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Internal Energy
11.2.4 Internal energy Relation between temperature and molecular kinetic
energy. The Boltzmann constant
In Summary…
 Internal Energy is a fancy Physics word for the heat
contained in an object. When we talk about an
object having a high temperature can also say that it
has a great deal of internal energy.
 At an atomic level this energy is kinetic and the
vibrational energy of the atoms or molecules. At AS /
A2 level we don’t consider any rotational energies
which also exist.
 A key point is that the internal energy of an ideal gas
depends only on temperature.
Key Assumptions I – Newton's Laws
1. Every object in a state of uniform motion tends to remain
in that state of motion unless an external force is applied
to it. (mv – momentum)
2. The relationship between an object's mass m, its
acceleration a, and the applied force F is F = ma.
3.
For every force there is an force equal acting in the
opposite direction. A ↔B
Key Molecular Assumptions II
1.
A gas consists of point molecules (i.e.
volume particles << volume container
2.
Gas molecules behave randomly
3.
Molecules collide elastically
4.
Except when in collision molecules have
negligible forces on each other
5.
Duration of collisions is << time between
collisions themselves
Results….
•
Now we can equate our two derived
formulae to find a useful and neat
construct for the Kinetic Energy of
particles.
1 2
p  c
3
and
pV  nRT
nRT
p
V
so
nRT 1 2
 c
V
3
Equate
Sub  = (m/v)
xV
x3
/2N
answer
nRT 1 2
 c
V
3
nRT 1 Nm 2

c
V
3 V
1
2
nRT  Nmc
3
3nRT  Nmc 2
3nRT 1 2
 mc
2N
2
1 2 3nRT
KE  mc 
2
2N
Alternative Results….
So we can now see that KE of a gas
particle as;
3nRT 1 2
KE 
 mc
N
2
But as we also know that;
N  n NA
N
n
NA
1
n

NA N
n = number of moles of a gas
N = number of particles of a gas
NA=number of particles in a mole
We can work out this equation in a different form and
create a substitution factor
1 2 3nRT
mc 
2
N
 n  3RT
KE    
N 2
KE 
 1  3RT
 
KE  
 NA  2
3RT
KE 
2N A
An other useful constant….
We can also actually describe our formulae
not in respect of lots of particles but only
one particle. There is a simple
conversion of the molar gas constant;
R
k
NA
R = 8.31 J K-1 mol-1 (molar gas constant)
k= 1.38 x 10-23 JK-1( Boltzmann constant)
NA= 6.022 x 1023 particles per mole
Internal Energy….
So from our new equation it also makes
sense that the internal energy of a
gas or sum of all the individual KE’s
will simply be;
So by substitution;
3RT
KE 
2N A
 R  3T

KE  
 NA  2
3kT
KE 
2
3
KE  kT
2
3
U  RT
2
Example 1
What is the total kinetic energy of 2 moles of Oxygen molecules at a temperature
of 400K
3RT 1 2
 mc  KE
2N A 2
3RT
KE 
2N A
KE =
3 x 8.31 J K-1 mol-1 x 400K
2 x 6.022 x 1023 molecules mol-1
= 8.28 x 10-21 J per molecule
But that is per atom so need to multiply by Avogadro's number and x2 to get for
the whole gas;
KE = 9972J
Example 1 – Alternate Method
What is the total kinetic energy of 2 moles of Oxygen gas at a temperature of
400K
3RT
U
n
2
KE = 1.5x 8.31 J K-1 mol-1 x 400K x 2 moles
KE = 9972J
Example 2
What is the kinetic energy of 1 Molecule of Oxygen gas at a temperature of 310K
3RT 1 2
KE 
 mc
2N A 2
or use
3
KE  kT
2
KE = 1.5 x 8.31 J K-1 mol-1 x 310K
6.022 x 1023 atoms per mole
KE = 6.4 x 10-21J per atom
Example 3
What is the temperature of a canister of CO2 gas if each molecule has a
kinetic energy of 1.23 x 10-20J
3
KE  kT
2
2 KE
T
3k
T=
2 x 1.23 x 10-20 J
3 x 1.38 x 10-23 JK-1
T = 594K
Example 4
What temperature is this monatomic gas at if a container contains 5 moles and
has a kinetic energy of 58 x 10-21J for each atom
3
KE  kT
2
2 KE
T
3k
T=
2 x 58 x 10-21J atom-1
3 x 1.38 x 10-23 J K-1 mol-1
T = 2802K