge;
• the molecules are tiny compared to their average separation (point masses);
• the molecules do not interact with each other;
• the molecules obey Newton’s laws of motion, their motion is random;
• collisions between the molecules and the container walls are elastic.
The Equation of State of Ideal Gases
An equation of state - an equation that relates macroscopic
variables (e.g., P, V, and T) for a given substance in
thermodynamic equilibrium.
In equilibrium ( no macroscopic motion), just a few
macroscopic parameters are required to describe the state of
a system.
The ideal gas
equation of state:
P
V
n
T
–
–
–
–
pressure
volume
number of moles of gas
the temperature in Kelvins
R – a universal constant
PV  nRT
[Newtons/m2]
[m3]
[mol]
[K]
R  8.315
J
mol  K
The Ideal Gas Law
In terms of the total number of molecules, N = nNA
Avogadro’s number
NA  6.022045×1023
PV  NkBT
the Boltzmann constant kB = R/NA  1.3810-23 J/K
(introduced by Planck in 1899)
The equations of state cannot be derived within the frame of
thermodynamics: they can be either considered as experimental
observations, or “borrowed” from statistical mechanics.
Avogadro’s Law: equal volumes of different
gases at the same P and T contain the same
amount of molecules.
The P-V diagram – the projection of the surface
of the equation of state onto the P-V plane.
isotherms
Connection between KEtr and T for Ideal Gases
?
T of an ideal gas  the kinetic energy of molecules
Pressure – the result of collisions between the molecules and walls of the
container.
Momentum
Strategy: Pressure = Force/Area = [p / t ]/Area
px = 2 m vx
Intervals between collisions: t = 2 L/vx
For each (elastic) collision:
Piston
area A
vx
Volume = LA
L
N
For N molecules -
2mvx 1
1
2 1
Pi 
 p x v x  mvx
2 L / vx A
V
V
PV   mvx2  N m v x2
i
no-relativistic
motion
Connection between KEtr and T for Ideal Gases (cont.)
N
PV   mvx2  N m v x2
i
PV  NkBT
Average kinetic energy of
the translational motion of
molecules:
KEtr
3
 k BT
2
3
U  KEtr  Nk BT
2
2
PV  U
3
KEtr
m v x2  k BT
1
1
3
2
2
2
2
 m v  m v x  v y  v z  m v x2
2
2
2
- the temperature of a gas is a direct measure of the
average translational kinetic energy of its molecules!
The internal energy U of a monatomic ideal gas
is independent of its volume, and depends only on
T (U =0 for an isothermal process, T=const).
- for an ideal gas of non-relativistic particles,
kin. energy  (velocity)2 .
Comparison with Experiment
dU/dT(300K)
(J/K·mole)
3
U  Nk BT
2
Monatomic
Helium
12.5
Argon
12.5
Neon
12.7
Krypton
12.3
Diatomic
H2
20.4
N2
20.8
O2
21.1
CO
21
Polyatomic
H20
27.0
CO2
28.5
- for a point mass
with three degrees
of freedom
Testable prediction: if we put a known
dU into a sample of gas, and measure
the resulting change dT, we expect to
get
dU 3
 Nk B
dT 2
3
 6 10 23 mole -1 1.38 10  23 J/K
2
 12.5 J/K  mole



Conclusion: diatomic and polyatomic
gases can store thermal energy in forms
other than the translational kinetic
energy of the molecules.
Degrees of Freedom
The degrees of freedom of a system are a
collection of independent variables required to
characterize the system.
Polyatomic molecules: 6
(transl.+rotat.) degrees of freedom
Diatomic molecules: 3 + 2 = 5
transl.+rotat. degrees of freedom
Degrees of Freedom (cont.)
Plus all vibrational degrees of freedom. The one-dimensional vibrational
motion counts as two degrees of freedom (kin. + pot. energies):
K  U x  
1
1
m x 2  k x 2
2
2
For a diatomic molecule (e.g., H2), 5 transl.+rotat. degrees of freedom
plus 2 vibrational degrees of freedom = total 7 degrees of freedom
Among 7 degrees of freedom, only 3 (translational) degrees correspond to a
continuous energy spectrum (classical) , the other 4 – to a discrete energy
spectrum (Quantum).
4
U(x)
U(x) 3
E4
Energy
2
1
0
Approx.
-1
E3
E2
-2
E1
-3
1.5
2.0
2.5
3.0
x
distance
3.5
4.0
x
“Frozen”
degrees of
freedom
U /kBT
one mole of H2
7/2N
Vibration
5/2N
Rotation
For an ideal gas
3/2N
PV = NkBT
U = f/2 NkBT
Example of H2:
Translation
10
100
1000
T, K
An energy available to a H2 molecule colliding U(x)
E4
with a wall at T=300 K: 3/2kBT ~ 40meV. If the
E3
difference between energy levels is >> kBT,
E2
then a typical collision cannot cause transitions kBT
E1 x
to the higher (excited) states and thus cannot
transfer energy to this degree of freedom: it is
“frozen out”.
The rotational energy levels are ~15 meV apart, the difference between
vibrational energy levels ~270 meV. Thus, the rotational degrees start
contributing to U at T > 200 K, the vibrational degrees of freedom - at T >
3000 K.
Equipartition of Energy
“Quadratic” degree of freedom – the corresponding energy = f(x2, vx2)
[ translational motion, (classical) rotational and vibrational motion, etc. ]
Equipartition Theorem:
At temperature T, the average energy of any
“quadratic” degree of freedom is 1/2kBT.
- holds only for a system of particles whose kinetic energy is a quadratic
form of x2, vx2 (e.g., the equipartition theorem does not work for photons, E =
cp)
Piston – a mechanical system with one degree of
freedom. Thus,
vx
m v 2x
2

M u2
2

1
k BT
2
M – the mass of a piston, u2 the average u2,
where u is the piston’s speed along the x-axis.
Thus, the energy that corresponds to the one-dimensional translational
motion of a macroscopic system is the same as for a molecule (in this
respect, a macrosystem behaves as a giant “molecule”).
Questions to solve in class: Think about them beforehand.
a. What are the molecules in the air in this room? Give % estimates.
b. How many molecules are in the air in this room?
c. On average, how far apart are the molecules from each other?
d. What is the total internal energy of the air in this room?
Questions Continued
e. Assuming (bad assumption) that we could use all this energy to
run a hair dryer (1 kW), how long could we run the air dryer on
energy?
f. Assuming the walls, floor, and ceiling of the room are perfect
blackbodies, how much power is emitted by them all?
Questions Continued
g. Use the idea of the deBroglie wavelength and the average kinetic
energy to calculate the thermal deBroglie wavelength for the
molecules in this room as a function of temperature. How does the
thermal deBroglie wavelength compare with the inter particle spacing
at room temperature? As T goes to zero? Interpret.
http://en.wikipedia.org/wiki/Matter_wave
http://en.wikipedia.org/wiki/Thermal_de_Broglie_wavelength
The root-mean-square speed
vrms 
v
2
3k BT

m
- not quite the average speed, but close...
For H2 molecules (m ~21.710-27 kg ) at 300K: vrms~ 1.84 103 m/s
For N2 – vrms (Pr. 1.18), for O2 – vrms= 461 m/s
This speed is close to the speed of sound in the gas – the sound wave
propagates due to the thermal motion of molecules.
D(v)
vrms
v
Problem 1.16
The “exponential” atmosphere
Consider a horizontal slab of air whose thickness is dz. If this slab is at rest,
the pressure holding it up from below must balance both the pressure from
above and the weight of the slab. Use this fact to find an expression for the
variation of pressure with altitude, in terms of the density of air, . Assume
that the temperature of atmosphere is independent of height, the average
molecular mass m.
P(z+dz)A
P( z  dz )  A  Mg  P( z )  A
area A
z+dz
z
P(z)A
Mg
Mg
P( z  dz )  P( z )  
A
dP
M  Adz 
  g
dz
M Nm 
PV  Pm

 N 
the density of air:  

V
V
k BT  k BT

Assuming T is independent of z
dP
mg


P

dz
kT
 mgz 
P( z )  P(0) exp 
 kT 
The First Law of Thermodynamics (Ch.1)
Outline:
1. Internal Energy, Work, Heating
2. Energy Conservation – the First Law
3. Quasi-static processes
4. Enthalpy
5. Heat Capacity
Internal Energy
The internal energy of a system of particles, U, is the sum of the kinetic
energy in the reference frame in which the center of mass is at rest and the
potential energy arising from the forces of the particles on each other.
Difference between the total energy and the internal energy?
system
boundary
system
U = kinetic + potential
“environment”
B
P
T
A
V
The internal energy is a state function – it depends only on
the values of macroparameters (the state of a system), not
on the method of preparation of this state (the “path” in the
macroparameter space is irrelevant).
In equilibrium [ f (P,V,T)=0 ] :
U = U (V, T)
U depends on the kinetic energy of particles in a system and an average
inter-particle distance (~ V-1/3) – interactions.
For an ideal gas (no interactions) :
U = U (T) - “pure” kinetic
Internal Energy of an Ideal Gas
The internal energy of an ideal gas
with f degrees of freedom:
f
U  Nk BT
2
f  3 (monatomic), 5 (diatomic), 6 (polyatomic)
(here we consider only trans.+rotat. degrees of freedom, and neglect
the vibrational ones that can be excited at very high temperatures)
How does the internal energy of air in this (not-air-tight) room change
with T if the external P = const?

f
PV  f
U  N in roomk BT   N in room 
  PV
2
k BT  2

- does not change at all, an increase of the kinetic energy of individual
molecules with T is compensated by a decrease of their number.
Work and Heating (“Heat”)
We are often interested in U , not U. U is due to:
Q - energy flow between a system and its
environment due to T across a boundary and a finite
thermal conductivity of the boundary
WORK
HEATING
– heating (Q > 0) /cooling (Q < 0)
(there is no such physical quantity as “ heat ” ; to
emphasize this fact, it is better to use the term
“heating” rather than “heat”)
W - any other kind of energy transfer across
boundary
- work
Work and Heating are both defined to describe energy transfer
across a system boundary.
Heating/cooling processes:
conduction: the energy transfer by molecular contact – fast-moving
molecules transfer energy to slow-moving molecules by collisions;
convection: by macroscopic motion of gas or liquid
radiation:
by emission/absorption of electromagnetic radiation.
The First Law
The first law of thermodynamics: the internal energy of a system can be
changed by doing work on it or by heating/cooling it.
U = Q + W
conservation of energy.
Sign convention: we consider Q and W to be positive if energy
flows into the system.
P
For a cyclic process (Ui = Uf)  Q = - W.
If, in addition, Q = 0 then W = 0
T
V
An equivalent formulation:
Perpetual motion machines of the first type do not exist.
Quasi-Static Processes
Quasi-static (quasi-equilibrium) processes – sufficiently
slow processes, any intermediate state can be considered
as an equilibrium state (the macroparamers are welldefined for all intermediate states).
Advantage: the state of a system that participates in a quasi-equilibrium
process can be described with the same (small) number of macro
parameters as for a system in equilibrium (e.g., for an ideal gas in quasiequilibrium processes, this could be T and P). By contrast, for nonequilibrium processes (e.g. turbulent flow of gas), we need a huge number
of macro parameters.
Examples of quasiequilibrium processes:
isochoric:
isobaric:
isothermal:
adiabatic:
V = const
P = const
T = const
Q=0
For quasi-equilibrium processes, P, V, T are
well-defined – the “path” between two states
is a continuous lines in the P, V, T space.
P
T
2
1
V
Work
A – the
The work done by an external force on a gas
enclosed within a cylinder fitted with a piston:
piston
area
W = (PA) dx = P (Adx) = - PdV
force
x
P
The sign: if the volume is decreased, W is positive (by
compressing gas, we increase its internal energy); if the
volume is increased, W is negative (the gas decreases
its internal energy by doing some work on the
environment).
V2
W1 2    P(T , V )dV
V1
W = - PdV - applies to any
shape of system boundary
dU = Q – PdV
The work is not necessarily associated with the volume changes – e.g.,
in the Joule’s experiments on determining the “mechanical equivalent
of heat”, the system (water) was heated by stirring.
W and Q are not State Functions
- we can bring the system from state 1 to
state 2 along infinite # of paths, and for each
path P(T,V) will be different.
V2
W1 2    P(T , V )dV
V1
P
T
2
Since the work done on a system depends not
only on the initial and final states, but also on the
intermediate states, it is not a state function.
V
1
U = Q + W
U is a state function, W - is not 
thus, Q is not a state function either.
B
Wnet  WAB  WCD   P2 V2  V1   P1 V1  V2 
P
P2
P1
A
 P2  P1 V2  V1   0
D
V1
PV diagram
C
V2
V
- the work is negative for the “clockwise” cycle; if
the cyclic process were carried out in the reverse
order (counterclockwise), the net work done on
the gas would be positive.
Comment on State Functions
U, P, T, and V are the state functions, Q and W are not. Specifying an initial and final
states of a system does not fix the values of Q and W, we need to know the whole
process (the intermediate states). Analogy: in classical mechanics, if a force is not
conservative (e.g., friction), the initial and final positions do not determine the work, the
entire path must be specified.
In math terms, Q and W are not exact differentials of some functions
of macroparameters. To emphasize that W and Q are NOT the state
functions, we will use sometimes the curled symbols  (instead of d)
for their increments (Q and W).
d U  T d S  P dV
y
U
V
S
- an exact differential
dz  Ax  x, y  dx  Ay  x, y  dy
z(x1,y1)
z(x2,y2)
x
- it is an exact differential if it is
the difference between the values of some (state) function
dz  z  x  dx, y  dy   z  x, y 
z(x,y) at these points:
Ax x, y  Ay x, y 

y
x
 z 
 z 
d z    dx    dy
 x  y
 y  x
A necessary and sufficient condition for this:
If this condition
holds:
e.g., for an ideal gas:
z x, y 
y
T
f

Q  dU  PdV  Nk B  dT  dV 
V
2

Ax x, y  
z x, y 
x
Ay x, y  
- cross derivatives
are not equal
Problem
Imagine that an ideal monatomic gas is taken from its initial state A to state
B by an isothermal process, from B to C by an isobaric process, and from
C back to its initial state A by an isochoric process. Fill in the signs of Q,
W, and U for each step.
P,
105 Pa
2
A
T=const
1
Step
Q
W
U
AB
+
--
0
BC
--
+
--
CA
+
0
+
B
C
1
2
f
U  Nk BT
2
V, m3
PV  NkBT
The Enthalpy
Isobaric processes (P = const):
dU = Q - PV = Q -(PV) 

Q =  U + (PV)
H  U + PV - the enthalpy
The enthalpy is a state function, because U, P,
and V are state functions. In isobaric processes,
the energy received by a system by heating equals
to the change in enthalpy.
isochoric:
Q=U
isobaric:
Q=H
in both cases, Q
does
not
depend on the
path from 1 to 2.
Consequence: the energy released (absorbed) in chemical reactions at constant
volume (pressure) depends only on the initial and final states of a system.
f
f

The enthalpy of an ideal gas: H  U  PV 
Nk BT  Nk BT    1 Nk BT
2
2 
(depends on T only)
Heat Capacity
The heat capacity of a system - the amount of energy
transfer due to heating required to produce a unit
temperature rise in that system
T
C is NOT a state function (since Q is not a
state function) – it depends on the path
between two states of a system

f1
Q
T
f2
f3
T1+dT
T1
i
V
( isothermic – C = , adiabatic – C = 0 )
The specific heat capacity
C
C
c
m
Quasistatic Processes in an Ideal Gas
isochoric ( V = const )
P
W12  0
2
PV= NkBT2
PV= NkBT1
1
V1,2
V
Q12
3
 Nk B T2  T1   0
2
P
(see the last slide)
dU  Q12
isobaric
 CV T 
( P = const )
2
W12    P(V , T )dV   PV2  V1   0
2
1
V1
1
PV= NkBT2
PV= NkBT1
V2
V
5
Q12  Nk B T2  T1   0
2
dU  W12  Q12
 CP T 
Isothermal Process in an Ideal Gas
P
isothermal ( T = const ) :
PV= NkBT
W
V2
Wi  f
dU  0
V2
V1
V
V
 NkBT ln i
Vf
Wi-f > 0 if Vi >Vf (compression)
Wi-f < 0 if Vi <Vf (expansion)
V2
dV
V
  NkBT ln 2
V
V1
V1
W12    P(V , T )dV   Nk BT 
V1
Q12  W12
Adiabatic Process in an Ideal Gas
Q12  0
adiabatic (thermally isolated system)
dU  W12
The amount of work needed to change the state of a thermally isolated system
depends only on the initial and final states and not on the intermediate states.
V2
W12    P(V , T )dV
P
V1
2
V2
1
PV= NkBT2
PV= NkBT1
V1
V
to calculate W1-2 , we need to know P (V,T)
for an adiabatic process
U

2  dP
0
1   
f
P




 dU 
f
Nk B dT   PdV
2
( f – the # of “unfrozen” degrees of freedom )
PV  NkBT  PdV  VdP  NkB dT
dV
V
f
Nk BT
2
,
  1
2
PdV
 PV
f
V
P
dV
dP


0
V
P
V1
P1
PdV  VdP  
2 Adiabatic
f exponent
V 
P 

ln    ln  1   PV   P1V1  const
P
 V1 
Adiabatic Process in an Ideal Gas (cont.)

PV   P1V1  const
P
2
V2
1
PV= NkBT2
PV= NkBT1
V1
V
An adiabata is “steeper” than an isotherma:
in an adiabatic process, the work flowing
out of the gas comes at the expense of its
thermal energy  its temperature will
decrease.
V2
V2
V1
V1
W12    P (V , T )dV   

V2
PV
1

  1
1 1
dV


PV
V
1
1
V
  1
V1
1  1
1 
 PV
  1   1 
1 1
  1  V2
V1 


 1+2/31.67 (monatomic), 1+2/5 =1.4 (diatomic), 1+2/6 1.33 (polyatomic)
(again, neglecting the vibrational degrees of freedom)
Prove W12 
f
f
  PV   Nk B T  U
2
2
Summary of quasi-static processes of ideal gas
U  U f  U i
Quasi-Static
process
isobaric
(P=0)
isochoric
(V=0)
isothermal
(T=0)
adiabatic
(Q=0)
U
U 
f
f
Nk B T  PV
2
2
f
f
U  Nk B T   P V
2
2
0
f
f
U  Nk B T    PV 
2
2
Ideal gas
law
Q
W
f 2
P V
2
 PV
Vi V f

Ti T f
0
Pi Pf

Ti T f
f
 P V
2
W
0
 Nk BT ln
U
Vf
Vi
PV
i i  Pf V f


PV
i i  Pf V f
Problem
Imagine that we rapidly compress a sample of air whose initial pressure is
105 Pa and temperature is 220C (= 295 K) to a volume that is a quarter of
its original volume (e.g., pumping bike ’ s tire). What is its final
temperature?
Rapid compression – approx. adiabatic, no time for the energy
exchange with the environment due to thermal conductivity
P1V1  Nk BT1
P2V2  Nk BT2
P2 
P1V1  P2V2

P1V1
V2
P1V1
P1V1

Nk
T

T2
B 2
V2 1
T1
For adiabatic processes:
 V1 
T2  T1  
 V2 
 1

 1
 V1 
  
 V2 
 1
T1 V1
 T2 V2
also
P 1 / T   const
 1

 const
 295 K  40.4  295 K 1.74  514 K
- poor approx. for a bike pump, works better for diesel engines
T2
T1
Non-equilibrium Adiabatic Processes
Free expansion
1.
2.
TV  1  const
V – increases
 T – decreases (cooling)
On the other hand, U = Q + W = 0
U ~ T  T – unchanged
(agrees with experimental finding)
Contradiction – because
approach
#1 cannot be justified – violent
expansion of gas is not a quasistatic process. T must remain the
same.
TV  1  const
- applies only to quasi-equilibrium processes !!!
CV and CP
Q
dU  PdV
C

dT
dT
 U  the heat capacity at
CV  

 T V constant volume
 H 
CP  

 T  P
the heat capacity at
constant pressure
To find CP and CV, we need f (P,V,T) = 0 and U = U (V,T)
 f

f
H


1
For an ideal gas U 
Nk BT

 Nk BT
2

2
f
f
 f

CP    1 nR
CV  Nk B  nR
2

2
2
# of moles
For one mole of a
monatomic ideal gas:
CV 
3
5
R CP  R
2
2
P
Another Problem
During the ascent of a meteorological helium-gas filled balloon,
its volume increases from Vi = 1 m3 to Vf = 1.8 m3, and the
pressure inside the balloon decreases from 1 bar (=105 N/m2) to
0.5 bar. Assume that the pressure changes linearly with volume
between Vi and Vf.
(a) If the initial T is 300K, what is the final T?
(b) How much work is done by the gas in the balloon?
(c) How much “heat” does the gas absorb, if any?
Pi
Pf
Vi
Vf
V
PV   0.625 bar/m 3 V  1.625 bar
(a)
PV  Nk B T
PV
T
NkB
Vf
(b) WON    P(V )dV
Vi
0.5bar 1.8m3
T f  Ti
 300K
 270K
Pi Vi
1bar 1m3
Pf V f
Vf
- work done on a system WBY   P (V )dV - work done by a system
Vi
Vf
WON  WBY WBY   P(V )dV  0.5  0.8 bar  m3  0.5  0.4 bar  m3   0.6bar  m3  6 10 4 J
(c)
U  Q  WON
Vi
T

3
3
Q  U  WON  NkB T f  Ti   WON  Pi Vi  f  1  WBY  1.5 105 J   0.1  6 104 J  4.5 104 J
2
2
 Ti
