Molecular Speed Distribution in Thermal Physics
The Distribution of Molecular Speeds
The root-mean-square speed
v
rms
gives us a general idea of
molecular speeds in a gas at a given temperature.
But, we often want to know more about how the possible values of
speed are distributed among the molecules.
The speed distribution for oxygen molecules at room temperature
(
T
= 300 K) is shown below.
Chapter 19
Kinetic Theory
The Distribution of Molecular Speeds
Chapter 19
Kinetic Theory
In 1852, Scottish physicist James Clerk Maxwell first solved the
problem if finding the speed distribution of gas molecules.
The resulting
Maxwell’s speed distribution
law can be written
as:
M
: molar mass of the gas
R
: gas constant
T
: gas temperature
v
: molecular speed
The quantity
P
(
v
) is a probability distribution
function.
For any speed
v
, the product
P
(
v
)
dv
(an area,
dimensionless) is the fraction of molecules
with speed in the interval
dv
centered on
speed
v
.
The Distribution of Molecular Speeds
Chapter 19
Kinetic Theory
As shown in the figure, the fraction of molecules with speeds in the
interval
dv
is equal to the area of a strip with height
P
(
v
) and
width
dv
.
The total area under the distribution curve corresponds to the
fraction of the molecules whose speeds lie between zero and
infinity.
Since all molecules fall into this category, the value of this total
area is unity:
The fraction (frac) of
molecules with speeds
in an interval of
v
1
and
v
2
is then:
The Distribution of Molecular Speeds
Chapter 19
Kinetic Theory
As the speed is a function of temperature, the speed distribution
also varies with temperature. The distribution at
T
= 300 K is
compared with the one at
T
= 80 K in the following figure.
Average, RMS, and Most Probable Speeds
Chapter 19
Kinetic Theory
In principle, we can find the average speed
v
avg
of the molecules in
a gas by evaluating:
After substituting
P
(
v
) and performing the integral,
Average Speed
Similarly, we can find the average of the square of the speeds
v
2
avg
with
After substituting
P
(
v
) and performing the integral,
Average, RMS, and Most Probable Speeds
Chapter 19
Kinetic Theory
Thus,
RMS Speed
The most probable speed
v
P
is the speed at which
P
(
v
) is
maximum. To calculate
v
P
, we set
dP
/
dv
=
0 and then solve for
v
.
Doing so, we find:
Most Probable Speed
A molecule is more likely to have speed
v
P
than any other speed,
but some molecules will have speeds that are many times
v
P
.
These molecules lie in the
high-speed tail
of a distribution curve.
Problem
A container is filled with oxygen gas maintained at room temperature
(300 K). What fraction of the molecules have speeds in the interval
599 to 601 m/s? The molar mass
M
of oxygen is 0.0320 kg/mol.
Chapter 19
Kinetic Theory
The interval
Δ
v
= 2 m/s is very small compared to the center speed
v
=
600 m/s. Thus the integration can be approximated through:
The Molar Specific Heats of an Ideal Gas
Chapter 19
Kinetic Theory
In this section, we want to derive an expression for the internal
energy
E
int
of an ideal gas from molecular consideration.
In other words, we want an expression for the energy associated
with the random motions of the atoms or molecules in the gas.
We shall then use that expression to derive the molar specific
heats of an ideal gas.
The Molar Specific Heats of an Ideal Gas
Chapter 19
Kinetic Theory
Internal Energy
E
int
Let us firs
t
assume that the ideal gas is a
monatomic gas
such as
helium, neon, or argon.
The internal energy
E
int
of the ideal gas is simply the sum of the
translational kinetic energies of its atoms.
The average translational kinetic energy of a single atom depends
only on the gas temperature and is given as
K
avg
=3/2·
kT.
A sample of
n
moles of such a gas contains
nN
A
atoms. The
internal energy of the sample is then:
Since
k
=
R
/
N
A
, we can rewrite this as:
Monatomic Ideal Gas
The internal energy
E
int
of an ideal gas is a function of the gas
temperature
only
.
The Molar Specific Heats of an Ideal Gas
Chapter 19
Kinetic Theory
Molar Specific Heat at Constant Volume
The figure on the right shows
n
moles of an
ideal gas at pressure
p
and temperature
T
,
confined to a cylinder of fixed volume
V
.
This initial state is denoted as
i
.
Suppose a small amount of energy to the gas
as heat
Q
is added to the gas.
The gas temperature rises a small amount
T
+
Δ
T
, and its pressure rises to
p
+
Δ
p
.
This final state is denoted as
f
.
We would find that the heat
Q
is related to the
temperature change
Δ
T
by:
Constant
Volume
C
V
is a constant called the
molar specific heat
at constant volume
[J/mol·K].
The Molar Specific Heats of an Ideal Gas
Chapter 19
Kinetic Theory
The first law of thermodynamics can now be written as:
Monatomic Gas
With the volume held constant, the gas cannot do any work,
W
=
0.
This yields:
Thus,
We can now generalize the equation for the internal energy as:
Any Ideal Gas
The Molar Specific Heats of an Ideal Gas
Chapter 19
Kinetic Theory
When an ideal gas that is confined to a container undergoes a
temperature change
Δ
T
, then we can write the resulting change in
its internal energy as:
Any Ideal Gas, Any Process
The change in the internal energy
E
int
of a confined ideal gas depends
on the change in the gas
temperature
only
. It does not
depend on what type of process
produces the change in the
temperature.
The Molar Specific Heats of an Ideal Gas
Chapter 19
Kinetic Theory
Molar Specific Heat at Constant Pressure
We now assume that the temperature of our
ideal gas is increased by the same small
amount
Δ
T
as previously but now the
necessary energy (heat
Q
) is added with the
gas under constant pressure.
From such experiments we find that the heat
Q
is related to the temperature change
Δ
T
by:
Constant
Pressure
C
P
is a constant called the
molar specific heat
at constant pressure
[J/mol·K].
C
P
>
C
V
, because energy must now be supplied
not only to raise the temperature of the gas but
also for the gas to do
work
.
The Molar Specific Heats of an Ideal Gas
Chapter 19
Kinetic Theory
To relate the molar specific heats
C
P
and
C
V
, we start with the first
law of thermodynamics:
We will find that:
The Molar Specific Heats of an Ideal Gas
Chapter 19
Kinetic Theory
The Equipartition of Energy
Chapter 19
Kinetic Theory
The figure shows the
ratio
C
v
/
R
as a function
of temperature for
H
2
gas:
Below 80
K,
C
v
/
R
=
3/2,
gas behaves as though it is mon
o
atomic
.
Only
translational modes of motion are excited.
At 2
00-
3
00
K,
C
v
/
R
=
5/2,
rotational modes become excited.
At above 3000
K,
C
V
/
R
=
7/2, v
ibrational modes become excited
.
Further Problem
Chapter 19
Kinetic Theory
(a)
What is the average translational kinetic
energy of a molecule of
an ideal gas at a temperature of
27°C ?
(b)
What is the total random translational kinetic energy of
the
molecules in 1 mole of this gas?
(a)
(b)
Checkpoint
The figure here shows five paths traversed by a gas on a
p
-
V
diagram. Rank the paths according to the change in internal energy
of the gas, greatest first.
5
,
t
h
e
n
t
i
e
o
f
1
,
2
,
3
,
a
n
d
4
.
Chapter 19
Kinetic Theory
Problem
A bubble of 5.00 mol of helium is submerged at a certain depth in
liquid water when the water (and thus the helium) undergoes a
temperature increase
Δ
T
of 20.0°C at constant pressure. As a result,
the bubble expands. The helium is monatomic an ideal.
(a)
How much energy is added to the helium as heat during the
increase and expansion?
Chapter 19
Kinetic Theory
(b)
What is the change
Δ
E
int
in the internal energy of the helium
during the temperature increase?
(c)
How much work W is done by the helium as it expands against
the pressure of the surrounding water during the temperature
increase?
Class Group Assignments
1.
The rms velocity of oxygen molecules at 27°C is 318 m/s. Thus, the rms velocity of
hydrogen molecules at 127 °C is:
(a) 1470 m/s
(b) 1603 m/s
(c) 1869 m/s
(d) 2240 m/s
(e) 3211 m/s
2.
An ideal gas of
N
monoatomic molecules is in thermal equilibrium with an ideal gas
of the same number of diatomic molecules and equilibrium is maintained as the
temperature is increased. The ratio of the changes in the internal energies
Δ
E
dia
/
Δ
E
mon
is:
(a) 1/2
(b) 3/5
(c) 1/1
(d) 5/3
(e) 2/1
3.
An ideal gas has molar specific heat
C
P
at constant pressure. When the temperature
of
n
moles is increased by
Δ
T
the increase in the internal energy is:
(a)
nC
P
Δ
T
(d)
n
(
C
P
+R
)
Δ
T
(b)
n
(
C
P
–R
)
Δ
T
(e)
n
(2
C
P
+R
)
Δ
T
(c)
n
(2
C
P
–R
)
Δ
T
Chapter 19
Kinetic Theory
No Homework This Week
Chapter 19
Kinetic Theory
Prepare well for the midterm exam.
Exploring the distribution of molecular speeds in gases at different temperatures through the Maxwell-Boltzmann speed distribution law. This lecture covers the concepts of probability distribution functions for speed intervals, comparison of speed distributions at varying temperatures, and calculations for average, root-mean-square (RMS), and most probable speeds of gas molecules.
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Presentation Transcript
Thermal Physics Lecture 5 Dr.-Ing. Erwin Sitompul President University http://zitompul.wordpress.com 2 0 1 8 President University Erwin Sitompul Thermal Physics 5/1
Chapter 19 Kinetic Theory The Distribution of Molecular Speeds The root-mean-square speed vrms gives us a general idea of molecular speeds in a gas at a given temperature. But, we often want to know more about how the possible values of speed are distributed among the molecules. The speed distribution for oxygen molecules at room temperature (T = 300 K) is shown below. President University Erwin Sitompul Thermal Physics 5/2
Chapter 19 Kinetic Theory The Distribution of Molecular Speeds In 1852, Scottish physicist James Clerk Maxwell first solved the problem if finding the speed distribution of gas molecules. The resulting Maxwell s speed distribution law can be written as: 3 2 2 2 ( ) 4 2 RT M R T v M 2 = Mv RT P v v e : molar mass of the gas : gas constant : gas temperature : molecular speed The quantity P(v) is a probability distribution function. For any speed v, the product P(v)dv (an area, dimensionless) is the fraction of molecules with speed in the interval dv centered on speed v. President University Erwin Sitompul Thermal Physics 5/3
Chapter 19 Kinetic Theory The Distribution of Molecular Speeds As shown in the figure, the fraction of molecules with speeds in the interval dv is equal to the area of a strip with height P(v) and width dv. The total area under the distribution curve corresponds to the fraction of the molecules whose speeds lie between zero and infinity. Since all molecules fall into this category, the value of this total area is unity: = ( ) 1 P v dv 0 The fraction (frac) of molecules with speeds in an interval of v1 and v2 is then: frac = v 2 ( ) P v dv v 1 President University Erwin Sitompul Thermal Physics 5/4
Chapter 19 Kinetic Theory The Distribution of Molecular Speeds As the speed is a function of temperature, the speed distribution also varies with temperature. The distribution at T = 300 K is compared with the one at T = 80 K in the following figure. President University Erwin Sitompul Thermal Physics 5/5
Chapter 19 Kinetic Theory Average, RMS, and Most Probable Speeds In principle, we can find the average speed vavg of the molecules in a gas by evaluating: = ( ) v vP v dv avg 0 After substituting P(v) and performing the integral, 8RT v M = Average Speed avg Similarly, we can find the average of the square of the speeds v2avg with 2 2 avg 0 After substituting P(v) and performing the integral, 3RT v M = ( ) v v P v dv = 2 avg President University Erwin Sitompul Thermal Physics 5/6
Chapter 19 Kinetic Theory Average, RMS, and Most Probable Speeds Thus, 3RT v M = RMS Speed rms The most probable speed vP is the speed at which P(v) is maximum. To calculate vP, we set dP/dv=0 and then solve for v. Doing so, we find: 2 P v M RT = Most Probable Speed A molecule is more likely to have speed vP than any other speed, but some molecules will have speeds that are many times vP. These molecules lie in the high-speed tail of a distribution curve. President University Erwin Sitompul Thermal Physics 5/7
Chapter 19 Kinetic Theory Problem A container is filled with oxygen gas maintained at room temperature (300 K). What fraction of the molecules have speeds in the interval 599 to 601 m/s? The molar mass M of oxygen is 0.0320 kg/mol. The interval v = 2 m/s is very small compared to the center speed v=600 m/s. Thus the integration can be approximated through: 3 2 M 2 = 2 2 Mv RT frac 4 v e v 2 RT 3 2 0.0320 ( ) 2 = (601 599) (0.0320)(600) 2(8.31)(300) 2 4 (600) e 2 (8.31)(300) 2.31 = (1.321 10 )( ) (2) 2 e 2.622 10 = 3 = 0.2622% President University Erwin Sitompul Thermal Physics 5/8
Chapter 19 Kinetic Theory The Molar Specific Heats of an Ideal Gas In this section, we want to derive an expression for the internal energy Eint of an ideal gas from molecular consideration. In other words, we want an expression for the energy associated with the random motions of the atoms or molecules in the gas. We shall then use that expression to derive the molar specific heats of an ideal gas. President University Erwin Sitompul Thermal Physics 5/9
Chapter 19 Kinetic Theory The Molar Specific Heats of an Ideal Gas Internal Energy Eint Let us first assume that the ideal gas is a monatomic gas such as helium, neon, or argon. The internal energy Eint of the ideal gas is simply the sum of the translational kinetic energies of its atoms. The average translational kinetic energy of a single atom depends only on the gas temperature and is given as Kavg =3/2 kT. A sample of n moles of such a gas contains nNA atoms. The internal energy of the sample is then: ( ) ( )( E nN K nN kT = = ) 3 2 int A avg A Since k = R/NA, we can rewrite this as: 3 int 2 E nRT = Monatomic Ideal Gas The internal energy Eint of an ideal gas is a function of the gas temperature only. President University Erwin Sitompul Thermal Physics 5/10
Chapter 19 Kinetic Theory The Molar Specific Heats of an Ideal Gas Molar Specific Heat at Constant Volume The figure on the right shows n moles of an ideal gas at pressure p and temperature T, confined to a cylinder of fixed volume V. This initial state is denoted as i. Suppose a small amount of energy to the gas as heat Q is added to the gas. The gas temperature rises a small amount T+ T, and its pressure rises to p+ p. This final state is denoted as f. We would find that the heat Q is related to the temperature change T by: Q nC T = Constant Volume V CV is a constant called the molar specific heat at constant volume [J/mol K]. President University Erwin Sitompul Thermal Physics 5/11
Chapter 19 Kinetic Theory The Molar Specific Heats of an Ideal Gas The first law of thermodynamics can now be written as: E nC T W = int V With the volume held constant, the gas cannot do any work, W=0. This yields: E C n T n T 3 2nR T = = int V Thus, = = 12.5 J mol K C R 3 2 Monatomic Gas V We can now generalize the equation for the internal energy as: E nC T = Any Ideal Gas int V President University Erwin Sitompul Thermal Physics 5/12
Chapter 19 Kinetic Theory The Molar Specific Heats of an Ideal Gas When an ideal gas that is confined to a container undergoes a temperature change T, then we can write the resulting change in its internal energy as: = E nC T Any Ideal Gas, Any Process int V The change in the internal energy Eint of a confined ideal gas depends on the change in the gas temperature only. It does not depend on what type of process produces the change in the temperature. President University Erwin Sitompul Thermal Physics 5/13
Chapter 19 Kinetic Theory The Molar Specific Heats of an Ideal Gas Molar Specific Heat at Constant Pressure We now assume that the temperature of our ideal gas is increased by the same small amount T as previously but now the necessary energy (heat Q) is added with the gas under constant pressure. From such experiments we find that the heat Q is related to the temperature change T by: = Q nC T Constant Pressure P CP is a constant called the molar specific heat at constant pressure [J/mol K]. CP> CV, because energy must now be supplied not only to raise the temperature of the gas but also for the gas to do work. President University Erwin Sitompul Thermal Physics 5/14
Chapter 19 Kinetic Theory The Molar Specific Heats of an Ideal Gas To relate the molar specific heats CP and CV, we start with the first law of thermodynamics: E Q W = = int T nC nC T nR T V P We will find that: C = = + C R V P C C R P V President University Erwin Sitompul Thermal Physics 5/15
Chapter 19 Kinetic Theory The Molar Specific Heats of an Ideal Gas President University Erwin Sitompul Thermal Physics 5/16
Chapter 19 Kinetic Theory The Equipartition of Energy The figure shows the ratio Cv/R as a function of temperature for H2 gas: Below 80 K, Cv/R = 3/2, gas behaves as though it is monoatomic. Only translational modes of motion are excited. At 200-300 K, Cv/R = 5/2, rotational modes become excited. At above 3000 K, CV/R = 7/2, vibrational modes become excited. President University Erwin Sitompul Thermal Physics 5/17
Chapter 19 Kinetic Theory Further Problem (a) What is the average translational kinetic energy of a molecule of an ideal gas at a temperature of 27 C ? (b) What is the total random translational kinetic energy of the molecules in 1 mole of this gas? (a) = = 23 = 21 3 2(1.38 10 )(300) K kT 3 2 6.21 10 J avg (b) = = (1)(6.02 10 )(6.21 10 = = 23 21 3738.4 J ) ( ) K E nN K total int avg A President University Erwin Sitompul Thermal Physics 5/18
Chapter 19 Kinetic Theory Checkpoint The figure here shows five paths traversed by a gas on a p-V diagram. Rank the paths according to the change in internal energy of the gas, greatest first. 5, then tie of 1, 2, 3, and 4. = E nC T int V President University Erwin Sitompul Thermal Physics 5/19
Chapter 19 Kinetic Theory Problem A bubble of 5.00 mol of helium is submerged at a certain depth in liquid water when the water (and thus the helium) undergoes a temperature increase T of 20.0 C at constant pressure. As a result, the bubble expands. The helium is monatomic an ideal. (a) How much energy is added to the helium as heat during the increase and expansion? = = = 2077.5J (5)( )(20) R Q nC T 5 2 P (b) What is the change Eint in the internal energy of the helium during the temperature increase? = 2 (5)( )(20) R = = 1246.5 kJ E nC T 3 int V (c) How much work W is done by the helium as it expands against the pressure of the surrounding water during the temperature increase? = = nR T = = 2077.5 1246.5 831kJ W Q E int President University Erwin Sitompul Thermal Physics 5/20
Chapter 19 Kinetic Theory Class Group Assignments 1. The rms velocity of oxygen molecules at 27 C is 318 m/s. Thus, the rms velocity of hydrogen molecules at 127 C is: (a) 1470 m/s (b) 1603 m/s (c) 1869 m/s (d) 2240 m/s (e) 3211 m/s 2. An ideal gas of N monoatomic molecules is in thermal equilibrium with an ideal gas of the same number of diatomic molecules and equilibrium is maintained as the temperature is increased. The ratio of the changes in the internal energies Edia/ Emon is: (a) 1/2 (b) 3/5 (c) 1/1 (d) 5/3 (e) 2/1 3. An ideal gas has molar specific heat CP at constant pressure. When the temperature of n moles is increased by T the increase in the internal energy is: (a) nCP T (d) n(CP+R) T (b) n(CP R) T (e) n(2CP+R) T (c) n(2CP R) T President University Erwin Sitompul Thermal Physics 5/21
Chapter 19 Kinetic Theory No Homework This Week Prepare well for the midterm exam. President University Erwin Sitompul Thermal Physics 5/22
