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Unformatted text preview: Chapter
Chapter 19
Thermodynamics II
kinetic theory of gases Thermodynamics II
Ideal
Ideal gas
Kinetic
Kinetic theory calculations of pressure and
temperature
Internal
Internal energy and equipartition of energy
Specific
Specific heat of ideal gas The
The kinetic theory of gases
Treating
Treating thermodynamics from a
microscopic viewpoint.
Clarify
Clarify the microscopic basis of pressure and
temperature
temperature
Macroscopic
Macroscopic properties can be found by
averaging over individual particles Mole and Avogadro’s number
One
One mole = 6.02 × 1023 elements
= the number of atoms in a 12 g
sample carbonsample of carbon12
Avogadro’s
Avogadro’s number NA = 6.02 × 1023 mol1
10
A.
A. Avogadro: suggested that all gases
occupying the same volume under the same
temperature and pressure with the same
number of atoms or molecules. What
What is ideal gas?
All
All gases approaches the ideal state at low
density
No
No interaction between molecules
Experiments
Experiments showed that 1 mol different
gases
gases (hydrogen, oxygen, methane…) exhibit
nearly the same pressure under identical
volume and temperature Ideal gases
pV = nRT
p: absolute pressure
V: volume
n: number of moles of gas Macroscopic
R: gas constant = 8.31 J/mol．K Microscopic Boltzmann
Boltzmann constant k = R/NA = 1.38 × 1023 J/K
R/N
pV = NkT , N: number of molecules T: temperature in Kelvins
Please refer to Feynman lecture on Physics Ch 39 for derivation Work
Work done at constant temp
Isothermal
Isothermal processes
Vf nRT
dV
Vi
V
Vf
1
dV = nRT ln
V
Vi
Vf W = ∫ pdV = ∫
Vi Vf = nRT ∫ Vi Which are isothermal processes? Work done at constant volume or
constant pressure
Constant
Constant pressure i ↔ a
Vf W = ∫ pdV = p (V f − Vi ) = p∆V
Vi Constant
Constant volume a ↔ f
W =0 Pressure
Pressure and molecular speeds
What
What is the connection between pressure
and the speeds of molecules inside?
Ignore
Ignore molecule interaction and consider only
elastic collisions with the wall
Momentum delivered to the wall by
the molecule during a collision ∆p x = 2mv x
The average rate of momentum delivery
2
∆p x
2mv x
mv x
=
=
∆t
2 L / vx
L Force on the wall Pressure and molecular speeds
2
2
2
pressure on
Fx mv x1 / L + mv x 2 / L + ... + mvxN / L
p= 2 =
2
the wall L L ( m2
2
2
= 3 v x1 + v x 2 + ... + v xN
L ) () 2
= N vx avg 2
= nN Av x ,rms 2
= nN Avrms / 3 2
2
nmN Avrms nMvrms
⇒ p=
=
3L3
3V Rootmeansquare Link between microscopic
and macroscopic properties vrms 3RT
=
M Kinetic
Kinetic energy
Average
Average translational kinetic energy ( Kavg = 1 mv2
2 ) () = 1 m v2
2
avg 2
= 1 mvrms
avg
2 = 3RT 3RT
=
M 2NA
Kavg = 3 kT
2
Kavg = 1 m
2 3RT
M At
At a given temperature T, all ideal gas
molecules—
molecules—no matter what their mass—
mass—
have the same average translational
kinetic energy.
Measuring
Measuring T = measuring Kavg MaxwellMaxwellBoltzmann distribution
What
What is the speed distribution of gas
molecules?
Probability
Probability function:
3/ 2 3/ 2 M 2 −Mv / 2RT m 2 −mv2 / 2kT
P(v) = 4π = 4π ve ve 2πRT 2πkT 2 ∫ ∞ 0 P(v)dv = 1 Derivation MaxwellDerivation of MaxwellBoltzmann distribution
Boltzmann distribution P Normalization P For ideal gas in 1D P P In 3D space To find out the number (probability) of particle at v, we have to integrate over a shell dv = dvxdvydvz
3/ 2 m 2 −mv2 / 2kT
P(v) = 4π ve
2πkT P http://hyperphysics.phyastr.gsu.edu/hbase/kinetic/maxspe.html#c3 Molecular speeds
Average
Average speed ∞ 8RT
πM vavg = ∫ vP(v)dv =
0 RMS
RMS speed vrms = ∞ ∫ 0 Most
Most probable speed v2 P(v)dv = 3RT
M dP(v)
2RT
= 0; vP =
dv
M
What is vavg, vrms, and vP, for
oxygen and hydrogen gases?
vrms 483 m/s 1932 m/s vavg 445 m/s 1780 m/s vP 395 m/s 1580 m/s The
The internal energy
Kavg = 3 kT
2 ⇒ Eint = NKavg = 3 NkT = 3 nRT
2
2 The
The internal energy of an ideal gas is a
function
function of the gas temperature only.
only Molar
Molar specific heat
At
At constant volume Q = nCV ∆T
∆Eint = Q
3
2 nR∆T = nCV ∆T ⇒ CV = 3 R
2 At
At constant pressure Q = nC p ∆T
∆Eint = Q − W W = p∆V = nR∆T ∆Eint = nCV ∆T ⇒ C p = CV + R Quick
Quick quiz
Rank
Rank ∆Eint Equipartition of energy
Each
Each degree of freedom has associated
with an energy of 1 kT per molecule (or RT
RT
2 kT
per mole). As Table 192 shows, the prediction that Cv=R agrees
As
19with experiment for monatomic gases but fails for
diatomic and polyatomic gases. Let us try to explain the
gases.
discrepancy ( 差異 ) by considering the possibility that
molecules with more than one atom can store internal
energy in forms other than translational kinetic energy.
Fig. 1911 shows common models of He (monatomic),
Fig. 19oxygen (diatomic), and methane ( 甲 烷 , polyatomic).
polyatomic).
From such models, we would assume that all 3 types of
molecules can have translational motions (x,y,z(x,y,zdirections) and rotational motions.
motions.
We would assume that the diatomic and polyatomic
We
molecules can have oscillatory motions (振動), with the
atoms oscillating slightly toward and away from one
another, as if attached to opposite ends of a spring.
spring. To account of the various ways in which energy can be
To
stored in a gas, Maxwell introduced the theorem of the
equipartition of energy (能量均分原理):
Every kind of molecule has a certain number f of
degrees of freedom ( 自由度 ), which are independent
ways in which the molecule can store energy. Each such
degree of freedom has associated with it — on average
— an energy of kT per molecule (or RT per mole).
Let us apply the theorem to the translational and
Let
Fig. 19rotational motions of the molecules in Fig. 1911.
For the translational motion, the molecules will have
velocity components along x, y, and z axes. Thus, gas
axes.
molecules of all types have 3 degrees of translational
freedom and, on average, an associated energy of kT
per molecule.
molecule. For the rotational motion, imagine the origin of our xyz
For
Fig.
coordinate system at the center of each molecule (Fig.
1911). In a gas, each molecule should be able to rotate
19with an angular velocity component along each of the 3
axes, so each gas should have 3 degrees of rotational
freedom and, on average, an additional energy of kT
per molecule. However, experiment shows this is true
molecule.
only for the polyatomic molecules.
According
According to quantum theory, the physics dealing with
the allowed motions and energies of molecules and
atoms, a monatomic gas molecule does not rotate and
so has no rotational energy.
energy.
A diatomic molecule can rotate only about axes
perpendicular to the line connecting the atoms (Fig. 19Fig. 1911b) and not about that line itself. Therefore, a diatomic
11b
itself.
molecule can have only 2 degrees of rotational freedom
and a rotational energy of only 2(kT) per molecule.
molecule. We replace Eq. 1938 (Eint=nRT) with Eint=(f)nRT,
We
Eq. 19where f is the number of degrees of freedom listed in
important Table 193.
19 Doing
Doing so leads to the prediction which agrees with Eq. 1943 for monatomic gases (f
=3). As Table 192 shows, this prediction also agrees
with experiment for diatomic gases (f=5), but it is too
low for polyatomic gases (f=6 for molecules
comparable to CH4). f=3 (三個移動
自由度) f=5 (三個移動、
三個移動、
二個轉動自由度) f=6 (三個移動、
三個移動、
三個轉動自由度) We can improve the agreement of kinetic theory with
We
experiment by including the oscillations of the atoms in
a gas of diatomic or polyatomic molecules.
molecules.
For example, the two atoms in the O2 molecule of Fig.
For
Fig.
1911b can oscillate toward and away from each other,
with the interconnecting bond acting like a spring.
spring.
However,
However, experiment shows that such oscillations occur
only at relatively high temperatures of the gas — the
motion is “turned on” only when the gas molecules have
relatively large energies. Rotational motion is also
energies.
subject to such “turning on,” but at a lower temperature.
temperature. Fig. 1912 CV/R v.s. temperature for (diatomic) H2 gas. Because
rotational and oscillatory motions begin at certain energies, only
translation is possible at very low temperatures. As the temperature
increases, rotational motion can begin. At still higher temperatures,
oscillatory motion can begin. Figure 19Figure 1912: Below about 80 K, we find that CV/R=1.5.
/R=1
This result implies that only the three translational
degrees of freedom of hydrogen are involved in the
specific heat.
heat.
As the temperature increases, the value of CV/R
As
gradually increases to 2.5, implying that two additional
degrees of freedom have become involved. Quantum
involved.
theory shows that these two degrees of freedom are
associated with the rotational motion of the hydrogen
molecules and that this motion requires a certain
minimum amount of energy. At very low temperatures
energy.
(below 80 K), the molecules do not have enough energy
to rotate.
rotate.
As the temperature increases from 80 K, first a few
As
molecules and then more and more obtain enough
energy to rotate, and CV/R increases, until all of them
are rotating and CV/R=2.5.
/R=2 Similarly,
Similarly, quantum theory shows that oscillatory motion
of the molecules requires a certain (higher) minimum
amount of energy. This minimum amount is not met
energy.
until the molecules reach a temperature of about 1000 K,
Fig. 19as shown in Fig. 1912.
As the temperature increases beyond 1000 K, more
As
molecules have enough energy to oscillate and CV/R
increases, until all of them are oscillating and CV/R=3.5.
In Fig. 1912, the plotted curve stops at 3200 K because
In Fig. 1912,
there the atoms of a hydrogen molecule oscillate so
much that they overwhelm (戰勝) their bond, and the
molecule then dissociates ( 分 解 ) into two separate
atoms.
atoms. Gas
Gas processes 定壓 等溫 絕熱
定容 Homeworks Ans: Ans: Ans: Ans: Ans: Ans:
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This note was uploaded on 12/27/2011 for the course LSCI 103 taught by Professor K.y.liao during the Fall '11 term at National Cheng Kung University.
 Fall '11
 K.Y.Liao

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