Thermodynamics: Chapter 01
Transcript of Thermodynamics: Chapter 01
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Thermodynamics: Chapter 01
September 3, 2013
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§1. Basics concepts in thermodynamics
We start this course from some fundamental con-
cepts. By introducing a system that consists of many
particles and is also simply enough for us to investi-
gate the major thermal properties, the ideal gas sys-
tem, we will introduce more fundamental concepts,
laws of thermodynamics, and apply them in solving
practical thermal problems.
• Temperature:1
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1. operational definition - how we measure the tem-
perature of an object;
2. theoretical definition - relation to zeroth law of
thermodynamics, energy, entropy; absolute tem-
perature; absolute zero degree; (Ideal Gas sys-
tem)
3. different scales and units: pay attention to them
in applications!
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Celsius (oC): water boiling point at 1 atm pressure
- 100oC, water freezing point at 1 atm pressure -
0oC
Fahrenheit (oF): water boiling point at 1 atm
pressure - 212oF , water freezing point at 1 atm
pressure - 32oF
Kelvin (K, SI unit): 0K = −273oC, the size of
1K = the size of 1oC
• Pressure: The ratio of force to the area over which
that force is distributed.
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Pressure and mechanical equilibrium:
Unit: SI unit is Pascal, 1Pa = 1N/m2. Other units
see the table below.
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• Equilibrium
It is a condition of a system in which competing
influences are balanced. Examples are
Equilibrium ExchangesThermal Thermal energy
Mechanical VolumeDiffusive ParticlesChemical Chemical reactions
H2O ↔ H+ +OH−
H2SO4↔ 2H+ + SO2−4
The zeroth law of thermodynamics: Systems
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that are in thermal equilibrium with one system,
these systems are in thermal equilibrium with each
other. The theoretical basis for the concept of
”Temperature”. We will explore this more in future
(Section 3.1 in Chapter 3)
• Systems in thermal physics
Isolated system: No interaction with environment,
no matter and energy exchange.
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Closed system: No matter, but has energy ex-
change.
Open system: Has matter and energy exchange.
• Energy
Kinetic energy: energy associated with motion
(and mass), of a ”particle”, of a system.
Potential energy: energy associated with interac-
tions (various fields), of a ”particle”, of a system.
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Chemical energy: energy associated with EM in-
teration, of a ”particle”, of a system. Unit: SI
unit is Joule 1J = 1kg ·m2/s2
• Heat (Q): Flow of energy from one system to an-
other caused by a difference in temperature be-
tween the two systems.
Heat transfer: conduction - at microscopic level;
convection - bulk motion of medium; radiation -
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electromagnetic waves, photons.
Unit: Calories, not SI unit !! 1C ≈ 4.2J
• Work (W): Any other transfer of energy (except
heat) into or out of a system.
The first law of thermodynamics: ∆U = Q+W ,
the system’s energy change (∆U) is the sum of the
changes in heat and work into or out of a system.
This is the law of energy conservation.
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Aug. 29
*********
§2. The Ideal Gas: A simple thermal system that explains
a lot about thermal physics
Ideal Gas: Particles have no size; There is no interactions among
particles; Particles have elastic collisions against the wall of the
container.
This is a not-so-bad approximation for gas at low pressure/low
density and when the temperature is not so low.
In the study of ”bulk” properties:
2
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. . .
. . .
.
.
1664 and 1676: R. Boyle and E. Mariotte discovered that ata fixed temperature T the pressure and volume satisfy therelation: pV = p0V0. Or, pV = constant!
1802: Gay-Lussac obtained V = TT0V0, at a fixed pressure p.
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(He credited to the unpublished work from the 1780s by Jacques
Charles. Charles’ Law)
How p, V, T correlate to each other if we let the system change
from state (p0, V0, T0) to (p, V, T )?
(1) (p0, V0, T0) to (p, V ∗0 , T0): Fixed T
p0V0 = pV ∗0(2) (p, V ∗0 , T0) to (p, V, T ): Fixed p
V = TT0V ∗0
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By canceling V ∗0 in these two equations, one has
pV
T=p0V0
T0= constant.
This constant should be proportional to the number of
particles (N) in the system, therefore we can writepV
T= Nk (1)
where k is called Boltzmann constant.
k = 1.380658× 10−23J ·K−1.
We can re-write it as
pV = nRT (2)
where n is the number of moles of gas, R is a constant that
has empirical value of R = 8.31 Jmol·K when all variables are in
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SI units: Pressure in N/m2 = Pa, volume in m3, and
temperature on Kelvin,K
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In-Class Exercise: A mole of molecules is Avogadro’s number
(NA) of them. Given the empirical values of R and k, what is
the value of NA?
pV = nRT
pV = NkT
nRT = NkT
NA =1.0 ·Rk
=8.31 J
mol·K1.380658× 10−23J ·K−1
NA = 6.017× 1023 mol−1
3
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Example: Equation of State: Relation between state parame-
ters (such as P, V, T), or f(P, V, T ) = 0.
Or, generally, f(X1, X2, ..., Xn, T ) = 0. Such as f(m,H, T ) = 0
for paramagnetism.
Different materials have different equation of state. Equation of
state exists for a system that reaches equilibrium state.
A famous equation of state that better describes gas (and even
dense fluids!) is the van der Waals equation:(P +
an2
V 2
)(V − nb) = nRT
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n is mole number. a and b are constants that depend on thetype of matter.an2
V 2 : correction to the attraction between particles.nb: correction to the size of the particles.Material a/(Pa ·m6 ·mol−1) b/(m3 ·mol−1)H2 0.02476 0.02661N2 0.1408 0.03913CO2 0.3639 0.04267H2O 0.5535 0.03049
We define coefficient of expansion as α = 1V
(∂V∂T
)P
(fixed pres-sure). Calculate the coefficient of expansion for van der WaalsGas.
If we can solve for V from the van der Waals Gas equation, itwould be great...
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Sept. 3
*********
Homework policy revisit:
• Due date and how to submit.
• Posting of answers.
• Questions and discussions.
5
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§3. The Ideal Gas: Connection between bulk property
and microscopic picture
§3.1 Temperature and kinetic energy of particles
Vz dt Z
V
Side wall area = A
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Number of particles that have velocity within [~v,~v + d~v]
dN = Nf(~v)d3~v
f(~v) =1
N
dN
d3~v, velocity density distribution∫
f(~v)d3~v = 1
N of particles (with velocity in [~v,~v + d~v]) in dV is
dN = NdV
Vf(~v)d3~v , dV: volume by motion in dt
dV = Avzdt , A: area perpendicular to vz
Assuming elastic scattering on the wall
dpz = mvz − (−mvz) = 2mvz : for one particle
dFAdt = dpzdN = 2mvzdN = 2Nmv2z f(~v)d3~v
Adt
V
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So, dFA = 2Nmv2z f(~v)d3~vAV , the force by all particles with veloc-
ity in [~v,~v + d~v]!!
By the definition of pressure:
p =1
A
∫dFA . Integration over all velocity. Be very careful here:
p =N
V
∫ +∞
−∞dvx
∫ +∞
−∞dvy
∫ +∞
0dvzf(~v)2mv2
z
Assuming f(~v) is independent from the direction of ~v,∫ +∞
0dvz =
1
2
∫ +∞
−∞dvz
Therefore,
pV = mN∫ +∞
−∞f(~v)v2
z d3~v (3)
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The integration is the mean value of v2z !
Since the motion of gas particles is isotropic (in direction), the
mean value of the projection of velocity in all directions should
be the same. That is∫+∞−∞ f(~v)v2
z d3~v =< v2
x >=< v2y >=< v2
z >
and v2 = v2x + v2
y + v2z , we have < v2
z >= 13 < v2 >. So, Eq. (3)
can be written as
pV = mN1
3< v2 >=
2
3N < εkin > (4)
where < εkin >= 12m < v2 > is the mean kinetic energy of the
particle.
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Using Ideal Gas Law, let’s take a look at the temperature:
pV = kNT
pV =2
3N < εkin >
kNT =2
3N < εkin >
T =2
3< εkin > /k
§3.2 Equipartition theorem
Let’s look at the relation between average translational kinetic
energy of a particle and the temperature of the system: From
T = 23 < εkin > /k, we get:
< εkin >= 32kT . Since < εkin >=< 1
2mv2 >= 1
2m < v2x +v2
y +v2z >,
we can have identity: 32m < v2
x >= 32kT , or for three degrees of
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freedom,12m < v2
x >= 12m < v2
y >= 12m < v2
z >= 12kT .
This is a special case when the particles have only three degreesof freedom. This result can be extended to systems (not allthermal systems!) that have arbitrary degrees of freedom - theEquipartition theorem: At temperature T, the average energyof any quadratic degree of freedom is 1
2kT . - This can be provedbased on principles in Statistical Physics which we will learn laterin this semester.
For a system with N particles, each with f DoF, and there is NOother non-quadratic temperature-dependent forms of energy, thetotal thermal energy in the system is
Uthermal = Nf1
2kT (5)
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Remarks:
• It only applies to systems in which the energy is in the formof quadratic degree of freedom: E(q) = cq2.
• It is about ”thermal energy” of the system - those changeswith temperature, not the total energy.
• Degree of freedom: different systems require specific analy-sis: vibration, rotation, .... The NDoF of a system may alsovary as temperature changes.
§3.3 Compression work
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Let’s deal with the energy change in a thermal system. There
are different ways to do so, one of which is by doing work to the
system:
δx
Piston area = A Force = F
F P
V
Area = work !
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The work done by the pushing force:
δW = Fδx, −Aδx = δV
δW = −F/AδVdW = −PdV (6)
W = −∫ V2
V1
P (V )dV (7)
Two processes:
1. Isothermal compression: temperature stays constant - slow
process / closed system.
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Using Eq. (7) and (1) pV = NkT , one has
W = −∫ V2
V1
P (V )dV
= −∫ V2
V1
NkT
VdV when T is constant, we have:
W = −NkT lnV2
V1(= NkT ln
V1
V2) (8)
The first law of thermodynamics takes the following form:
∆U = Q+W
∆(Nf1
2kT ) = Q−NkT ln
V2
V1, when T is constant:
Q = NkT lnV2
V1(9)
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where Q as the heat input or output depending on the work doneto or by the gas.
2. Adiabatic compression: no heat escape from the system.Examples: fast process or isolated system.
The first law of thermodynamics takes the following form:
∆U = Q+ ∆W
d(Nf1
2kT ) = Q− PdV , Q=0 for isolated system:
1
2NfkdT = −PdV For ideal gas PV = NkT (10)
1
2NfkdT = −
NkT
VdV
fdT
T= −2
dV
VOr, after the integration: (11)
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flnT1
T2= −2ln
V1
V2(12)
This can be simplified further:
{T1
T2}f = {
V2
V1}2
{T1
T2}f/2 =
V2
V1
Tf/21 V1 = T
f/22 V2 = ... , which is
V T f/2 = constant. (13)
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§3.4 Heat capacityWhen we talk about the energy change in a thermal system, itis convenient to define a quantity that measures the correlationbetween heat exchange and the temperature change caused bythe heat flow.
Heat capacity: of an system is the amount of heat needed toraise the temperature by one degree: C = Q
∆T . Or, for one mass
unit as c = Qm∆T .
Heat capacity at constant volume CV , Heat capacity at constantpressure CP , and Partial Derivative: (remember Q = ∆U −W )
CV =(∂Q
∂T
)V
=(∂U
∂T
)V
(no V change → no work) (14)
CP =(∂Q
∂T
)P
=(∂U
∂T
)P
+ P
(∂V
∂T
)P
(15)
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In-class exercise:
• Prove the heat capacity at constant volume for ideal gas is
CV = 32Nk.
• Prove the heat capacity at constant pressure for ideal gas is
CP = CV + nR.
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§3.5 The Carnot cycle: Ideal gas case
Four step of Carnot cycle: a: Isothermal expansion at Th b: adiaba:c expansion to Tc c: isothermal compression at Tc d: adiaba:c compression back to Th
PV diagram for an ideal monoatomic gas undergoing a Carnot cycle.
Nicolas Léonard Sadi Carnot (06/01/1796 – 08/24/1832)
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The Carnot cycle is very important in thermal physics because
it explains how to make a system (like engine) that can achieve
the maximum possible efficiency when it works within a given
temperature range.
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Let’s see how energy changes in the cycle:
Step 1: isothermal expansion, (V1, P1, Th)→ (V2, P2, Th)
∆Q1 = −∆W1 = −(−∫ V2
V1
PdV ) =∫ V2
V1
NkThV
dV
= NkThlnV2
V1Step 2: adiabatic expansion, (V2, P2, Th)→ (V3, P3, Tc)
∆Q2 = 0,∆W2 = ∆U2 = CV (Tc − Th)
Step 3: isothermal compression, (V3, P3, Tc)→ (V4, P4, Tc)
∆Q3 = −∆W3 = NkTclnV4
V3Step 4: adiabatic compression, (V4, P4, Tc)→ (V1, P1, Th)
∆Q4 = 0,∆W4 = ∆U4 = CV (Th − Tc)
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The overall energy budget is
∆Utotal = (∆Q1 + ∆W1) + ∆W2 + (∆Q3 + ∆W3) + ∆W4
∆Utotal = (0) + CV (Tc − Th) + (0) + CV (Th − Tc)∆Utotal ≡ 0
This is exactly what needed for a cycle - the state of the system
comes back to its original state!!!
For adiabatic expansion/compression, we have:
dU = dQ− PdVdU = −PdV
That is, all work done to the system is stored as thermal energy.
Using heat capacity:
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CV =(∂Q∂T
)V
=(∂U∂T
)V,
we have:
CV dT = −PdV, PV = NkT → P =NkT
V
CVdT
T= −Nk
dV
Vwhich integrates to:
CV lnTc
Th= −Nkln
V3
V2
We know for Ideal Gas, CV = 32Nk. The last equation can be
re-written as:
3
2Nkln
Tc
Th= −Nkln
V3
V2(Tc
Th
)3/2
=V2
V3
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So, we have the following additional relations:
Step-2:V3
V2=(ThTc
)3/2
Step-4:V1
V4=
(Tc
Th
)3/2
These give:V3
V2=V4
V1or
V1
V2=V4
V3
Since∆Q1
Th= Nkln
V2
V1(see what obtained in Step-1)
(16)
and∆Q3
Tc= Nkln
V4
V3= Nkln
V1
V2= −Nkln
V2
V1Therefore, we have
∆Q1
Th+
∆Q3
Tc≡ 0 (17)
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If we divide one big cycle into many micro-Carnot cycles, wehave
∮ δQT = 0. It turns out this is true for all reversible cycles..
A cycle of an arbitrary closed path in P‐V space can be achieved by a large number of small Carnot cycles.
A B C D
Isothermal lines
Adiaba=c lines
The quantity ∆QT plays a very important role in thermodynamics
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and statistical physics. It closely connects with another impor-
tant concept entropy and the second law of thermodynamics.