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Sunday, March 22, 2009

THERMAL PHYSICS II

1. Let us agree to call whatever we are studying the "system" (a mixture of ice and water, a
hot gas, etc). The state of this system is specified by giving its pressure, volume,
temperature, etc. These are called "thermodynamic variables". The relation between these
variables is called the "equation of state". This equation relates P,V,T. So, for example:
P= f( , ) (knowing and gives you )
( , ) (knowing and gives you )
( , ) (knowing and gives you )
For an ideal gas, t
V T V T P
V g PT P T V
T hPV P V T
=
=
he equation of state is . Here is the number of molecules
in the gas, is the temperature in degrees Kelvin, and is called the Boltzman cosntant.
This EOS is easy to derive, altho
B
B
PV Nk T N
T k
=
ugh I shall not do it here. In principle it is possible to
mathematically derive the EOS from the underlying properties of the atoms and molecules
in the system. In practice, however, this is a very difficult task (except for an ideal gas).
However the EOS can be discovered experimentally.
2. Thermodynamics is the study of heat and how it flows. The First Law of Thermodynamics
is actually just an acknowledgment that heat is a form of energy (and, of course, energy
is always conserved!). Mathematically, the First Law states that:
where: internal energy of the system
heat transf
E q w
E
q
Δ = Δ + Δ
=
= erred to the system from the surroundings
work done on the system by the surroundings.
In words, what the above formula says is this: if you do an amount of work and also
w
w
=
Δ
transfer an amount of heat , then the sum of these two quantities will be the additional
amount of energy that is stored in the system. There could be nothing simpler! Note:
and
q
E
q
Δ
Δ
Δ Δ are positive if heat is added to or work is done on a system, and negative if
heat is removed from the system, or if the system does work on the environment.
3. Work and heat are called
w
path var - their values depend on the steps leading from
one state to another. To give an obvious example: suppose you drag a heavy box on the
floor in a straight line from point A to point B, an
iables
d then in a very roundabout way from
A to B again. Obviously the work done by you will be very different. On the other hand,
the internal energy does not depend upon the path. Example: a gas has internal energy
proportional to its temperature. It makes no difference whether the gas had been slowly
heated or rapidly; it will have the same internal energy. Since the internal energy doe
2
1 2 1
s not
depend on the path we can write . State
State
∫ dE = E −E =ΔE
Gas Vacuum
P (P1,V1)
( ) 2 2 P,V
( ) 0 0 P P ,V
V
P
ΔV
W=PΔV
4. Let us calculate the work done by an expanding gas. In the diagram below, suppose the
piston moves out by distance dx, then the work done is where is the force
exerted on the gas.
dw = Fdx F
But . (The minus sign is important to understand;
remember that the force exerted by the gas on the piston is the negative of the force that the
piston exerts on the gas). So
ext F P A P A
dw Fdx
= × = − ×
=
2
1
1 2
. The work done by the
expanding gas is where V and V are the initial and final volumes.
ext ext
V
V ext
P A dx P dV
w PdV
= − × × = −
= −∫
Note that depends on the path. Let's consider 3 different paths.
a)Suppose that the gas expands into the vacuum, 0. Then 0. So, if in the figure
below, the valve is opened, then
ext
w
P = w=
gas will flow into the vacuum without doing work.
2 1 b)Suppose the pressure outside has some constant value. Then, ( ) or,
. You can see that this is just the area of the rectangle below.
ext
ext
w P V V
w P V
= × −
= Δ
c)We can also let the gas expand so that the internal and
external pressures are almost the same (this is called
reversible expansion). Then the work done the
system is: ex
on
dw = −P
2
1
2
2 1
1
2 1
. Using gives,
(ln ln ) ln
Of course, the work done the gas as it expands is
positive since is larger than . Remember that the
l
t B
V
B
V B B
dV PV Nk T
w Nk TdV Nk T V V Nk T V
V V
by
V V
=
= −∫ = − − = −
og function is positive if its argument is bigger than 1.
Compressed
ideal gas
Pint Pext
Closed system
5. The internal energy of a gas depends only on the number of molecules it contains and
on the temperature, 3 . In a free expansion of gas (it doesn't do any work in this
2
case), the initia
B E= NkT
f i l and final internal energies are equal, E = E .
6. Imagine that a substance is heated while keeping its volume constant. Obviously, you have
to supply more heat as you raise the temperature higher, , where is given
the name "specific heat at constant volume". No work is done since there is no expansion,
0. Hence, using the First Law, , and so .
7. Now tak
V V V
V V
Q C T C
W Q E W E E C T
Δ = Δ
Δ = Δ = Δ +Δ = Δ Δ = Δ
e the same system as above, but allow it to expand at constant pressure as it is
heated. Then you will have to supply an amount of heat, . Here, is called
the specific heat at const
P P P ΔQ =CΔT C
ant pressure, . Hence,
. If we supply only a small amount of heat then, .
This gives, . If we consider the special case of an ideal gas, then
P
P P
P V
Q E w E PV
C T U P V C dT dU PdV
C dT C dT PdV
Δ = Δ + Δ = Δ + Δ
Δ = Δ + Δ = +
= +
. Hence, and . Hence, .
Since the internal energy is 3 , it follows that 3. From this,
2 2
the specific heat at constant pressure is
B B P V B P V B
B V B
PV Nk T PdV Nk dT C dT C dT Nk dT C C Nk
E NkT C dU Nk
dT
= = = + = +
= = =
: 3 5 . From the equation
2 2
, it follows that . Let us define a new symbol, .
8. We can find the EOS of an ideal gas when it expands without losing any heat
P B B B
B P V P
P V B
V V V
C Nk Nk Nk
C C Nk Nk C C C
C C C
γ
= + =

= + = ≡
( )
(this is
called adiabiatic expansion). In this case, 0. Hence, 0,
or 0. Dividing by gives, 0 or 1 0.
Integrating this gives, ln
V
B
V B
V
dQ dE dW C dT PdV
C dT Nk T dV dT dT Nk dV dT dV
V T C V T V
T
γ
γ
= + = + =
+ = + = + − =
+( ) ( 1)
1
1 ln Constant. Equivalently, ln Constant.
A more convenient form is: Constant. This holds for the entire adiabatic expansion.
9. Almost nothing beats the importance of the Second Law Of
V TV
TV
γ
γ


− = =
=
Thermodynamics. There are
many equivalent ways of stating it. The one I like is: "There can be no process whose only
final result is to transfer thermal energy from a cooler object to a hotter object". This seems
extremely simple (and almost useless), but in fact it tells you, among other things, that no
perpetual motion machine (such as that which generates electricity without any fuel input)
can ever be built. If someone could build such a machine, then it could also be used to
transfer heat from a cold object to a hot object - a contradiction with the Second Law.
Hot reservoir at temperature Th
h Q
c Q
Heat engine W
Cold reservoir at temperature Tc
10. Before we consider some implications of the Second Law, I want to introduce the
concept of a . Suppose you put a thermometer in your mouth. You will
transfer a small am
thermal reservoir
ount of heat to the thermometer, but because your body is so big this
will make no measureable difference to your body temperature. Your body therefore is
a thermal reservoir as far as the thermometer is concerned. More generally, any large
mass of material will act as a thermal reservoir if it is at constant temperature. A
(e.g. a car engine, refrigerator, etc) ope
heat
engine rates between reservoirs at two different
temperatures.
,
,
11. A heat engine works between a high temperature
and a low temperature . It aborbs heat from
the hot reservoir and rejects heat into the cold
reservoir. Since the inte
H
C in h
out c
T
T Q
Q
, ,
rnal energy of the engine does
not change, 0 and the First Law gives simply,
. The work done by the engine is
. Now define the efficiency of the
machine as
in h out c
U
Q U W W
W Q Q
ε
Δ =
= Δ + =
= −
,
, , ,
, ,
work done by engine . Then
heat put in
1 . Now, the heat that is
in h
in h out c out c
in h in h
W
Q
Q Q Q
Q Q
ε
= =

= = −
transferred out of or into a reservoir is proportional to its absolute temperature. Hence we
arrive at the important result that 1 . This number is always less than one, showing
C
H
T
T
ε = −
0 0
that no machine can convert all the input heat into useful work. As an example, a nuclear
reactor has temperature 300 at the core and rejects heat into a river at 30 . The
maximum effic
C C
iency it can have is 1 30 273 0.471
300 273
ε
+
= − =
+
12. A refrigerator is a heat engine working in reverse.
Work is done (by an electric motor) to pump heat
from a cold reservoir (inside of refrigerator) to the
hot exterior. The Second Law can be shown to imply
the following: it is impossible for a refrigerator to
produce no other effect than the transfer of thermal
energy from a cold object to a hot object. Again, the
efficiency of a refrigerator is always less than one.
h Q
c Q
Refrigerator W
Hot reservoir at temperature Th
Cold reservoir at temperatureTc
v
v
Δp

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