Type:
Double
Date:___________
Objective:
Kinetic Energy of an Ideal Gas I
Kinetic Energy of an Ideal Gas II
Homework: Read 14.3, Do Concept Q. # (15),
Do Problems # (28, 29, 31, 37)
AP Physics
Mr. Mirro
Date ___________
Kinetic Energy of an Ideal Gas I
Three physical properties – pressure (P), volume (V) and temperature (T) – describe a gas. At low
densities, all gases approach an ideal behavior. This means that these three variables are related by
the Ideal Gas Law.
PV = nRT
Where n is the number of moles of gas and
R is the Universal Gas Constant (8.31 J/mol K)
An important consequence of this equation is that, for a fixed volume of gas, an increase in pressure (P)
gives a proportional increase in temperature (T).
The pressure increases when the gas molecules strike the
walls of their container with more force, which occurs
if they move more rapidly.
Consider the macroscopic container shown
at right. The container which is filled with an
Ideal Gas at Standard Temperature and Pressure
(STP) contains an enormous amount of particles.
These particles are in constant, random motion, colliding
with each other and with the walls of the container.
In the course of one second, a particle undergoes many collisions and each one changes the
particle’s speed and direction of motion.
As a result, the atoms or molecules have many different speeds.
It is possible, however, to speak about an average particle speed. At any given instant, SOME particles
have speeds less than, SOME near and SOME greater than the average.
For conditions of low gas density, the
various speeds within a large collection
of molecules held at a constant temperature
possess a uniform distribution of speeds.
Therefore, an average value of v2 must be
used rather than the speed of any individual
particle to determine the kinetic energy
of the distribution.
For example, consider oxygen gas (O2) at two different absolute temperatures. When the gas is at an:
‰
Absolute temperature of 300 K, the average speed in the curve is approximately 400 m/s.
‰
Absolute temperature of 1200 K, the distribution shifts to the right and speed increases to 800 m/s.
Therefore, if we wish to discuss the speed of the “distribution”, statistically we must use the
square-root of the square of the average particle speed.
FTBO ⇒
The square-root of the quantity v2avg is called the “root-mean-square speed,” can be written as:
vrms = √ (v2) avg
which is NOT equal to √ (vavg)2
We can develop a relationship between the r-m-s speed (vrms) and the average kinetic energy (KEavg) for
an Ideal Gas, in terms of P, V, T and R, by analyzing the momentum and energy of collisions occurring
in the sample.
First, we can use Newton’s Second Law, in the form of momentum, to find the force exerted on
the particle.
rate of change of momentum = force on particle
ΔP = F on-partical
t
Fon-partical = pF – pi =
t
⇒ ma
(-mvF) – mvi
t
Since the collision is elastic and the molecule travels there and back (d = 2L)
Therefore, velocity
Yeilds,
vrms = d = 2L
t
t
[ Solving for time (t) in terms of vrms and (L) ]
t = 2L
vrms
By substituting t = 2L/v and realizing that the magnitude of the initial and final velocity is (v) we can
write:
Fon-partical = -m vrms – m vrms = - (2) mvrms =
2L
(2) L/vrms
vrms
-mvrms2
L
According to Newton’s Third Law (action-reaction), the
force applied to the wall by the particle is equal in magnitude
but opposite in direction yielding:
Fon-wall = +mvrms2
L
Since the (N) particles move randomly in three dimensions,
one-third of them on-average strike the right wall during
the time (t). Therefore, the total force is:
Fon-wall = 1 N mvrms2
3
L
AP Physics
Mr. Mirro
Date ___________
Kinetic Energy of an Ideal Gas I (cont.)
Now, let’s consider the force (F) that the molecule exerts per unit area (A).
P on-wall = Fon-wall = 1/3 N (mvrms2)/L = 1/3 N m vrms2
A
L2
L3
Since the volume of the box is V = L3
P on-wall =
1/3 N m vrms2
V
Therefore, it can be shown that the pressure – exerted by N molecules of gas in a container of volume (V)
is related to the average kinetic energy of the molecules by the equation:
PV = 1 N m vrms2 = 1 2 N [m vrms2]
3
2 3
PV = 2 N [½ m vrms2] =
3
2 N KEavg
3
Finally, comparing this to the Ideal Gas Law, [PV = nRT], we can further develop our relation in terms of
gas variables.
n RT = 2 N KEavg
3
Since Avogadro’s number NA = N , we can rewrite the equations as:
n
RT = 2
3
N
n
KEavg = 2 NA KEavg
3
KEavg = 3 R T
2 NA
Where the ratio, R is a fundamental constant of nature
NA
called Boltzmann’s constant (kB), where:
B
kB = R =
8.31 J/mol K
=
6.022 x 1023 atm/mol
NA
B
1.38 x 10-23 J/K
Therefore, the equation for the average kinetic energy becomes:
KEavg = 3
2
R T =
NA
3 kB T
2
FTBO ⇒
Similarly, several equations for the r-m-s speed emerge in terms of gas variables as follows:
Case #1: Relating rms speed and kinetic energy.
2
KEavg = 1 m vrms
[ solving for the r-m-s velocity in terms of mass and energy ]
2
vrms =
2 (KE)
√ mparticle
Case #2: Relating rms speed and Boltzmann’s constant.
Since
2
KEavg = 3 kB T = 1 m vrms
2
2
3 1 kB T = 1 m vrms2
2
2
3 kB T =
vrms =
m vrms2
3 kB T ⇒
3 kBT
√ mparticle
√
μ
B
Case #3: Relating rms speed and the Universal Gas constant.
vrms =
vrms =
3 kB T =
√ mparticle
√
B
√
3 R T
Mper mol
3 R T____
NA mparticle
⇒
3 RT
√ M
Recall that the mass (m = μ) of a substance expressed in grams or kilograms, is related to the number
of moles (n) of that substance by the expression [ m = n M = M/NA ], where (M) is the molar mass,
usually expressed in grams per mole (g/mol) or kilograms per mole (kg/mol).
AP Physics
Mr. Mirro
Date ___________
Kinetic Energy of an Ideal Gas Law I
Ex 1: Given that helium has a molecular mass (M) of 4.0026 g/mol (kg/mol = ??).
[PrincetonReview9.8mod]
a. Convert the molecular mass (M) of helium into mass (m) in kilograms.
b. Determine the rms speed of the helium molecules at a room temperature ( ≈ 25 °C),
i.
ii.
Using the rms speed and the Universal Gas constant (R).
Using the rms speed and Boltzmann’s constant (kB).
c. How much faster do the helium molecules move on a comfortably warm day (≈ 30 °C)
Ex 2: Determine the average kinetic energy of the helium molecules from the previous example at
TRoom = 25 °C,
a. Using the rms speed and mass (m) in kilograms.
b. Using the rms speed and Boltzmann’s constant ( kB).
c. What is the average kinetic energy of these helium molecules at TRoom = 25 °C ?
AP Physics “B”
Mr. Mirro
Date: ________
Kinetic Energy of an Ideal Gas II
Ex 1: Nitrogen gas surrounds us all, yet we barely notice that this benign substance exists ! It turns out
that air is primarily a mixture of 78 % nitrogen N2 (M = 28 u), 15 % oxygen O2 (M = 32 u)
plus negligible amounts of several other gases. If the absolute temperature of the gas is 293 K
(20 °C ≈ 68 °F) and each particle behaves as an ideal gas, determine: [Cutnell14.6]
a. The average kinetic energy per particle for both nitrogen and oxygen.
b. The mass (in kg) for both a nitrogen (N2) and oxygen (O2) molecule.
c. The rms speed for both nitrogen and oxygen.
<i> Nitrogen (N2)
<ii> Oxygen (O2)
FTBO ⇒
Ex 2: Two moles of Argon gas (Ar) [M = 39.948 u or g/mol] are placed in a container whose volume
is 5 x 10-3 m3. [CutnellP14.28sim]
a. How many Ar molecules are in this sample of gas ?
b. If the pressure of the gas is 3 x 105 Pa, what is the average translational kinetic energy of
an Argon molecule ?
Ex 3: Near the surface of Venus, the rms speed of carbon dioxide molecules (CO2) is 650 m/s. If the atomic
masses for carbon and oxygen gas are C = 12 u and O2 = 2 (16u), respectively: [CutnellP14.30]
a. What is the mass (in kg) for carbon dioxide (CO2) ?
b. What is the temperature (in Kelvins) of the atmosphere at that point ?