Thermal Physics
Thermodynamics and Temperature
•Thermodynamics deals with the internal energy of systems, and the key
physical quantities in this context are:
Temperature
Heat
Entropy
Internal energy
It is temperature that is most fundamental to quantifying the thermal state
of a system and the scale of measurement humans can best relate to in
terms of defining “hot” and “cold”.
•Absolute zero: the temperature of a body can be raised without limit, but
it cannot be lowered without limit. If the pressure of a gas is kept constant,
its volume is found to decrease linearly with temperature. Extrapolation
shows that zero volume is reached at a temperature of -273.15°C. Similarly
if the volume of gas is kept constant, the pressure is seen to cross the
zero-pressure axis at -273.15°C. This temperature is known as absolute
zero and is used to define the Kelvin scale: Tk=Tc + 273.15.
•In practice, absolute zero can never be reached, with the lowest temperature
being achieved in the laboratory being 2 x 10-9 K.
The Zeroth Law of Thermodynamics
•Many bodies are said to be thermometric in their behaviour, since their
properties change in response to changes in temperature. Examples are:
Volume of liquid increases with increasing temperature
Metal rod expands with increasing temperature
Electrical resistance of a wire increases with increasing
temperature
This behaviour can be used as the basis of an instrument that measures
temperature (i.e., a thermometer). But how do we deal with the
situation when we are trying to measure the temperature of one body
which has a different temperature to the body we are trying to
measure it with? This is where the zeroth law and the concept of
thermal equilibrium are important.
The Zeroth Law of Thermodynamics contd...
•In formal language the zeroth law states:
“If two bodies are each in thermal equilibrium with a third body, then they
are in thermal equilibrium with each other.”
•In plain language the zeroth law states:
“Every body has a property called temperature. When two bodies are
found to be in thermal equilibrium, their temperatures are equal.”
•How do we use this in an everyday situation?
If we want to know whether the liquids in two different containers are
at the same temperature, we measure each with a thermometer. We do
not need to bring the two liquids into contact and observe whether or
not they are in thermal equilibrium, since the zeroth law tells us they
will be if at the same temperature.
Temperature scales and measuring temperature
Having established how we can measure temperatures, we now need to
define a reproducible temperature scale in order that thermometers can be
calibrated.
The triple point of water: this is used to define a standard fixed point,
since liquid water, solid ice, and water vapor can coexist in thermal
equilibrium at only one set of temperature and pressure values. By
international agreement (in 1967), the triple point of water has been
assigned a value of 273.16K, and is denoted T3.
The constant volume gas thermometer: this has been adopted as the
standard against which all other thermometers are to be calibrated, and is
based on the pressure exerted by a gas confined to a fixed volume.
The international temperature scale: measuring temperatures precisely
with a constant volume gas thermometer is arduous work, and so in
practice, this thermometer is used only to establish certain “primary” fixed
temperature points. These points can then be used to calibrate other
thermometers, such as the every-day mercury in glass thermometer.
Temperature scales and measuring temperature
The International temperature scale for such calibration purposes. It
consists of a set of recipes for providing the best possible approximations
to the Kelvin temperature scale. A set of primary fixed points is adopted,
and a set of instruments is specified for interpolating between these fixed
point temperatures and for extrapolating beyond the highest temperature
and below the lowest temperature.
Substance
Fixed-point state
Temperature (K)
Hydrogen
Triple point
13.81
Hydrogen
Boiling point (high p)
17.042
Hydrogen
Boiling point
20.28
Neon
Boiling point
27.102
Oxygen
Triple point
54.361
Argon
Triple point
83.798
Oxygen
Boiling point
90.188
Water
Boiling point
373.125
Tin
Melting point
505.074
Zinc
Melting point
692.664
Silver
Melting point
1235.08
Gold
Melting point
1337.58
Temperature scales and measuring temperature
The Celsius and Fahrenheit Scales – the relationship between these two
scales is:
TF = 9/5 TC + 32°
Thermal Expansion
Most materials expand when heated. Over a limited range of temperatures,
it is found that the linear expansion of a material, L, is proportional to the
length:
L = L  T
in which  is a constant called the coefficient of linear expansion (whose
value depends on the material and on the temperature range of interest). 
is a measure of the fractional change in length per unit change in
temperature. As a general rule, the higher the melting point of the material,
the lower the value of .
Volumetric Expansion: solids & liquids
•The thermal expansion of a solid is like a three-dimensional enlargement:
the previous equation applies to every linear dimension of the object.
Hence the volume of the solid will also increase.
•For liquids, volume expansion is the only meaningful expansion
parameter.
Hence we can write the equation for the volumetric expansion of a solid or
liquid as:
V = V  T
where V is the original volume, T is the temperature increase, and  is the
coefficient of volume expansion.
Consider a solid block of dimensions a x b x c. The new volume of the
block when the temperature is increased T is:
V = a (1+T) x b (1+T) x c (1+T)
=abc (1 + 3T + 2 terms + 3 terms)
V0 (1 + 3T)
   3.
The expansion of water
Water has the unusual property that its coefficient of volume expansion
becomes zero at 3.98°C, and is negative between 0°C and 3.98°C.
Ice is also unique in that it is significantly less dense than water – most
solids are more dense than their liquid form.
Heat
•A hot body when placed in cooler surroundings (e.g. casserole dish taken
out of an oven) will undergo a decrease in temperature – rapidly at first and
then increasingly slowly – until its temperature equals that of its
surroundings. Similarly, a cool body placed in a warmer environment will
increase in temperature until its temperature equals that of its surroundings.
Q: What is being exchanged between the body and its surroundings
that causes this temperature change?
A: It is THERMAL or INTERNAL energy  collective kinetic and
potential energies associated with the random motions of the atoms,
molecules, and other microscopic bodies within an object.
We call this internal or thermal energy: HEAT
Facts about heat:
• denoted by the symbol Q and measured in units of Joules (J)
• Q is positive when internal energy transferred to a system from its
environment
• Q is negative when internal energy transferred from a system to its
environment
A further definition of heat
“Heat is the energy that is transferred between a system
and its environment because of a temperature difference
that exists between them.”
Work: another form of energy transfer
Energy can be transferred between a system and its environment by
means of work (symbol W). In this case it requires a force to act upon the
system and to displace it (work = force x displacement).
Heat capacity
Heat capacity (C) is the proportionality constant between the amount of
heat either gained or lost by an object, and the temperature change that
this produces:
Q = C (Tf – Ti)
where Ti and Tf are the initial and final temperatures of the object.
Specific heat
Two objects made of the same material will have heat capacities
proportional to their masses, and so it is convenient to define a “heat
capacity per unit mass” or specific heat (c) that refers not to an object but
to a unit mass of the material from which it is made. Hence the relationship
between heat and temperature change (previous equation) can be written:
Q = cm (Tf – Ti)
Heats of transformation
When a material undergoes a phase change – that is, it changes from a
solid to a liquid (or vice-versa), or from a liquid to a gas (or vice-versa) –
heat will be either absorbed or released by the material without a change
in its temperature:
Melting – heat absorbed
Vapourisation – heat absorbed
Freezing – heat released
Condensation – heat released
Heats of transformation
The amount of heat per unit mass, m, that must be transferred when a
sample completely undergoes a phase change is called the heat of
transformation, L:
Q=Lm
Liquid  Gas phase: L is called heat of vapourisation (absorbed)
Gas  Liquid phase: L is called heat of vapourisation (released)
Solid  Liquid phase: L is called heat of fusion (absorbed)
Liquid  Solid phase: L is called heat of fusion (released)
Worked example:
Q: How much heat is needed to take ice of mass m=0.72kg at -10°C to a
liquid state at 15°C?
A: Need to consider 3 steps: (i)raising the temperature of the ice from
-10°C to the melting point at 0°C, (ii)the ice melting at 0°C, and (iii)raising
the temperature of the melted ice from 0°C to 15°C.
Worked example contd…
Step (i): Q1 = cice m (Tf – Ti)
= (2220 J/kg.K)(0.72)(0-)-10)
= 15,984 J
Step (ii): Q2 = LF m = (333000J/kg)(0.72kg) = 239,800 J
Step (iii): Q3 = cliq m (Tf – Ti)
= (4190J/kg.K)(0.72)(15-0)
= 45,252 J
Therefore the total heat required is the sum of the heat required for each
of the 3 steps:
Qtot = Q1 + Q2 + Q3 = 15984 + 239800 + 45252 = 301,036J
Heat and work for a gas
W
piston
insulated walls
gas
Q
heat reservoir
Key point: all changes
occur slowly, such that
system is always in
thermodynamic
equilibrium!
Consider a cylinder of gas with movable
piston at the top and a heat reservoir at the
bottom (whose temperature T can be
controlled). We call this “the system”.
•The system starts in an initial state i, with
pressure pi, volume vi, and temp Ti.
•The system is changed, ending up in a final
state f, with pressure pf, volume vf, and temp
Tf.
•The procedure by which the system is
changed from its initial state to its final state
is called a THERMODYNAMIC PROCESS.
•During this process, the system can absorb
or release heat (from and to the reservoir),
and it can do work or have work done on it
(via raising or lowering the piston).
Heat and work for a gas
W
Suppose the gas is allowed to push the
piston up through a differential displacement
ds, with an upward force F.
piston
insulated walls
gas
Q
heat reservoir
•The magnitude of the force F can be written
as F=pA, where p is the pressure of the gas
and A is the area of the piston.
•The differential work dW done by the gas
during the displacement is
dW = F.ds = (pA)(ds) = pdV
Key point: all changes
occur slowly, such that
system is always in
thermodynamic
equilibrium!
where dV is the differential change in volume
of the gas owing to the movement of the
piston.
•The total work done in going from Vi to Vf is:
Vf
W = ∫ dW = ∫ pdV
Vi
(courtesy of Halliday, Resnick, Walker, 4th Edition)
There are many ways to take the gas from
state i to state f, and this is best shown in
a p-V diagram:
(a) Simple drop in pressure with increase
in volume: W = area under (green) curve,
which is positive  gas did positive work.
(b) Two steps, one at constant p and one
at constant V: Only the first (ia) involves
work, which is positive, and would require
heating the gas (from Ti to Ta, so it
expands), while maintaining a constant
pressure on the piston. Step af requires
keeping the gas at a constant volume (by
keeping the piston fixed), and reducing the
pressure by reducing the temperature
(from Ta to Tf). Heat is lost from the gas.
(c) Two steps of (b) are carried out in
reverse order: W is smaller, as is the net
heat absorbed.
(d) W can be as small (icdf) or as large
(ighf) as you want depending on path.
(e) Negative work is done by the system
through some external force compressing
it.
(f) A thermodynamic cycle where the
system goes from initial state i to final
state f and then back to i again. The net
work done by the system is the sum of the
positive work done during the expansion
and the negative work done during the
compression. Wnet > 0.
(courtesy of Halliday, Resnick, Walker, 4th Edition)
The First Law of Thermodynamics
•We have just seen that the various paths taken on the p-V diagram in going
from an initial state i to final state f, can involve both changes in Q (heat) and
W (work done by the system).
•Experimentally we find that the quantity Q-W is the same for all processes
•In other words, Q-W depends only on the initial and final states and not on
the path taken between!
•The quantity Q-W must therefore represent some intrinsic property of the
system:
We call this the Internal Energy (Eint) of the system
Eint = Eint,f – Eint,I
Eint = Q-W
1st Law of Thermodynamics
Note: W = the work done by the system (not on the system)
Special cases of the 1st Law
Process
Restriction
Consequence
Adiabatic
Q=0
Eint = -W
Constant
volume
W=0
Eint = Q
Closed cycle
Eint = 0
Q=W
Free expansion
Q=W=0
Eint = 0
The transfer of heat - conduction
•Transfer of heat via conduction can be ascribed qualitatively to the
vibrations of atoms and electrons within a material, which can be passed
along its length (in direction of hot to cold).
•Consider a slab with faces of area A and thickness L, whose faces are
maintained at temperatures TH and TC.
•Let Q be the heat that is transferred through the slab
from its hot face to its cold face in time t.
L
TH
Q
k
TC
•Experiment shows that the rate of heat transfer, H, is
given by:
H = Q/t = kA (TH – TC)/L
where k is called the thermal conductivity, which
depends on the material the slab is made of. A large
value of k implies the material is a good heat conductor.
slab
Thermal resistance to conductance (R-value)
•The opposite to good conduction is good insulation and we can think of
this in terms of thermal resistance R. The R-value of a slab of thickness L
is defined as:
R = L/k
•Thus the lower the thermal conductivity of the material (i.e. the lower the
value of k), the higher its R-value is.
•Substituting the above definition of R into the equation for H gives:
H = A (TH – TC) / R
Some k and R-values for various materials:
Material
k (W/m.K)
R-value
Material
k (W/m.K)
R-value
Copper
401
0.00036
Lead
35
0.0041
Silver
428
0.00034
Aluminium
235
0.00061
Air
0.026
5.5
Poly-foam
0.024
5.9
Fiberglass
0.048
3.0
Window glass 1.0
0.14
Conduction through a composite slab
TH
Q
L2
L1
k2
k1
TX
TC
Consider a composite slab, consisting of two materials
having different thickness, L1 and L2, and different
thermal conductivities, k1 and k2. The outer surfaces of
the slab are in contact with a heat reservoir of
temperature TH (left end) and TC (right end). Each face
of the slab has area A. Temperature at the boundary of
two materials is TX.
Assumption: heat transfer through the slab is a
steady-state process – temperatures throughout the
slab and the rate of heat transfer do not change with
time.
In the steady-state, the rates of heat transfer through the two materials are
the same, hence:
H = Q/t = k2A(TH – TX)/L2 = k1A(TX-TC)/L1
Solving for TX then gives: TX = (k1L2TC+k2L1TH)/(k1L2+k2L1) and
substituting back in to previous equation:
H = A(TH – TC)/ [(L1/k1) + L2/k2)]
Conduction through a multi-composite slab
The previous equation for the rate of heat transfer through a composite slab
of two materials:
H = A(TH – TC)/ [(L1/k1) + L2/k2)]
can be generalised for any number of materials to:
H = A(TH – TC)/  (L/k) = A(TH – TC)/  R
Kinetic Theory of Gases – Heat at the Microscopic Level
•Up until now we have been concerned only with the macroscopic
behaviour of bodies when talking about their thermal properties and the
transfer of heat between them. Moreover, the main physical quantities we
have been using in this context – pressure, volume, temperature, work,
internal energy – have all been understood and applied in a macroscopic
sense.
•The kinetic theory of gases attempts to understand these quantities in a
microscopic way: for example, Q. why does gas exert a pressure? A.
because of the constant collisions the atoms/molecules of the gas are
having with the wall of the container. Also these same motions give the gas
temperature and internal energy.
Ideal Gases
•A problem in attempting to describe the microscopic behaviour of gases via
general laws is that they are all slightly different in their behaviour. However,
it has been found by experimentation that as you go to lower and lower gas
densities, all gases tend towards having the same behaviour. An ideal gas
is one that exhibits this behaviour in the low-density limit.
Ideal Gases
So what is this behaviour?? An ideal gas exhibits the following relation
between pressure, p, volume, V, and temperature, T:
pV = nRT (Ideal Gas Law)
where n is the number of moles of gas present (1 mole = Avagadro’s
Number of molecules = 6.02 x 1023), and R is the gas constant (=8.31
J/mol.K).
Note: T must be expressed in degrees Kelvin!!
Work done by an Ideal Gas at Constant Temperature
Suppose that a sample of n moles of an ideal gas confined to a cylinder
with a piston at one end, is allowed to expand from an initial volume Vi to a
final volume Vf. The temperature, T, is held constant throughout the
process  an isothermal expansion. The work W done by the gas is:
Vf
W = ∫ pdV
Vi
V
Vi
And we can subsitute in for p from the Ideal Gas Law: W = ∫ f(nRT/V)dV
Work done by an Ideal Gas at Constant Temperature
Vf
W = ∫V (nRT/V)dV
i
and since T is constant:
Vf
W = nRT ∫V dV/V
i
 W = nRT ln (Vf/Vi)
Worked example:
A cylinder contains 12 litres of oxygen at 293K and 15 atm. The
temperature is raised to 308K and the volume reduced to 8.5 litres.
What is the final pressure of the gas (assuming it is ideal)?
The ideal gas law can be written as (piVi/Ti) = (pfVf/Tf)
Rearranging and plugging in data: pf=[(15)(308)(12)]/[(293)(8.5)]
= 22atm
A Molecular View of Pressure & Temperature
y
•Let n moles of ideal gas be confined in a
cubical box of volume V (=L3), whose walls are
at temperature T. The molecules of the gas are
moving in all directions at varying speeds, and
are regularly colliding with the walls (elastically).
v
L
m
L
L
z
•Consider a typical particle of mass m and
x velocity v, that is about to collide with the righthand wall. Upon collision, the only component of
its momentum which will change is in the x
direction:
momentum before collision = mvx
momentum after collision = m(-vx)
Change in momentum = -mvx – mvx = -2mvx
Hence the momentum delivered to the wall by
the molecule during the collision = +2mvx
A Molecular View of Pressure & Temperature
y
•The molecule will hit the wall repeatedly, and
we can calculate how often it will do this:
•Time taken to travel from right-hand wall to lefthand wall and back again = 2 x (L/vx)
v
L
m
L
L
z
x
•Thus the rate at which momentum is delivered
to the right-hand wall is given by:
p/t = 2mvx/2L/vx = mv2x/L
•From Newton’s 2nd law, the rate at which
momentum is delivered to the wall equals the
force, F, exerted on the wall by the molecule.
•We need to add up the contributions from all
molecules to get the total force on the wall, and
divide it by the area of the wall to get the
pressure, p:
p=F/L2 = [mvx12/L+mvx22/L+….+mvxN2/L]/L2 = (m/L3)(vx12+vx22+…+vxN2)
(N = number of molecules in box)
A Molecular View of Pressure & Temperature
y
•Since N = nNA and <vx2> =  (vxn2)/N, we can
write the previous equation as:
p =(m/L3)N<vx2>
=(nmNA/L3)<vx2>
v
L
m
L
L
z
•But mNA is the molar mass M of the gas (that
is, the mass of 1 mole of the gas), and L3 is the
x
volume of the box, so
p = nM<vx2>/V
•For any molecule v2 = vx2 + vy2 + vz2, and
because they are all moving in random
directions, the average values of the squares of
their velocity components are equal so that
vx2 = ⅓ v2
and so the previous equation can be written as:
p = nM<v2>/3V
A Molecular View of Pressure & Temperature
y
•Now the square root of <v2> is called the root
mean square speed, and is symbolised by
vrms. So we can express the pressure in terms
of this quantity:
p = nMvrms2/3V
v
L
m
L
L
z
x
or pV = ⅓ nMvrms2
•Finally, if we take the ideal gas law:
pV = nRT
and combine it with the previous expression, we
get:
vrms = 3RT/M
Values of vrms (in km/s) for various molecules at room temperature (300K):
H2=1920, He=1370, water vapour = 645, N2=517, O2=483, CO2=412, S02=342
Translational Kinetic Energy
•Consider the single gas molecule (of mass m) moving inside the box as per
the previous section. Its translational kinetic energy at any instant is ½mv2.
Its average kinetic energy over time is:
<K> = <½mv2> = ½m<v2> = ½mvrms2
Note the assumption that is made here: the average speed of the molecule
is the same as the average speed of all the molecules at any given time.
•Now we have shown that vrms = 3RT/M, so we can write <K> as:
<K> = ½m (3RT/M)
But M/m is the molar mass divided by the mass of the molecule, which is
simply Avagadro’s Number, NA, so:
<K>=(3RT)/2NA
or <K> = (3/2)kT
where k = Boltzmann’s constant = R/NA (= 1.38 x 10-23 J/K)
Translational Kinetic Energy
The fact that the average kinetic energy of a gas is given by the simple
equation <K> = (3/2)kT reveals an important property of gases:
• At a given temperature, T, all gas molecules – no matter what their
mass! – have the same average translational kinetic energy, namely
(3/2)kT. When we measure the temperature of a gas, we are
measuring the average translational kinetic energy of its molecules.
vrms revisited
We have shown that vrms = 3RT/M. We can also express this in terms of
Boltzmann’s constant, k, and the mass of the molecule m:
The ratio M/m = NA  M = mNA.
Therefore vrms = (3RT)/(mNA) = 3kT/m.
vrms = 3kT/m
version used for
exoplanets
Heat Capacities of an Ideal Gas
We now take a microscopic approach to understanding the heat capacity of
an ideal gas – that is, the proportionality relationship between heat (or
internal energy) and temperature change, per unit mass of gas.
•Our previous derivation of mean kinetic energy, <K>, leads naturally into
defining what the internal energy, Eint, for an ideal gas is: it is the sum of
the translational kinetic energies of its molecules. Hence a sample of n
moles of such as gas contains nNA molecules, so Eint is simply:
Eint = (nNA) <K> = (nNA)(3/2)(kT)
And since NAk = R, the gas constant, then:
Eint = (3/2)nRT (monatomic ideal gas).
Molar specific heat at constant volume, Cv
Consider the cylinder and piston arrangement again, enclosed in which is n
moles of an ideal gas at pressure p and temperature T. The volume of the
cylinder is V. Suppose the gas is taken from this initial state to a final state
by adding a small amount of heat Q to the gas by slowly increasing the
temperature of the reservoir.
Molar specific heat at constant volume, Cv
The gas temperature rises a small amount to
T+T and its pressure to p+p. The volume
of the gas does not change!
•Remembering that Cv is the constant of
proportionality in the equation:
Q = nCv T
•And the 1st Law of Thermodynamics has:
Eint = Q – W
= nCv T – W
•With the volume held constant, W = 0, and so
Cv is given by:
Cv = Eint / n T
and since Eint = (3/2)nR T, we have
Cv = (3/2)R
Hence Cv = 12.5 J mol-1K-1 (monatomic gas)
Molar specific heat at constant pressure, Cp
In this case we assume that the temperature
of the gas is increased by the same small
amount T, but that the necessary heat Q is
added with the gas under constant pressure.
Cp is then given by:
Q = nCp T
and it will be greater than Cv, since not only is
energy supplied to raise the temperature, but
also to raise the piston and hence do work W.
•The 1st Law has Eint = Q – W
 nCvT = nCpT – W
And since the pressure remains constant, we
can write W = pV = nRT, and so:
nCvT = nCpT - nRT
(dividing by nT)
Cv = Cp - R
pins removed
The Equipartition of Energy
How does our kinetic theory and its predictions on the values of Cv and Cp
stack up against the experimental values?
Molecule
Example
Cv (J/mol.K)
Monatomic
Ideal
12.5
He
12.5
Ar
12.6
Ideal
12.5
N2
20.7
O2
20.8
Ideal
12.5
NH4
29.0
CO2
29.7
Diatomic
Polyatomic
Cv = (3/2)R
The Cv = (3/2)R prediction is correct for monatomic gases, but not for
diatomic and polyatomic gases!!
The Equipartition of Energy
The Cv = (3/2)R prediction is correct for monatomic gases, but not for
diatomic and polyatomic gases!! – WHY IS THIS SO??
Monatomic molecules, which are point like, can only possess one type of
energy and that is translational kinetic energy.
Diatomic and polyatomic molecules can possess other forms of energy
by rotating and vibrating!
Maxwell’s theorem on the equipartition of energy:
“Every kind of molecule has a certain number f of degrees of freedom
which are independent ways in which it can store energy. Each such
degree of freedom has associated with it, on average, an energy of
½kT per molecule.”
For translational motion, there are 3 degrees of freedom, corresponding to
the x, y, z directions.
For rotational motion, a monatomic molecule has 0 degrees of freedom, a
diatomic molecule has 2 degrees of freedom (along the 2 axes that are
perpindicular to the axis that joins the two atoms).
The Equipartition of Energy
A molecule with more than two atoms has 6 degrees of freedom: 3
translational and 3 rotational.
How does this carry over into the derivation of specific heat:
Instead of Eint = (3/2)nRT
Eint = (f/2)nRT
where f = number of degrees of feedom.
This then gives:
Cv = (f/2)R = 4.16f J/mol.K
Molecule
Example
f(trans) f(rot)
f(total) Cv (pred)
Cv
(obs)
Monatomic
He
3
0
3
(3/2)R=12.5
12.5
Diatomic
O2
3
2
5
(5/2)R =20.8
20.8
Polyatomic
CH4
3
3
6
3R= 25.0
29.4 X
The Equipartition of Energy
As is seen from the previous table, the predicted values of Cv agree with
the experimental values for monatomic and diatomic molecules, but it is too
low for polyatomic molecules. While it would be gratifying to try and get
better agreement, the classical theory has reached its limits in terms of
explaining the way energy is stored by gas molecules, and quantum
theory is what is required to reconcile the theoretical and observed values.
However, experimental measurements do provide an interesting hint as to
how energy is quantized. The figure shows the ratio Cv/R as a function of
temperature for H2 gas:
The Equipartition of Energy
•Below 80K, Cv/R = 1.5  gas
behaves as though it is monatomic (even though H2 is diatomic). Only translational modes
of motion are excited.
•At T=100-200K, Cv/R = 2.5 and
clearly rotational modes become
excited.
•At above T~3000K, Cv/R=3.5
and vibrational modes become
excited; however, H2 molecule
dissociates at about T=3200K.
Adiabatic Expansion of an Ideal Gas
W
Once again we consider an ideal gas
contained in a cylinder with piston at one
end. In this case, all walls are insulated.
piston
insulated walls
gas
Initially, some lead shot is keeping the piston
in place. Then some is removed, allowing the
has to expand and push up the piston and
remaining lead shot. The change in volume is
dV, and we assume for this infinitesimal
change, p remains constant.
•From the 1st Law of Thermodynamics:
dEint = Q – pdV
and since the gas is thermally insulated, Q=0. Also dEint = nCvdT, so:
nCvdT = -pdV
And from the ideal gas law pV = nRT, which in differential form is:
pdV + Vdp = nRdT
Adiabatic Expansion of an Ideal Gas
W
Replacing R with Cp-Cv leads to:
piston
ndT = (pdV+Vdp)/(Cp-Cv)
Equating the two expressions for ndT:
insulated walls
gas
-(pdV)/Cv = (pdV+Vdp)/(Cp-Cv)
-[(Cp-Cv)/Cv]pdV = pdV + Vdp
pdV[1+(Cp-Cv)/Cv] = -Vdp
(dV/V)(Cp/Cv) = -dp/p
dp/p + (Cp/Cv)(dV/V) = 0
and by defining  to be the ratio of the molar specific heats,  = Cp/Cv
dp/p +  dV/V = 0
and integrating gives:
ln p +  ln V = constant
or
pV = constant
Tut question: show that this can also be written as TV-1 = constant