SME
University Physics I: Mechanics
Temperature, heat and kinetic theory
Use this for “Find the rms speed and the average kinetic energy of a nitrogen molecule at 300 K.”
Chart
Most probable vₚ = 422.1 m/s < mean ⟨v⟩ = 476.3 m/s < rms vᵣₘₛ = 516.9 m/s; the area under the curve is 1
This is an example. Type your own values to start a new problem.
vᵣₘₛ (m/s)
=
3
×
8.314
×
T (K)
×
10
−3
kg/mol
⟨K⟩
Average kinetic energy of a molecule
calculated · J
n
Amount of gas
example · mol
Kₜᵣ
Kinetic energy of the gas
calculated
Formulas
vᵣₘₛ
=
√(3
×
8.314
×
T
M
÷ 1,000
)
516.9
=
√(3
×
8.314
×
300
28 ÷ 1,000
)
⟨K⟩
=
3
2
×
1.381 ×
10
-23
×
T
6.2145 ×
10
-21
=
3
2
×
1.381 ×
10
-23
×
300
Kₜᵣ
=
3
2
×
n
×
8.314
×
T
3,741
=
3
2
×
1
×
8.314
×
300
Assumptions
•
An ideal gas in equilibrium: point molecules colliding elastically, with T in kelvins.
•
R = 8.314 J/(mol·K) and kB = 1.381 × 10⁻²³ J/K; M in g/mol is divided by 1000 for kg/mol.
•
At one temperature every gas has the same average kinetic energy, but not the same speed: lighter molecules move faster.
•
Kₜᵣ = (3/2)nRT is the energy of motion of n moles; it is the whole internal energy U only for a monatomic gas.
Step-by-step
WE KNOW
T = 300 K, M = 28 g/mol, n = 1 mol
FIND
root-mean-square speed (vᵣₘₛ), average kinetic energy of a molecule (⟨K⟩), kinetic energy of the gas (Kₜᵣ)
Step 1 · Find root-mean-square speed (vᵣₘₛ)
Use vᵣₘₛ = √(3 × 8.314 × T ÷ (M ÷ 1,000))
Each
molecule
averages
1
2
m⟨
v
2
⟩
=
(
3
2
)kBT,
so
⟨
v
2
⟩
=
3R
T
M
per
mole;
vᵣₘₛ
is
its
square
root.
vᵣₘₛ
=
√(3
×
8.314
×
300
28 ÷ 1,000
)
vᵣₘₛ
=
√(3
×
8.314
×
300
0.028
)
vᵣₘₛ
=
√(24.942
×
300
0.028
)
vᵣₘₛ
=
√(
7,482.6
0.028
)
vᵣₘₛ
=
√
267
,200
vᵣₘₛ = 516.9 m/s
Step 2 · Find average kinetic energy of a molecule (⟨K⟩)
Use ⟨K⟩ = 3/2 × 1.381 × 10⁻²³ × T
Temperature
measures
the
average
energy
of
motion:
(
3
2
)kBT
for
each
molecule,
whatever
its
mass.
⟨K⟩
=
3
2
×
1.381 ×
10
-23
×
300
⟨K⟩
=
(2.0715
×
10
-23
)
×
300
⟨K⟩
=
6.2145 ×
10
-21
J
Step 3 · Find kinetic energy of the gas (Kₜᵣ)
Use Kₜᵣ = 3/2 × n × 8.314 × T
Add
up
(
3
2
)kBT
over
all
the
molecules:
there
are
nNA
of
them,
and
NA·kB
=
R.
Kₜᵣ
=
3
2
×
1
×
8.314
×
300
Kₜᵣ = 12.471 × 300
Kₜᵣ = 3,741 J
Check
Put the numbers back in. Both sides should match.
516.9
=
√(3
×
8.314
×
300
28 ÷ 1,000
)
✓
6.2145 ×
10
-21
=
3
2
×
1.381 ×
10
-23
×
300
✓
3,741
=
3
2
×
1
×
8.314
×
300
✓
Topic check
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