Classical Physics I
PHY131
Lecture 32
Gases at the Atomic Scale
Lecture 32
1
Recap: Equation of State
First: remember mass and density:
– mass: m = n M [kg], with n the number of moles [mol] and M the molar mass
[kg/mol]
– density: ρ ≡ m/V [kg/m3]
– 1 mol of a stuff is defined as NA = 6.022×1023 molecules of it (or as the
number of atoms in exactly 12 grams of Carbon-12)
Equation of State for an “IDEAL GAS”:
pV = nRTK
– where p is pressure, V is volume, T is the Kelvin temperature, n is # moles,
and R = 8.3145 J/mol/K is the “GAS CONSTANT”
– for 1 mol of an (ideal) gas at STP (Standard Temperature T = 0 °C, and
Pressure p = 1 atm = 1.013×105 Pa):
V = 1×8.3145×273.15/1.013×105 = 22.41 L
– Î pV/T=nR i.e. is constant for fixed amount of gas!
for not-so-ideal gas (van der Waals): [p – a(n/V)2]V[1 – b(n/V)] = nRTK
– e.g. CO2: a =0.364 Nm4/mol2, b =4.27×10–5 m3/mol
– Average distance between molecules at STP:
V/NA = 22.4×10–3 / 6.022×1023 = 3.7×10–26 m3 = L3 Î L = 3.3 nm ( ~30-60× atomic
dimension)
Lecture 32
2
State and
Phase
Diagrams
Substances exist in three
phases:
– solid,
– liquid,
– gas/vapor
Changing volume V,
pressure p, and or
temperature T allows us to
“scan” the whole phase
space, which is a surface in
3-DIMENSIONAL space!
Lecture 32
4
Kinetic Gas Theory
Derivation of pV=nRT: Assume a gas of nNA molecules, all of mass m.
Assume these to be point-like, and colliding elastically with the walls of a
box with V = L3, and with each other…
– the molecules bounce off the wall (wall’s mass >>> m)
and thereby exert a minute force…
– Let’s calculate: consider a molecule i with velocity vi;
then the force on the molecule by the wall is:
−vxi ) − vxi
(
Δp xi
Δvxi
2mvxi
Fxi =
=m
=m
=−
Δt
Δt
Δt
Δt
x
vyi
vi
vxi
– the reaction force on the wall is thus 2mvxi/Δt. We find the TIME-AVERAGED
force on the wall when we take Δt as the time between successive collisions
with the (same) wall:
Fxi = mvxi2/L.
Δt = 2L/vxi Î
– there are nNA molecules in the volume; for each molecule: vi2 = vix2 + viy2 + viz2,
but the averages of the squared velocity components of the molecules are
equal:
average
over bar = average
2
2
2
2
2
kinetic energy
vi = vix + viy + viz 2 =2 2 3vix
vix = viy = viz
of a molecule
– the total force by nNA molecules is:
mvx2
Fx
2
mv 2
mv 2
mv 2
= nNA
⇒ p = 2 = nNA 3 ⇒ pV = nNA
= n NA K
Fx = nNA
3L
3L
3
3
L
L
Lecture 32
5
Kinetic Gas Theory
Comparing this to pV = nRT we conclude:
2
pV = nNA K = nRT ⇒ K =
3
3
2
R
3
T = kT
2
NA
– i.e. the molecules’ average kinetic energy is proportional to the (Kelvin)
temperature, with proportionality constant kB≡ R/NA =1.38×10–23 J/K
(Boltzmann constant)
In terms of molecular speed:
1
3
K = mv 2 = kT ⇒ 3kT m = v 2 = vroot-mean-square ≡ vrms
2
2
– e.g. air at room temperature:
32 g/mol
3kT
−26
m(O 2 ) =
5.31
10
kg
v
=
×
⇒
(O
)
=
= 478 m/s
rms
2
23
m
6.022 ×10 /mol
Molecular speeds cover a RANGE of values, which can be DERIVED
from statistical principles:
3
T1< T2 < T3
8π ⎛ m ⎞ 2 − kTK
f (v ) =
⎜
⎟ Ke
m ⎝ 2π kT ⎠
(Maxwell speed distribution). The most probable value:
vmp = 2 kT m
Lecture 32
6
Molecular Origin of Specific Heat
The heat added to a substance generally raises its temperature (if the
phase doesn’t change!):
dQ = m c dT = nM c dT = nCV dT,
– where c is the SPECIFIC HEAT CAPACITY, i.e. heat required per unit mass
(kg) and per unit temperature increase (Kelvin)
– where CV is the “MOLAR HEAT CAPACITY” at CONSTANT VOLUME, i.e. the
specific heat on a molar basis:
CV =
(
)
1 dQ
d
3
3
=
NA kT = kNA =
2
2
n dT dT
3
2
R = 12.47 J/K/mol
– Checking with actual data on Gases:
• monatomic: CV(He) = 12.47, CV(Ar) = 12.47, …
• di-atomic: CV(H2) = 20.42, CV(N2) = 20.76, …
• tri-atomic: CV(CO2) = 28.46, CV(H2S) = 25.95, …
So, what is this? ÎHeat energy also going into rotations and vibrations!
– EQUIPARTITION THEOREM: each molecule’s “DEGREE OF FREEDOM”
(DoF) absorbs on average dQ/dT = ½k of energy/K, e.g.:
• translations: 3 DoF Î 3/2k Î CV = 3/2R = 12.47
• rotations: di-atomic molecules +2 DoF Î +2/2k Î CV = 5/2R = 20.78;
• vibrations (only at high T): di-atoms: + 2 DoF (SHM kinetic + potential)
Lecture 32
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Molecular Origin of Specific Heat
Molar Heat capacity CV for Solids:
consider only monatomic solids
– the translational and rotational DoF don’t contribute, only
vibrations Î +6 DoF (3 directions for kinetic and potential
energies) Î CV = 3R = 25.0 J/K
– explains the empirical rule by Dulong and Petit
Lecture 32
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