SME
General Chemistry I · Problem type
Kinetic theory: rms speed and kinetic energy
Use this for “Find the root-mean-square speed and the average kinetic energy per mole of N₂ molecules at 300 K.”
Chart
Most probable vₚ = 422 m/s < mean ⟨v⟩ = 476.2 m/s < rms vᵣₘₛ = 516.9 m/s; the area under the curve is 1
Dashed: He at the same T, vₚ = 1,116 m/s
This is an example. Type your own values to start a new problem.
u (m/s)
=
3R
T (K)
÷
Mkg (kg/mol)
M
Molar mass
example · g/mol
KE
Average kinetic energy per mole
calculated · J/mol
uHe
Helium’s rms speed at the same temperature
worked out (not typed) · m/s
Formulas
Mkg
=
M
1,000
0.02801
=
28.01
1,000
u
=
√(3
×
8.314
×
T
Mkg
)
516.9
=
√(3
×
8.314
×
300
0.02801
)
KE
=
3
2
×
8.314
×
T
3,741
=
3
2
×
8.314
×
300
uHe
=
√(3
×
8.314
×
T
0.004003
)
1,367
=
√(3
×
8.314
×
300
0.004003
)
Assumptions
•
An ideal gas: the molecules are points that only bounce off each other and the walls, and their speeds follow the Maxwell distribution.
•
R = 8.314 J/(mol·K). A joule is kg·m²/s², so the molar mass goes in kg/mol for the speed to come out in m/s.
•
The average kinetic energy per mole, (3/2)RT, depends on T only. At the same T a lighter gas moves faster: helium is dashed beside yours.
Step-by-step
WE KNOW
T = 300 K, M = 28.01 g/mol
FIND
molar mass in kg/mol (Mkg), root-mean-square speed (u), average kinetic energy per mole (KE), helium’s rms speed at the same temperature (uHe)
Step 1 · Find molar mass in kg/mol (Mkg)
Use Mkg = M ÷ 1,000
R
is
in
joules,
and
a
joule
is
kg·
m
2
/
s
2
,
so
the
molar
mass
must
be
in
kg/mol:
divide
the
grams
by
1000.
Mkg
=
28.01
1,000
Mkg = 0.02801 kg/mol
Step 2 · Find root-mean-square speed (u)
Use u = √(3 × 8.314 × T ÷ Mkg)
Set
(
1
2
)Mu²
equal
to
(
3
2
)RT
and
solve
for
u
:
the
square
root
of
the
mean
of
the
squared
speeds.
u
=
√(3
×
8.314
×
300
0.02801
)
u
=
√(24.942
×
300
0.02801
)
u
=
√(
7,482.6
0.02801
)
u
=
√
267
,140
u
=
516.9
m/s
Step 3 · Find average kinetic energy per mole (KE)
Use KE = 3 ÷ 2 × 8.314 × T
The
average
kinetic
energy
of
a
mole
of
molecules
is
(
3
2
)RT,
whatever
the
gas.
KE
=
3
2
×
8.314
×
300
KE = 1.5 × 8.314 × 300
KE = 12.471 × 300
KE = 3,741 J/mol
Step 4 · Find helium’s rms speed at the same temperature (uHe)
Use uHe = √(3 × 8.314 × T ÷ 0.004003)
Helium (0.004003 kg/mol) at the same temperature has the same kinetic energy, so being lighter it moves faster (dashed).
uHe
=
√(3
×
8.314
×
300
0.004003
)
uHe
=
√(24.942
×
300
0.004003
)
uHe
=
√(
7,482.6
0.004003
)
uHe
=
√
1
,869,000
uHe = 1,367 m/s
Check
Put the numbers back in. Both sides should match.
0.02801
=
28.01
1,000
✓
516.9
=
√(3
×
8.314
×
300
0.02801
)
✓
3,741
=
3
2
×
8.314
×
300
✓
1,367
=
√(3
×
8.314
×
300
0.004003
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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