Clausius molecular speed
Solutions
1) The root mean square speed is
3RT
u=
,
M
where R is the ideal gas constant, T the absolute temperature, and M the molar mass of the gas.
If we express everything in MKS units, we will get a speed in MKS units, namely m/s. That
means we take the gas constant to be R = 8.3145 J mol–1 K–1 and we must express the molar
mass in kg/mol rather than the more usual g/mol. The temperature of melting ice is, of course,
0°C = 273.15 K. Putting in the numbers, we get:
for O2:
u=
for N2:
u=
for H2:
u=
3(8.3145 J mol -1 K-1 )(273.15 K)
-3
2 16.00  10
kg mol
-1
3(8.3145 J mol -1 K-1 )(273.15 K)
-3
2  14.0110
kg mol
-1
3(8.3145 J mol -1 K-1 )(273.15 K)
-3
-1
2 1.008  10 kg mol
It appears that Clausius’ estimates were dead on.
= 461 m s–1 ,
= 493 m s–1 ,
= 1838 m s–1 .
2) Begin with the textbook formula for root mean square speed,
3RT
u=
,
M
and compare it to Clausius’,
T
u = (485 m/s)
.
 (273 K)
We already have the correct temperature dependence. We must relate the molar mass factor to
the density factor in the denominator and reconcile the constants. First, let us relate  to M.  is
the density d of the gas divided by that of air:
 = d/dair .
Density is simply mass over volume, but we must have comparable conditions for both the gas
and the air. For the sake of convenience, let us consider a one-mole sample both of the gas and
of air. (The notion of a mole of air is a bit strange, since air is not a pure substance. A mole of
dry air is well defined, though. It is Avogadro’s number of gas molecules, 78% of which are
nitrogen, 21% oxygen, 1% argon, etc. The mass of a mole of dry air is 28.964 g/mol.) The
density of the gas is M/V and the density of the air is Mair/Vair . But V and Vair are the same, for
one mole of any ideal gas occupies the same volume under the same conditions of temperature
and pressure. The end result of these considerations is:
 = M/Mair .
Since we want to go from the textbook formula to Clausius’, we want to express M in terms of ,
not the other way around, so the form of the equation useful to us is:
M = Mair .
Putting this into the speed formula gives:
3RT
3R(273 K)
T
.

M air
M air
(273 K)
In the last equality, I wrote the temperature and density terms separate from the constants, and I
multiplied and divided by the square root of 273 K. All that remains is to plug numbers into the
first square root and hope that they work out to the overall constant Clausius has:
u=
u=
3(8.3145 J mol -1 K-1)(273 K)
3
28.964  10 kg mol
which is the expression to be derived.
-1
T
T
 (485 m/s)
,
(273 K)
(273 K)