Ch. 19: The Kinetic Theory of Gases

Ch. 19: The Kinetic Theory of Gases
In this chapter we consider the physics of gases. If the atoms or molecules
that make up a gas collide with the walls of their container, they exert a
pressure p on it. This p is a function of the quantity of gas n (the number of
moles) in volume V at (Kelvin) temperature T:
p = p (n, V, T)
An equation relating p, n, V, and T is called an equation of state. Such
relations exist for all materials and phases, not just gases. The simplest such
expression is the ideal gas law,
p = nRT/V,
where R is the (universal) gas constant, 8.31 J/mol K. The derivation of this
law assumes that the atoms or molecules of the gas (1) do not interact with
one another except for perfectly elastic collisions, and (2) are point particles,
i.e. have zero volume.
[If two containers of gas have the same volume, temperature and pressure all three of these variables are easily measured - then they automatically
have the same number of particles, regardless of the chemical composition!]
The number of particles N of any material is proportional to the number of
moles n,
N = N A n,
where the proportionality constant N A is Avogadro’s number, approximately
6.02 x 1023. [By definition, one mole is the number of atoms in a 12 gram
sample of C-12. The mass of one C-12 atom can be accurately determined
using a mass spectrometer. Dividing this atomic mass into 12 grams gives
Avogadro’s number. In other words, the more precisely one can measure
the atom’s mass, the more significant figures one has for Avogadro’s
number.]
Ex: Suppose one knows (measures) the temperature, volume, and pressure of
a container of gas: How many molecules does it contain?
Since n = N/N A, it follows that N = N ApV/RT.
Ex: What is the mass of the above sample of gas?
If each molecule has mass m, the sample’s mass is M sam = mN = mN ApV/RT
= MpV/RT = Mn, where M = mN A = molar mass of the material.
Note that since mN = Mn, m/M = n/N = 1/N A.
Since pV = N(R/N A)T from above, the ideal gas law can also be written as
pV = N k T
where k = R/N A = Boltzmann’s constant = 1.38 x 10 -23 J/K.
If the work is given by
W = mpdV,
(which is the only kind of work done by a gas that we will consider), then
for an isothermal process (T = constant)
W = n RT m(dV/V) = n RT ln (V f /V i).
Similarly, if the process is isobaric (p = constant),
W = p mdV = p(V f - V i).
Finally, for an isochoric process (V = constant),
W = 0.
The molecules of a gas at any instant are moving with a variety of speeds,
and in different directions. The probability of any given molecule having a
speed between v and v + dv is given by the Maxwell distribution (1852):
P(v)dv = 4 B (M/2 B RT)3/2v 2exp(-Mv2/2RT)dv
By integrating between 0 and 4 (why?), one can show that the distribution is
normalized to unity:
m
P(v)dv = 1.
All kinetic properties of the gas can be calculated from this distribution. For
example, the average kinetic energy is given by
m
(mv2/2)P(v)dv = 3kT/2,
which the interested student can confirm using integral 20) in Appendix E.
Note that this result is independent of the molecule’s mass!
From the previous result follows that the root mean square speed, v rms =
[mv 2P(v)dv]1/2 is equal to [3kT/m]1/2. Note the trend in rms speeds in Table
19-1.
How does one find the most probable speed vp? The average speed vave?
Since the energy of the atoms that make up a monatomic ideal gas is entirely
(translational) kinetic, the internal energy of N such molecules is
E int = N(3kT/2) = n(3RT/2).
With ) E int = Q - W (First Law of Thermodynamics) and Q = nC V) T for a
constant volume process (definition of molar specific heat for constant V;
unlike solids, the molar specific heat capacities of gases depend on the
process, e.g. constant V or constant p), we have
) E int = nC V) T - 0 = nC V) T = ) [n(3RT/2)] = 3nR ) T/2
since W = 0 if V = constant. It follows that
C V = 3R/2 = constant (= 12.5 J/mol K)
Similarly, for a constant pressure process,
W = p ) V = nR ) T,
and
) E int = nC p) T - nR ) T = 3nR ) T/2,
so
C p = 5R/2 (= 20.8 J/mol K).
Finally, if we replace 3R/2 -> C V in Eint = n(3RT/2), we obtain an expression
that holds for any ideal gas, monatomic, diatomic, or polyatomic:
E int = nC VT.
For a diatomic or polyatomic ideal gas, the value of CV will depend on
temperature. Fig. 19-12 shows this dependence for a diatomic gas:
Taking ) of both sides of the previous expression gives
) E int = nC V) T,
which holds for any ideal gas and any process provided only that C V is
constant during the process.
Table 19-3 lists the maximum possible values of C V (and Cp = C V + R) for
monatomic, diatomic, and polyatomic ideal gases for temperatures too low
to excite oscillatory degrees of freedom.
Finally, for any ideal gas undergoing an adiabatic process,
pV ( = constant,
where
( = C p/C V.
The derivation of this result is given on p. 527, and is interesting reading.