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Overview
• This module will introduce the key concepts that form a cornerstone of modern physics.
• We will develop an understanding of the generic properties of matter - Thermodynamics
• We will relate atomic scale events to macroscopic phenomena – Statistical Mechanics
• We will use all of our problem solving tools to extract real world information – it is not a purely theoretical subject but a methodology.
Books – recommended texts
• There is a need to read around this topic and the module is designed to encourage this.
• Carrington Basic Thermodynamics Oxford– Most closely follows the module in the
earlier stages covering only Thermodynamics but at exactly the right level. Good worked examples and questions.
• F. Mandl Statistical Physics Wiley ~£25– Most closely follows the module in the latter
(more difficult) stages. Best overall buy as it combines Thermodynamics and Statistical Mechanics from the outset.
Books – reference texts• R. Bowley and M. Sanchez Introductory Statistical
Mechanics Oxford Science Publications– A good all rounder with a Thermodynamics
introduction to the Statistical Physics. Many examples.
• C. Kittel and H. Kroemer Thermal Physics Freeman ~£lots– Detailed and comprehensive but probably a little
too advanced. A good reference book to turn to.• D.S. Betts and R.E. Turner Introductory Statistical
Physics Addison Wesley– A paperback on the Statistical Physics only. A bit
mathematical but has good introduction to topic.
Books – reference texts
• Also don’t forget:-• Feynman, Leighton and Sands The
Feynman Lectures on Physics vol I Addison Wesley which as a series are always readable and informative, especially at potential stumbling points.
• Thermodynamics is often covered at a good level in most general physics textbooks. The Statistical Physics aspects, however, often prove to be the most problematic.
Books – A Warning• Many books adopt subtly different symbols for
the same quantity. Take care when looking at different sources.
• I will standardise in questions posed and our discussions and provide a definitive list.
• There is much terminology to get to grips with. It is probably worth making your own list as you proceed.
• There are many simple equations – some are always true and some only true under certain conditions. Don’t just rely on remembering them.
The Module
• 2 lectures and 1 workshops per week for first 7 weeks of each term.
• Assessed problems in week 3 of each term. 10% each term
• Directed Reading and Independent Learning weeks 8-10 each term. 1 Summary lecture (Autumn term only). Exam question based on this task!
• Departmental Test week 8 of Spring Term. 20%
• 2 Hour end of year exam. 60%
What's it all about?
• We need to understand the properties of matter.
• It is far too complicated to start from classical mechanics – there are too many atoms involved in even the simplest of systems.
• We have to take averages and understand what the majority are doing. In essence we start by sacrificing detailed knowledge at the atomic scale to understand the macroscopic properties of the system – Thermodynamics.
Some famous quotes
• “A theory is the more impressive the greater the simplicity of its premises, the more different kinds of things it relates, and the more extended its area of applicability. Therefore the deep impression classical thermodynamics made upon me. It is the only physical theory of universal content which I am convinced will never be overthrown, within the framework of applicability of its basic concepts.”
Albert Einstein
Some famous quotes
• “But although, as a matter of history, statistical mechanics owes its origin to investigations in thermodynamics, it seems eminently worthy of an independent development, both on account of the elegance and simplicity of its principles, and because it yields new results and places old truths in a new light in departments quite outside thermodynamics.”
J.W. Gibbs
Some famous quotes
• “The Physics of desperate men.” Unknown, Blackett
Laboratory, Imperial College
Some famous quotes
• “It’s a funny subject. The first time you go through it, you don’t understand it all. The second time you go through it, you think you understand it, except for one or two small points. The third time you go through it, you know you don’t understand it, but by that time you are so used to it, it doesn’t bother you any more.”
Sommerfield
Why does matter heat up when compressed? A taster.
• Let us take the simplest case and investigate a gas and determine what is meant by pressure.
• Imagine a piston of volume V and cross sectional area A containing a monatomic gas (Ar, He etc).
V
x dx
FA
Vacuum
Why does matter heat up when compressed? A taster.
• To stop the piston being ejected we have to hold it in – i.e. apply force F
• The magnitude of the force depends on the area – we define pressure P = F / A
V
x dx
FA
Vacuum
Why does matter heat up when compressed? A taster.
• We compress the gas by pushing the piston through an elemental distance -dx.
• The work done on the gas is therefore:-dw = F (-dx) = -PA dx = -PdV
(the area times the distance is the volume change dV)
V
x dx
FA
Vacuum
Why does matter heat up when compressed? A taster.
• How is pressure described microscopically?• The force on the piston is due to reflection of
atoms as they scatter off of the piston - they impart momentum to the piston.
• The Force on the piston is the amount of momentum transferred per second, by definition.
V
x dx
FA
Vacuum
dw = -PdV
Why does matter heat up when compressed? A taster.
• Split the problem into two parts:– What is the momentum imparted per
collision?– How many collisions do we get per
second?
V
x dx
FA
Vacuum
dw = -PdV
Why does matter heat up when compressed? A taster.
• We must assume the piston reflects atoms perfectly – why?
• If v is the velocity of the atom of mass m, vx is the velocity in towards the piston, mvx is the momentum towards and away (perfect reflector) from the piston.
• The momentum transferred per collision is therefore:
2mvx
dw = -PdV
Why does matter heat up when compressed? A taster.
• In time t only molecules within vxt of the piston will hit it.
• Let us suppose there are n atoms in our volume V so the density is = n / V.
• So the number of collisions in t is the number of atoms in volume Avxt which is:
• No. of collisions = Avxt (in time t)
• No. of collisions per second = Avx
• Therefore force on piston F = Avx 2mvx
dw = -PdV
Why does matter heat up when compressed? A taster.
• Pressure P = F / A = 2 mvx2
• Uh Oh! Duh!• We have assumed all atoms have same
velocity! – need to take averages of the velocity. P = m<vx
2> where is the two? • What’s so special about x direction? Nothing!• <vx
2> = <vy2> = <vz
2>• <vx
2> = 1/3 <vx2 + vy
2 + vz2> = <c2>/3
• c is speed• P = 1/3 m <c2> = 2/3 <mc2/2> • PV = (2/3) n<mc2/2>
dw = -PdV
Why does matter heat up when compressed? A taster.
• PV = (2/3) n<mc2/2> • PV= (2/3)U where U is the internal
energy of the entire system.
• We now know how much work we do on the gas by compressing it a little and the relationship between volume, pressure and energy. We can link the two by considering how much work we do on the gas goes into changing its internal energy.
dw = -PdVPV = (2/3)U
Why does matter heat up when compressed? A taster.
• We shall assume that on compressing the gas all the work done on the gas goes into internal energy. This means there is no leakage of “heat”.
• Such a compression (or expansion) where there is no flow of heat through the walls of the piston or vessel is called adiabatic. From the Greek a (not) dia (through) bainein (to go).
• For generality with other systems:– PV = (2/3)U is more commonly written as– PV = (-1)U so = 5/3 in this example.
dw = -PdVPV = (2/3)U
Why does matter heat up when compressed? A taster.
• For the adiabatic compression only all the work goes into internal energy so
dU = dW = -PdV
U = PV / (-1)
So by product rule dU = (PdV + VdP) / (-1) Hence, PdV = - (PdV +VdP) / (-1)
Grouping terms gives:dV / V = -dP / P
Which we can all integrate – hopefully!???
dw = -PdVPV = (-1)U
Why does matter heat up when compressed? A taster.
• For the adiabatic compression( / V) dV = (-1/P) dP
( / V ) dV = (-1/P) dP
ln(V) = -ln(P) + ln(C)
PV = C• This is our result. It tells us that under adiabatic
conditions the pressure times the volume to the power 5/3 (for our example) is constant.
• We discovered this without knowing anything about our gas – it must be true in general for monatomic gases or more specifically a “perfect” or “ideal” gas.
dw = -PdVPV = (-1)UPV = C
Temperature• We can relate internal energy to pressure and
volume. How are these related to temperature?• Common sense tells us that when two bodies at the
differing temperatures are placed next to each other (in thermal contact) the temperatures rise and fall until both bodies reach the same temperature. When at the same temperature they are in thermal equilibrium.
• This is commonly referred to as the 0th Law of Thermodynamics:-– “If two bodies A and B are in thermal equilibrium
with a third body C then A and B are in thermal equilibrium with each other.”
• When in thermal equilibrium there is no net energy flow from one body to the other.
Temperature Scales
• To measure temperature in general we need a property that varies with temperature- X(T). – length of mercury in a capillary– Resistance of a wire– Pressure of a gas at constant volume– Volume of a gas at constant pressure
• We need a reference point which is taken to be the triple point of water. By definition set to 273.16K in 1954.
Temperature Scales
• Then:Tx = 273.16 (X / Xtp)
• But different methods of X give differing values. Worse still differing gases give different results in constant volume gas thermometer!
gasHg
System to be measured
Height of Hg = hPressure = gh
Apply pressure to fix volume
Temperature Scales
• However, works in the limit of low gas density for all gases.
Tcvgt = 273.16 limlow (P / Ptp)
Tcpgt = 273.16 limlow (V / Vtp)• Works best when X = PV product for a gasT = 273.16 limlow ((PV) / (PV)tp)• This defines the Ideal Gas Temperature Scale limlow (PV) = (limlow (PV)tp / 273.16) T
PV = NRT= nkTThis is the ideal gas law
The Ideal Gas Law PV = NRT = nkT• N is the number of moles of gas atoms or
molecules.• One mole is 6.02 × 1023 entities, this is
Avagadro’s number N0
• R is the molar gas constant 8.31 J mol-1 K-1
• n is the number of atoms• k is Boltzmann’s constant 1.381 × 10-23 J
K-1
• Chemists like to use N and R, Physicists tend to use n and k. You will see both in your reading.
The Ideal Gas Law PV = NRT = nkT• The fundamental assumption here is:-
– The gas behaves as “A non-interacting assembly of point masses”
• This is increasingly realistic of the nature of gasses as T increases and or P (or density) decreases.
• It has been experimentally confirmed.• PV = nkT is an example of an equation of state. P, V
and T are state variables or thermodynamic coordinates.
• Other equations of states can be defined to fit non-ideal gas behaviour for example: van der Waals equation of state (P + aN2/V2)(V-Nb) = NRT where a and b are constants correcting for potential energy and excluded volume of gas molecules respectively.
Temperature on the atomic scale
• We have found that – PV = nkT = (2/3) U = (2/3) n<mc2/2> – Average energy per molecule = 3/2kT
• How is this energy distributed in the gas?– We should attempt to find the
distribution of velocities in the gas. This means finding a result for of the order of N0 atoms.
