Download Chapter 4 The First Law - Physics | Oregon State University

“... there is in the physical world one agent only, and this is
called Kraft [energy or work]. It may appear, according to
circumstances, as motion, chemical affinity, cohesion,
electricity, light and magnetism; and from any one of these
forms it can be transformed into any of the others.”
Karl Friedrich Mohr - Zeitschrift für Physik (1837).
Chapter 4
The First Law
4.1
Work, “Heat” and The First Law
Taking to heart Professor Murray Gell-man’s observation that opened Chapter 3, a
respite from quantum mechanics is in order and the discussion begun in Chapter 1
is continued. Restating the First Law of Thermodynamics
∆U = Q − W ,
(4.1)
where ∆U is the change in U, the internal energy. U is a thermodynamic state function with quantum foundations. In this chapter the First Law’s “classical” terms W
(work) and Q (heat) are discussed, followed by a few elementary applications.
1. W (Work ): W is energy transferred between a system and its surroundings by
mechanical or electrical changes at its boundaries. Work is a notion imported
from Newton’s mechanics with the definition
W = ∫ F ⋅ dr
(4.2)
where F is an applied force and dr is an infinitesimal displacement. The
integral is carried out over the path of the displacement.
Work is neither stored within a system nor does it describe any state of the
?
system. Therefore ∆W = Wf inal − Winitial has absolutely no thermodynamic
meaning!
1
2
CHAPTER 4. THE FIRST LAW
Incremental work, on the other hand, can be defined as d− W, with finite work
being the integral
W = ∫ d− W
(4.3)
The “bar” on the “d ” is a warning that d− W is not a true mathematical differential.1
Note: Macroscopic parameters are further classified as:
A. extensive – proportional to the size or quantity of matter (e.g. energy,
length, area, volume, number of particles, polarization, magnetization)
B. intensive – independent of the quantity of matter (e.g. force, temperature,
pressure, density, tension, electric field, magnetic field.)
Incremental work has the generalized meaning of energy expended by/on the
system due to an equivalent of force (an intensive parameter) acting with an
infinitesimal equivalent of displacement (an extensive parameter) which, in
accord with Eq.4.2, is written as the conjugate pair,
extensive
dW = ∑
−
j
Fj
¯
«
dxj
.
(4.4)
intensive
Examples of conjugate “work” pairs are
1
1.
pressure (p) ● volume ( dV ) ,
2.
tension (τ ) ● elongation ( dχ) ,
3.
stress (σ) ● strain ( d) ,
4.
surface tension (s) ● area ( dA) ,
5.
electric displacement field (D) ● polarization ( dP) ,
6.
magnetic field (H ) ● magnetization ( dM).2
7.
chemical potential (µ) ● particle number ( dN )
Incremental work is not a function of independent variables. It is not an exact differential.
The proper fields used in representing electric and magnetic work will be further examined in
Chapter 12.
2
4.1. WORK, “HEAT” AND THE FIRST LAW
3
It is a helpful rule in thermodynamics to think of work as the equivalent of
raising and lowering weights external to the system.
Finite work, as expressed in Eq.4.3, depends on the the integration path – the
details of the process – and not just on the process end points.
The sign convention adopted here3 is
d− W > 0 ⇒ system performs work
gas
(d− Wby
= pdV )
d− W < 0 ⇒ work done on the system
gas
(d− Won
= −pdV )
2. Q “Heat”: Q is energy transferred between a system and any surroundings with
which it is in diathermic contact. Q is not stored within a system and is not a
?
state parameter of a macroscopic system. Therefore ∆Q = = Qf inal −Qinitial has
no thermodynamic meaning! On the other hand, incremental heat is defined
as d− Q, with its integral being finite heat Q
Q = ∫ d− Q
The “bar” on the “d ” is a warning that d− Q is not a true mathematical differd− Q
ential. Therefore “derivatives” like
have no meaning.4
dT
Finite heat depends on the details of the process, i.e. the integration path, and
not just its end points.
3. Temperature T : Temperature is a parameter intuitively associated with a subjective sense of “hotness” or “coldness.” Empirical ideas about temperature
have objective meaning through thermometers – the height of a column of
mercury, the pressure of an ideal gas in a container with fixed volume, the electrical resistance of a length of wire – all vary in a way that allows a temperature
scale to be defined. Since there is no quantum hermitian “temperature operator” a formal, consistent definition of temperature is elusive, but is achievable
in practice through thermodynamic results of the least bias (maximum uncertainty) postulate5 which shows, with remarkable consistency, how and where
temperature appears and that it measures the same property as a common
thermometer.
3
With system pressure and volume as the conjugate pair.
Q is not a function of T nor of any other independent variables.
5
This crucial subject will be discussed in Chapter 6.
4
4
CHAPTER 4. THE FIRST LAW
4. Internal Energy U: As discussed in Chapter 1, internal energy U is a state
of the system. Therefore ∆U does have meaning and dU is an ordinary mathematical differential6 with
B
∆U = U(B) − U(A) = ∫ dU .
(4.5)
A
where A and B represent different macroscopic equilibrium states. Clearly, ∆U
depends only on the end points of the integration and not on any particular
thermodynamic process (path) connecting them.
4.1.1
Exact Differentials
All thermodynamic state variables are true (exact) differentials with a change in
state defined as in Eq.4.5. Moreover, state variables are not independent and can
be functionally expressed in terms of other state variables. Usually only a few are
needed to completely specify any state of a system. The precise number required is
the content of an important theorem which is stated here without proof.
The number of independent parameters needed to specify a thermodynamic state is equal to the number of distinct conjugate pairs of quasistatic work [see Eqs.below (4.4)] needed to produce a differential change
in the internal energy – plus ONE7
For example, if a gas undergoes mechanical expansion or compression, the only quasistatic work is mechanical work, d− WQS = p dV . Therefore only two variables are
required to specify a state. We could then write the internal energy of a system as a
function of any two other state parameters. For example we can write U = U (T, V )
or U = U (P, V ). To completely specify the volume of a gas we could write V =
V (P, T ) or even V = V (P, U) . The choice of functional dependences is determined
by the physical situation and computational objectives.
If physical circumstances also allow particle number to change, say by evaporation
of liquid into a gas phase or condensation of gas into the liquid phase or chemical
reactions where molecular species are lost or gained, a thermodynamic description
requires “chemical work” d− WQS = −µ dN where µ is the chemical potential and dN
6
It is an exact differential.
The one of the plus one originates from a conjugate pair T dS, where S is Entropy, about which
much will be said throughout the remainder of this book.
7
4.1. WORK, “HEAT” AND THE FIRST LAW
5
is an infinitesimal change in particle number. Chemical potential µ is an intensive
variable associated with energy per particle while particle number N is obviously an
extensive quantity. In cases like this quasi-static work8
d− WQS = p dV − µ dN
(4.6)
has two kinds of contributions and therefore three variables are required, e.g. U =
U (T, V, N ) or U = U (P, V, µ).
Incorporating the definitions above, an incremental form of the First Law of Thermodynamics is
dU = d− Q − d− W
(4.7)
4.1.2
Equations of State
Equations of state are functional relationships among a system’s equilibrium state
parameters – usually those that are most easily measured and controlled in the laboratory. Determining equations of state, by experiment or theory, is often a primary
objective in practical research – for then thermodynamics can fulfill its role in understanding and predicting state parameter changes for macroscopic systems.
1. Equations of state for gases (often called gas laws) take the functional form
f (p, V, T ) = 0 where p is the equilibrium gas pressure, V is the gas volume, T
is the gas temperature and f defines the functional relationship. For example a
gas at low pressure and low particle density, n = N /V , where N is the number
of gas particles (atoms, molecules, etc.), has the ideal gas equation of state
p − (N /V ) kB T = 0
(4.8)
pV = N kB T
(4.9)
or, more familiarly,
where kB is Boltzmann’s constant.9
2. An approximate equation of state for two and three dimensional metallic materials is usually expressed as g(σ, , T ) = 0 where σ = τ /A is the stress,
8
With a gas as an example.
For gases under realistic conditions Eq.4.9 may be inadequate and more complicated empirical
or theoretical equations of state are used. Nevertheless, because of its simplicity and “zeroth order”
validity, the ideal gas “law” is used in a variety of models and exercises.
9
6
CHAPTER 4. THE FIRST LAW
= (` − `0 )/ `0 is the strain10 , T is the temperature, A is the cross-sectional
area and κ is a characteristic constant. An approximate equation of state for
a metal rod is the Hookean behavior
= κσ.
(4.10)
3. An equation of state for a uniform one-dimensional elastomeric material has
the form Λ(τ, χ, T ) = 0 where τ is the equilibrium elastic tension, χ is the
sample elongation, T is the temperature and Λ is the functional relationship.
For example, a rubber-like polymer that has not been stretched beyond its
elastic limit can have the simple equation of state
σ − K T = 0
(4.11)
τ − K ′T χ = 0 ,
(4.12)
or, alternatively
where K and K ′ are material specific elastic constants.11,12
4. Non-polar dielectrics in weak polarizing fields have the equation of state
Pi = ∑ αij (T ) Ej
(4.13)
j
where Pi is an electric polarization vector component, αij (T ) is the temperature
dependent electric susceptibility tensor and Ej is an electric field component.13
In stronger fields equations of state can be generalize by including non-linear
terms, e.g.
(1)
(2)
Pi = ∑ αij (T )Ej + ∑ αijk (T )Ej Ek .
(4.14)
j
10
j,k
` is the stretched length and `0 is the unstretched length
E. Guth and and H. M. James, “Theory of Elastic Properties of Rubber,” J. Chem. Phys. 15,
2941 (1943).
12
Comparing these with Eq.4.10 suggests a physically different origin for rubber elasticity which
will be explored in Chapter 11.
13
Is this an external electric field or some average local internal electric field? How do Maxwell’s
equations fit in here? This will be addressed in an Chapter 12. For the present, we assume that
the sample is ellipsoidal in which case the internal and external electric fields are the same.
11
4.1. WORK, “HEAT” AND THE FIRST LAW
7
5. Non-permanently magnetic materials which are uniformly magnetized in weak
magnetic fields are described by the equation of state
M = χM (T ) B
(4.15)
where M is the magnetization vector, χM (T ) is a temperature dependent
magnetic susceptibility tensor and B is the magnetic induction field prior to
insertion of the sample.14,15
4.1.3
Examples: I
h2
h1
Figure 4.1: Quasi-static, reversible work. A piston loaded down with a pile of fine
sand.
1. A gas is confined in a cylinder by a frictionless, massless piston of negligible
mass and cross-sectional area A. The gas pressure is counteracted by a pile
14
Magnetism and magnetic “work”, will be discussed in Chapter 12.
Solid state physics text books often express Eq.4.15 in terms of an external magnetic field H
in which a needle-like sample geometry is assumed in which the long axis is parallel to the external
field.
15
8
CHAPTER 4. THE FIRST LAW
of fine sand of total mass m resting on the piston [see Figure 4.1] so that
it is initially at rest. Neglecting any atmospheric (external) force acting on
the piston, the downward force exerted by the sand is, initially, F0sand = mg.
Therefore the initial equilibrium gas pressure, pgas
0 , is
pgas
0 =
mg
.
A
(4.16)
Sand is now removed, “one grain at a time” allowing the gas to expand in
infinitesimal steps, continually passing through equilibrium states. Moreover,
it is conceivable that at any point a single grain of sand could be returned
to the pile and exactly restore the previous equilibrium state. Such a real or
idealized infinitesimal process is called quasi-static. A quasi-static process,
by definition, takes place so slowly that the system (in this example a gas)
always passes through equilibrium states so that an equation of state always
relates system pressure, volume, temperature, tension, etc.
If, as in this case, the previous, infinitesimally displaced state can be accessed
– with no dissipation of energy – the process is also said to be reversible.16
Assuming the quasi-static expansion also takes place isothermally at temperature TR , what work is done by the gas in raising the piston from h1 to h2 ?
At all stages of the quasi-static expansion equilibrium is maintained with pressure pgas and temperature TR well defined. Therefore incremental work done
by the gas (assumed ideal) is
d− WQS = pgas dV
N kB TR
dV .
=
V
(4.17)
(4.18)
Integrating Eq.4.18 along the isothermal path [as pictured in Figure 4.4] gives
V2 =h2 A
WQS = N kB TR ∫
V1 =h1 A
dV
V
= N kB TR ln (h2 /h1 ) .
16
(4.19)
(4.20)
Not all quasi-static processes are reversible but all reversible processes are quasi-static. Reversibility implies no friction or other energy dissipation in the process.
4.1. WORK, “HEAT” AND THE FIRST LAW
9
2. If, instead of a restraining sand pile the piston is now pinned a distance h1
above the bottom of the cylinder [See Figure 4.2], while at the same time a
constant atmospheric pressure patm exerts an external force F atm on the piston,
F atm = patm A .
(4.21)
The equilibrium gas pressure pgas
initially exerted by the gas on the pinned
0
piston is
Fpin
+ patm
A
pgas
0 =
(4.22)
atm .
where Fpin is the pinning force, so that pgas
0 >p
,
atm
pin
atm
Figure 4.2: Irreversible work. A gas filled piston initially pinned at height h1 .
Quickly (and frictionlessly) extracting the pin frees the piston allowing the
gas to expand rapidly and non-uniformly. After things quiet down the gas
reaches an equilibrium state, with the piston having moved upward a distance
10
CHAPTER 4. THE FIRST LAW
` = h2 − h1 with a final gas pressure pgas
= patm . This isothermal process –
f
characterized by a finite expansion rate and system (gas) non-uniformity – is
neither quasi-static nor reversible.17,18
What isothermal work is done by the rapidly expanding gas in raising the
piston a distance `?
The gas expands irreversibly. Therefore during expansion a gas law does not
apply. But initially the gas has equilibrium pressure pgas
with volume V0
0
V0 =
N kB T
.
pgas
0
(4.23)
After the piston has returned to rest (equilibrium),
pgas
= patm
f
(4.24)
N kB T
.
patm
(4.25)
with volume
Vf =
While expanding irreversibly the gas lifts the piston against a constant force
F atm (equivalent to raising a weight mg = F atm in the earth’s gravitational
field.)19 The incremental, irreversible work done by the confined gas (the system) is
d− W = patm dV
(4.26)
which, with constant patm , is integrated to give
W = patm (Vf − V0 ) .
(4.27)
With Eqs. 4.23 and 4.25
W = N kB T [1 −
17
patm
].
pgas
0
(4.28)
All real processes in nature are irreversible. Reversibility is a convenient idealization for thermodynamic modeling.
18
“Irreversible” as used here, is not intended as an absolute, since surroundings can always
be altered to recover the initial equilibrium state. But this is not the same as the sequence of
infinitesimal steps associated with reversibility.
19
The piston starts at rest and returns to rest, so there is no change in piston’s kinetic energy.
4.1. WORK, “HEAT” AND THE FIRST LAW
11
3. As in Example 2, the gas is initially confined by a piston exposed to the atmosphere with its initial position secured by a “pin.” However the piston’s
movement is restricted by a frictional wedge which, together with the external
atmosphere, slows the piston’s upward movement, enabling it to be approximated as quasi-static but irreversible. [See Figure 4.3.]
Assuming the process is isothermal, what work is done by the expanding gas?
The expanding gas (the system) raises the piston against atmospheric force
Fatm and the force of friction Ff riction , both taken to have constant value.
Therefore the work performed is
d− Wsys =
(Fatm + Ff riction )
dV
A
(4.29)
which after integration becomes
Wsys = (patm + Ff riction /A) (Vf − V0 )
(4.30)
where V0 is the initial equilibrium volume
V0 =
N kB T
Fpin /A
patm (1 +
)
patm
(4.31)
and Vf is the final equilibrium volume
Vf =
N kB T
.
patm
(4.32)
Therefore
Wsys =
4.1.3.1
N kB T Fpin (patm + Ff riction /A)
(
)
patm
A
(patm + Fpin /A)
(4.33)
The p-V Diagram
In performing quasi-static work, a system passes through equilibrium states,
so that it is possible to draw a continuous curve on a p − V diagram describing
the process. An isothermal work “path”, is represented by the isotherm in
Figure 4.4, where the work is the green area under the isotherm.
12
CHAPTER 4. THE FIRST LAW
Figure 4.3: Quasi-static, irreversible work. A piston slowed by a frictional wedge.
4.1.4
Examples II
1. A mass 2 m affixed to the end of a vertically stretched massless rubber band,
is suspended a distance h1 (rubber band length `1 ) above a table. When half
the mass is suddenly removed the rubber band rapidly contracts causing the
remaining mass to rise. After oscillating a while it finally comes to rest a
distance h2 , (h2 > h1 ) (rubber band length `2 ) above the table. [See Figure 4.5.]
(a) What is the work done by the rubber band?
After returning to rest, the work (not quasi-static) done by the rubber
band (work done by the system) in raising the weight m (against gravity)
is
W = −mg∆`
(4.34)
where ∆` = `2 − `1 < 0.
(b) What is the change in the rubber band’s internal energy?
4.1. WORK, “HEAT” AND THE FIRST LAW
13
p=0
Figure 4.4: p − V diagram for an isothermal, quasi-static process.
Since the process takes place rapidly it is irreversible. Moreover there is
not enough time for the rubber band to lose heat to the surroundings and
the process can be regarded as adiabatic. Using the First Law
∆U = +mg ∆` .
(4.35)
The rubber band loses internal energy.
(c) This time, in counteracting rubber elasticity, the total mass 2 m is restrained to rise slowly (quasi-statically) but isothermally. What is the
work done by the contracting rubber band?
We can safely use an equilibrium tension (an elastic state parameter)
which acts to raise the weight. The incremental quasi-static elastic work
done by the system is
d− WQS = −τ dχ
(4.36)
where dχ is the differential of elastic extension. Using the “rubber”
14
CHAPTER 4. THE FIRST LAW
l2
l1
h2
h1
Figure 4.5: A loaded rubber band, when released, rapidly contracts.
equation of state, Eq.4.12, and integrating over the range of extension χ
0
WQS = −K T ∫ χ dχ
′
(4.37)
−∆`
2
= K T (∆`)
′
(4.38)
2. A metal rod, cross sectional area A, is suddenly struck by a sledgehammer,
subjecting it to a transient but constant compressive (external) force F which
momentarily changes the rod’s length by ∆` = `f − `0 < 0.
4.1. WORK, “HEAT” AND THE FIRST LAW
15
(a) What is the work done on the rod during the interval that its length
decreases by ∣∆`∣?
Sudden compression of the rod implies that it does not pass through
equilibrium states so an equilibrium internal stress in the bar σ = f /A is
not defined. But the mechanical work done on the rod by the constant
compressive force is, however
W = −F ∆` .
(4.39)
(b) On the other hand, if the rod could be compressed infinitely slowly (perhaps by a slowly cranked vise) the rod’s internal stress is a well defined
equilibrium property and the quasi-static incremental work done by the
system (i.e. the rod) has the standard form
d− WQS = −V σ d
(4.40)
where V is the rod’s volume, σ is the internal stress and d = (d`/`) defines
the differential strain. We might now apply an approximate equation of
state, such as Eq.4.10, and get
f
WQS = −V ∫ (/κ) d
(4.41)
0
= − (V /2κ) (2f − 20 )
4.1.5
(4.42)
Heat Capacity
Heat capacities have played an important role in exploring microscopic properties of
matter. Historically, heat capacity measurements have revealed unexpected behaviors of macroscopic systems which inspired important microscopic quantum discoveries. Equations of state, together with heat capacities, usually offer the basis for a
complete thermal picture of matter.
An infinitesimal of thermal energy (heat) quasi-statically entering or leaving a system
produces an infinitesimal change in temperature. (Exceptions are certain first-order
phase transitions, such as boiling of water or melting of ice.) The proportionality
constant between the two infinitesimals, together with a list of state variables, α, that
16
CHAPTER 4. THE FIRST LAW
are fixed during the quasi-static process, is called heat capacity.20 The complete
definition of heat capacity (an extensive quantity) is then
d− QQS = Cα dT
(constant α)
(4.43)
where Cα is the heat capacity at constant α. The units of heat capacity are energy/K.
Specific heats are alternative intensive quantities, having units energy/K/mass or
energy/K/mole.
1. If heat is transferred at constant volume, from the “definition” given in Eq.4.43
d− QQS = CV dT
(constant volume).
(4.44)
Heat capacities are equivalently – and more usefully – expressed in terms of
partial derivatives of state parameters. For example, if a gas performs only
quasi-static mechanical work, the infinitesimal First Law for the process is
dU = d− QQS − p dV
(4.45)
where p dV is quasi-static incremental work d− WQS done by the gas (system.)
Since U can here be expressed in terms of two other thermal parameters, choose
U = U (T, V ) so that its total differential is
dU = (
∂U
∂U
) dV + ( ) dT
∂V T
∂T V
(4.46)
and Eq.4.45 can then be rewritten
d− QQS = [(
∂U
∂U
) + p] dV + ( ) dT .
∂V T
∂T V
(4.47)
Thus, for an isochoric (constant volume) path, i.e. dV = 0,
d− QQS = (
∂U
) dT
∂T V
(4.48)
∂U
)
∂T V
(4.49)
and from Eq.4.44
CV = (
20
The term heat capacity is an anachronistic holdover from early thermodynamics when heat
was regarded as a stored quantity. Although we now know better, the language seems not to have
similarly evolved.
4.1. WORK, “HEAT” AND THE FIRST LAW
17
2. Thermodynamic experiments, especially chemical reactions, are often carried
out in an open atmosphere where the pressure is constant. In such cases we
define
d− QQS = Cp dT (constant pressure).
(4.50)
Expressing Cp as the partial derivative of a state parameter requires a bit more
ingenuity. Here again the infinitesimal First Law description of the process is
dU = d− QQS − dV .
(4.51)
d (pV ) = p dV + V dp
(4.52)
d− QQS = d (U + pV ) − V dp .
(4.53)
But this time, using
Eq.4.51 can be rewritten as
Now impose the definition
H = U + pV
(4.54)
where H – also a state function – is called Enthalpy,21 permitting recasting a
“new” differential “Law” –
d− QQS = dH − V dp .
(4.55)
Then, expressing H in terms of T and p, i.e. H = H (T, p) so that a total
differential is
dH = (
∂H
∂H
) dp + (
) dT .
∂p T
∂T p
(4.56)
Then for an isobaric (constant pressure) path, i.e. dp = 0,
d− QQS = (
∂H
) dT
∂T p
(4.57)
and we have the definition in terms of state variables
Cp = (
21
∂H
)
∂T p
(4.58)
Enthalpy is one of several defined state variables that are natural to and simplify specific
thermodynamic conditions. Others will be acknowledged as need requires.
18
CHAPTER 4. THE FIRST LAW
4.1.6
Examples III
1. An ideal monatomic gas, volume V0 , temperature T0 , is confined by a frictionless, movable piston to a cylinder with diathermic walls. The external face of
the piston is exposed to the atmosphere at pressure P0 . A transient pulse of
heat is injected through the cylinder walls causing the gas to expand against
the piston until it occupies a final volume Vf .
(a) What is the change in temperature of the gas?
At the beginning and end of the expansion the system is in equilibrium
so that an equation of state applies. Initially P0 V0 = N kB T0 and finally
P0 Vf = N kB Tf so that
P0 ∆V
(4.59)
∆T =
N kB
where ∆T = Tf − T0 and ∆ V = Vf − V0 .
(b) How much heat Q was in the pulse?
The gas will expand turbulently, i.e. irreversibly, against constant atmospheric pressure P0 so that incremental work done by the expanding gas
is
d− W = P0 dV .
(4.60)
Applying the First Law
Q = ∆U + P0 ∆V .
(4.61)
and using Eqs.4.46, 4.58 and 4.59 gives the internal energy change for this
ideal gas
∆U = CV ∆T
CV P0 ∆V
=
N kB
(4.62)
(4.63)
Thus the injected heat is
CV
+ 1) P0 ∆V
(4.64)
N kB
where Eqs.4.61 and 4.63 have been used. Using another ideal gas property22
Q=(
Cp − CV = N k B
22
This is stated, for now, without proof.
(4.65)
4.1. WORK, “HEAT” AND THE FIRST LAW
19
and the definition
γ = Cp /CV ,
(4.66)
γ
P0 ∆V .
γ−1
(4.67)
Eq.4.64 can be condensed to
Q=
L
Figure 4.6: A wire held between firmly planted rigid vertical posts.
2. A steel wire, length L and radius ρ is held with zero tension τ between two
firmly planted rigid vertical posts, as shown in Figure 4.6 . The wire, initially
at a temperature TH , is then cooled to a temperature TL .
(a) Find an expression for the tension τ in the cooled wire.
The thermodynamic state variables used for describing elastic behavior
in the steel wire (or rod) are stress σ = τ /A where A is the wire’s cross
sectional area, and strain = (` − `0 )/`0 where ` and `0 are the wire’s
stretched and unstretched length. These are the same variables referred
to in elastic work [see Eq.4.40] and in the elastic equation of state [see
Eq.4.15].
In the present situation the rigid posts constrain the length of the wire to
remain constant so that a decrease in temperature is expected to increase
the strain (tension) in the wire. Referring to Section 4.1.2, consider a
functional dependence of strain as = (T, σ), which are the physically
relevant state variables for the wire, and take the total differential
d = (
∂
∂
) dT + ( ) dσ .
∂T σ
∂σ T
(4.68)
20
CHAPTER 4. THE FIRST LAW
The partial derivatives represent physical properties of the system under
investigation for which extensive numerical tables can be found in the
literature.23 In Eq.4.68
∂
αL = ( )
(4.69)
∂T σ
where αL is the linear thermal expansivity, and
1
∂ε
=( )
ET
∂σ T
(4.70)
where ET is the isothermal Young’s modulus.
Thus we can write
1
dσ ,
ET
and since the length of the wire is constant, i.e. d = 0, we have
d = αL dT +
0 = αL dT +
1
dσ .
ET
(4.71)
(4.72)
Finally, assuming αL and ET are average constant values, integration gives
∆σ = σ(TL ) − σ(TH ) = −αL ET (TL − TH ) ,
(4.73)
and with τ (TH ) = 0
τ (TL ) = − AαL ET (TL − TH )
= πr2 αL ET (TH − TL )
(4.74)
(4.75)
3. A gas is allowed to rapidly expand from an initial state (p0 , V0 , T0 ) to a final
state f.
What is the final state (pf , Vf , Tf )?
This rapid process can be regarded as taking place adiabatically, Q = 0. Applying the infinitesimal First Law we have,
d− Q = 0 = dU + d− W .
(4.76)
Since U is a state variable the integral
∆U = ∫ dU
23
See for example; http://www.periodictable.com/Properties/A/YoungModulus.html
(4.77)
4.1. WORK, “HEAT” AND THE FIRST LAW
21
is path (i.e. process) independent. Therefore the integral
−
∫ d W = ∆U
(4.78)
depends only on its thermodynamic end points. So we connect these equilibrium
states with a unique, quasi-static (maximum work) process and write
pf ,Vf ,Tf
0=
[ dU + p dV ] .
∫
(4.79)
p0, V0, T0
Then with dU as in Eq.4.46
pf ,Vf ,Tf
0=
{[(
∫
p0, V0, T0
∂U
∂U
) + p] dV + ( ) dT } .
∂V T
∂T V
(4.80)
So far the result is general (any equation of state). But for simplicity, assume
the gas is ideal. Using Eq.4.48 and applying the ideal gas property24
(
∂U
) =0
∂V T
(4.81)
together with the ideal gas equation of state [see Eq.4.9] gives
pf ,Vf ,Tf
0=
{[
∫
p0, V0, T0
N kB T
] dV + CV dT }
V
(4.82)
which implies
N kB T
] dV = −CV dT .
(4.83)
V
This differential equation separates quite nicely to give upon integration
[
N kB ln (
Vf
T0
) = CV ln ( )
V0
Tf
(4.84)
or
CV
Vf N kB
T0
( )
=( )
V0
Tf
24
(4.85)
This useful ideal gas result is assigned as a problem. Note: it only applies to an ideal gas.
22
CHAPTER 4. THE FIRST LAW
which with Eqs.4.65 and 4.66 becomes
(
Vf γ−1
T0
) =( ).
V0
Tf
(4.86)
4. A perfectly insulated, rigid, very narrow necked vessel is evacuated and sealed
with a valve. The vessel rests in an atmosphere with pressure P0 and temperature T0 . [Figure 4.7.] The valve is suddenly opened and air quickly fills the
vessel until the pressure inside the vessel also reaches P0 . If the air is treated as
an ideal gas, what is the temperature Tf of the air inside the vessel immediately
after equilibrium is attained?
In this problem we introduce the subtle strategy of replacing an essentially
open system (the number of molecules inside the vessel is not constant) with
a closed system (the gas is conceptually bounded by a movable piston that
seals the neck and, ultimately the closed vessel below, as in Figure 4.7) This
notional replacement is a common trick in thermodynamics and we take the
opportunity to illustrate it here.
4.1. WORK, “HEAT” AND THE FIRST LAW
23
Figure 4.7: Evacuated, insulated vessel in an atmosphere with pressure P0 and temperature T0 .
Physically, the narrow neck separates the air inside the vessel from the air outside the vessel, the latter maintained at pressure P0 by the infinite atmosphere.
But we replace the external atmosphere with a piston that forces outside air
across the narrow neck into the vessel by applying the constant external pressure P0 , as shown in Figure 4.8.
24
CHAPTER 4. THE FIRST LAW
P0
N T0
v0
N Tf
vf
Figure 4.8: Evacuated vessel filling under pressure P0 .
Focus attention on a small, fixed amount of air being forced into the vessel
by the conjectured piston, say N molecules at pressure P0 and temperature
T0 occupying a small volume v0 . In forcing these molecules through the valve
the piston displaces a volume v0 . The work done on the N molecules by the
4.1. WORK, “HEAT” AND THE FIRST LAW
25
external atmosphere, pressure P0 , is
Wby the piston = ∫ d− W
= P0 ∆v
= P0 v 0 .
(4.87)
(4.88)
(4.89)
Therefore the work done by the N molecules of gas is
Wby the gas = −P0 v0 .
(4.90)
Once the N molecules get to the other side of the valve and enter the evacuated,
rigid walled vessel, they do NO work.25 Therefore the total work done by
the gas is
Wtotal = −P0 v0 + 0
(4.91)
Applying the First Law to the N molecules, while approximating this sudden
process as adiabatic 26
0 = ∆U + Wtotal
(4.92)
or
f
0 = ∫ dU − P0 v0 .
(4.93)
0
Choosing to write U = U (V, T )
dU = (
∂U
∂U
) dT + (
) dV .
∂T V
∂V T
(4.94)
Since
(
∂U
) = CV
∂T V
(4.95)
and for an ideal gas
(
25
26
∂U
) =0
∂V T
This is evident since no weights are raised or lowered.
Too fast to allow heat exchange with the external atmosphere.
(4.96)
26
CHAPTER 4. THE FIRST LAW
we have
P0 v0 = CV (Tf − T0 ) .
(4.97)
Then with the ideal gas equation of state
N kB T = CV (Tf − T0 )
(4.98)
which with Eqs.4.65 and 4.66, becomes
Tf = γT0 .
(4.99)
For an ideal monatomic gas (γ = 5/3) at room temperature (T0 = 300K ○ ) ,
Eq.4.99 gives Tf = 500K ○ , indicating a substantial rise in temperature for the
gas streaming into the vessel.
4.1.7
Concluding Remarks
In this hiatus from quantum mechanics27 we have returned to discuss timeless classical thermodynamics, but with an overriding confidence only quantum mechanics will
give in establishing the meaning of state variables such as internal energy. We have
also indicated how profoundly U differs from purely classical Work W and “Heat”
Q. In the next chapters we will complete the process of unifying quantum mechanics
and thermodynamics.
27
Not really!