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Thermodynamics: Chapter 01
September 2, 2014
§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
1. operational definition - how we measure the tem-
perature of an object;
2. theoretical definition - relation to zeroth law of
thermodynamics
The zeroth law of thermodynamics: Systems
that are in thermal equilibrium with one system,
these systems are in thermal equilibrium with
each other.
We will explore this more about temperature
Chapter 3: Its relation with the energy in the
system, entropy, absolute temperature, absolute
zero degree, etc.
3. different scales and units: pay attention to them
in applications!
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.
Pressure and mechanical equilibrium:
Unit: SI unit is Pascal, 1Pa = 1N/m2. Other units
see the table below.
• 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
• Systems in thermal physics
Isolated system: No interaction with environment,
no matter and energy exchange.
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.
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 -
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.
Aug. 28
*********
§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
. . .
. . .
.
. p = F/(Piston area)
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.
(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
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
SI units: Pressure in N/m2 = Pa, volume in m3, and
temperature on Kelvin,K
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
Example, Equation of State: Relation between state param-
eters (such as P, V, T), or f(P, V, T ) = 0.
Or, generally, f(X1, X2, ..., Xn, T ) = 0.
Example: For paramagnetism, we can write in general format
f(M,B, T,C) = 0, which in the form of Curie’s Law looks like
MT = CB
in which:
M : is the ”resulting” magnetisation, the total magnetic moment
of the material proptoµ(N↑ −N↓).
T : is absolute temperature (Kelvin).
C: is a material-specific Curie constant. (Will learn more in
Section 3.3).
B: is the magnetic field (Tesla).
4
Note: Different materials have different equation of state. Equa-
tion 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
n is mole number. a and b are constants that depend on the
type 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 Waals
Gas.
If we can solve for V from the van der Waals Gas equation, it
would be great ...
Homework policy revisit:
• Due date and how to submit.
TA: Emily Dvorak, [email protected]
• Posting of answers.
• Questions and discussions.
5
§3. The Ideal Gas: Connection between bulk propertyand microscopic picture
§3.1 Temperature and kinetic energy of particles
Number of particles that have velocity within [~v,~v + d~v]
dN = Nf(~v)d3~v
f(~v) =1
N
dN
d3~v, it’s called the Velocity Density Distribution∫
f(~v)d3~v = 1
Assuming particles are uniformly distributed in volume V , thenumber (N) of particles that have velocity within [~v,~v + d~v] indV is
dN =N
VdV f(~v)d3~v
6
Let’s look at these particles in the volume (dV ) of the motion
of all particles in dt. dV can be written as (see Fig below),
Vz dt Z
V
Side wall area = A
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
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)
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.
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
freedom,12m < v2
x >= 12m < v2
y >= 12m < v2
z >= 12kT .
This is a special case when the particles have only three degrees
of freedom. This result can be extended to systems (not all
thermal systems!) that have arbitrary degrees of freedom - the
Equipartition theorem: At temperature T, the average energy
of any quadratic degree of freedom is 12kT . - This can be proved
based on principles in Statistical Physics which we will learn later
in this semester.
For a system with N particles, each with f DoF, and there is NO
other non-quadratic temperature-dependent forms of energy, the
total thermal energy in the system is
Uthermal = Nf1
2kT (5)
Remarks:
• It only applies to systems in which the energy is in the form
of quadratic degree of freedom: E(q) = cq2.
• It is about ”thermal energy” of the system - those changes
with temperature, not the total energy.
• Degree of freedom: different systems require specific analy-
sis: vibration, rotation, .... The NDoF of a system may also
vary as temperature changes.
Sept. 2
*********
§3.3 Compression work
Thermal system: At macroscopic level, one cares the energy flow
into or out from the system.
There are different ways to cause energy change in a thermal
system. One of which is by doing work to/by the system.
The law of energy conservation, the first law of thermodynamics:
∆U = Q + W , the system’s energy change (∆U) is the sum of
the changes in heat Q and work W into or out of a system.
7
δx
Piston area = A Force = F
F P
V
Area = work !
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.
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)
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
V(11)
Or, after the integration: (using∫ dVV = lnV )
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)
§3.4 Heat capacity
When we talk about the energy change in a thermal system, it
is convenient to define a quantity that measures the correlation
between heat exchange and the temperature change caused by
the heat flow.
Heat capacity: of an system is the amount of heat needed to
raise the temperature by one degree: C = Q∆T . Or, for one mass
unit as c = Qm∆T .
Looking at Q = ∆U −W ≈ [U(Tf) − U(Ti)] −∫PdV , one needs
to consider
(1) Heat capacity at constant volume CV
8
(2) Heat capacity at constant pressure CPwhich are related to partial derivatives:
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)
In-class quiz:
1. Prove the heat capacity at constant volume for ideal gas is
CV = 32Nk.
2. Prove the heat capacity at constant pressure for ideal gas is
CP = CV + nR.
§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)
9
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.
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)
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:
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
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)
If we divide one big cycle into many micro-Carnot cycles, we
have∮ δQT = 0. It turns out this is true for all reversible cycles,
not just ideal gas system that undergoes the Carnot Cycle.
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
And more: ∆Q1Th
+ ∆Q3Tc≡ 0 is one of the most important relation
in thermal physics! The quantity ∆QT plays a very important role
in thermodynamics and statistical physics. It closely connects
with another important concept entropy and the second law of
thermodynamics.