Kinetic Theory

Imagine a fixed quantity of gas in a box with a movable piston at one end.  Suppose there is a vacuum on the outside of the piston.  The volume of the box is V.

The atoms or molecules move around inside the box with various velocities and bang against the piston.  Assume that these collisions are perfectly elastic.  What is the force on the piston due to the atoms in the box?

If nothing was holding the piston, each time a particle banged against it, it would pick up a little momentum and it would gradually get pushed out of the box.  In order to keep it from being pushed out of the box, a force F pointing inward has to be exerted on the piston.  What is the magnitude of the force needed to balance the banging of the molecules?

The piston receives from each collision a certain amount of momentum, therefore a certain amount of momentum is transferred to the piston per second.

F = Dp/Dt.

The amount of momentum per second that the atoms or molecules transfer to the piston is equal to the force needed to prevent it from moving outward.  If no inward force is acting on the piston it will pick up speed because of the bombardments.  With each collision its speed will increase by a small amount.  The rate at which the piston picks up speed, i.e. the acceleration of the piston, is proportional to the force on it.  How do we calculate the momentum transferred by the molecules to the piston every second?

Let v be the velocity of a particle, and let vx be the x-component of v.  Then mvx is the x-component of the momentum.  When a particle collides with the piston the x-component of its momentum changes sign.  The magnitude of the total momentum delivered to the piston by the particle in one collision is

Dpx = mvx - (-mvx) = 2mvx.

To find the momentum transferred by the particles per second we need the number of collisions made by the particles in a second.

Assume that there are N atoms or molecules in the volume V, or rparticle = N/V in each unit volume.  In a certain amount of time t only those atoms or molecules which are within a distance vxt from the piston are going to hit the piston.  Thus the number of collisions in a time t is equal to the number of atoms or molecules which are in the region within a distance vxt, moving towards the piston.  If A is the area of the piston, then the volume V occupied by the particles which are going to hit the piston is vxtA.  The number of particles that are going to hit the piston is that volume times 1/2 the number of atoms or molecules per unit volume, (1/2)rparticlevxtA.  (Only 1/2 of the particles are headed towards the piston.)  To get the number of particles that hit the piston per second we divide by t.

The force on the piston is equal to 2mvx times the number if hits per second.

F = rparticlemvx2A.

The pressure is

P = F/A = rparticlemvx2.

However, all the molecules do not have the same velocity, and they do not move in the same direction.  All the vx2 are different.  We must take the average of vx2, averaged over all molecules.

P = rparticlem<vx2>.

(The bracket denotes the average.)

There is nothing special about the x-direction.  The atoms can move up and down, back and forth, in and out.  The average velocity components in all directions are all going to be equal to each other.

<vx2> = <vy2> = <vz2>.

They are each equal to one-third of their sum, which is the square of the magnitude of the average velocity.

<v2> = <vx2>+<vy2>+<vz2>.

<vx2> = (1/3)<v2>.

We may therefore write

P = (1/3)rparticlem<v2> = (2/3)rparticle(m<v2>/2)

m<vx2>/2 is the kinetic energy of the center-of-mass or translational motion of an atom or molecule.  We find, therefore, that

PV = (2/3)N(m<v2>/2)

This equation relates the pressure to the kinetic energy of the atoms or molecules.  The ideal gas law states

PV = NkBT.

Comparing the two expressions we find

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The temperature is a direct measure of the average translational molecular kinetic energy.

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The root-mean-square speed of the atoms or molecules is

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Links:

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Particle Speeds and Temperatures

Welcome to the pressure chamber

Problems:

A cylinder contains a mixture of helium and argon gas in equilibrium at 150 oC.
(a) What is the average kinetic energy of each gas molecule?
(b) What is the root-mean-square speed of each type of molecule?
Solution:
(a) The average kinetic energy of each molecule is 
(3/2)kBT = (3/2)1.38´10-23 J/K(423 K) = 8.76´10-21 J.
The average kinetic energy is the same for both types of atoms,
(b) vrms2 = 2´8.76´10-21 J/m.
mHe = 4 u, mAr = 39.9 u. (Appendix A, Table A3, Serway)
1 u = 1 atomic mass unit = 1.66´10-27 kg.
vrms(He) = 1.62´103 m/s.
vrms(Ar) = 514 m/s.
The more massive molecules have a lower average speed.
A 5-liter vessel contains 0.125 moles of an ideal gas at a pressure of 1.5 atm.  What is the average translational kinetic energy of a single molecule?
Solution:
PV = nRT yields T.
T = PV/(nR)
T = (1.5´1.01´105 Pa)(5000 cm3´1 m3/106 cm3)/(0.125´8.31 J/K)
T = 729 K.

The average translational kinetic energy of a single molecule is 1.51´10-20 J.

Links to other Web Material:

The speed distribution of the particles in a gas
Kinetic Theory of Gases; A Brief Review

Heat

When you bring two objects of different temperature together, energy will always be transferred from the hotter to the cooler object.  The objects will exchange thermal energy, until thermal equilibrium is reached, i.e. until their temperatures are equal.  We say that heat flows from the hotter to the cooler object.  Heat is energy on the move.  Units of heat are units of energy.  The SI unit of energy is Joule.  Other often encountered units of energy are 1 Cal = 1 kcal = 4186 J, 1 cal = 4.186 J, 1 Btu = 1054 J.

Without an external agent doing work, heat will always flow from a hotter to a cooler object.  Two objects of different temperature always interact.  There are three different ways for heat to flow from one object to another.  They are conduction, convection, and radiation.

Conduction

The atoms in a solid vibrate about their equilibrium positions.  As they vibrate, they bump into their neighbors.  In those collisions they exchange energy with their neighbors.  If the different regions of a solid object or of several solid objects placed in contact with each other have the same temperature, then all atoms are just as likely to gain energy as to loose energy in the collisions.  Their average random kinetic energy does not change.  If, however, one region has a higher temperature than another region, then the atoms in the high temperature region will, on average, loose energy in the collisions, and the atoms in the low temperature region will, on average, gain energy.  In this way heat flows through a solid by conduction.

The stiffness of the springs (strength of the chemical bonds) determines how easily the atoms can exchange energy and therefore determines if the material is a good or bad conductor of heat.  Each atom has a nucleus, surrounded by electrons.  In a solid metal all nuclei are bound to their equilibrium positions.  But some electrons are free to move throughout the solid.  They can easily pick up kinetic energy in collisions with hot cores and loose it again in collision with cooler cores.  Since their mean free path between collisions is larger than the distance between neighboring atoms, thermal energy can move quickly through the material.  Metals are, in general, much better conductors of heat than insulators.

Convection

Convection transfers heat via the motion of a fluid which contains thermal energy.  In an environment where a constant gravitational force F = mg acts on every object of mass m, convection develops naturally because of changes in the fluids density with temperature.  When a fluid, such as air or water, is in contact with a hotter object, it picks up thermal energy by conduction.  Its density decreases.  For a given volume of the fluid, the upward buoyant force equals the weight of this volume of cool fluid. The downward force is the weight of this volume of hot fluid. The upward force has a larger magnitude than the downward force and the volume of hot fluid rises.  Similarly, when a fluid is in contact with a colder object, it cools and sinks.  When a volume of fluid such as air or water starts to move, the surrounding fluid has to rush in to fill the void.  Otherwise large pressure differences would develop.  This sets up a convection current and the looping path that follows is a convection cell.  Since fluid can not pile up at some point in space without creating a high-pressure area, it will flow in a closed loop.  Convection can be increased if the fluid is forced to circulate.  A fan, for example, forces the air to circulate.

Radiation

Nuclei and electrons are charged particles.  When charged particles accelerate, they emit electromagnetic radiation and loose energy.  Vibrating particles are always accelerating since their velocity is always changing.  They therefore always emit electromagnetic radiation.  Charged particles also absorb electromagnetic radiation.  When they absorb the radiation they accelerate.  Their random kinetic energy increases.  In thermal equilibrium, the amount of energy they loose to radiation equals the amount of energy they gain from radiation.  But hotter objects emit more radiation than they absorb from their cooler environment.  Radiation can therefore transport heat from a hotter to a cooler object.

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Electromagnetic radiation refers to electromagnetic waves, which travel through space with the speed of light.  We classify electromagnetic waves according to their wavelength.  A graphical representation of the electromagnetic spectrum is shown in the figure below.

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The visible part of the spectrum may be further subdivided according to color, with red at the long wavelength end and violet at the short wavelength end, as illustrated in the following figure.

visible.gif (13343 bytes)

Hot objects emit radiation with a distribution of wavelengths.  But the average wavelength of the radiation decreases as the temperature of the object increases.  Most thermal radiation lies in the infrared region of the spectrum.  We cannot see this radiation, but we can feel it warming our skin.  Different objects emit and absorb infrared radiation at different rates.  Black surfaces are generally good emitters and absorbers, while silvery metal surfaces are poor emitters and absorbers.

The Wood Stove

When a wood stove is used to heat the air in a room, conduction, convection, and radiation play a role.  When the wood burns, chemical energy stored in the wood is converted into thermal energy of the reaction products.  By conduction, these reaction products heat the surfaces and the air they are in contact with.  Convection draws the hot smoke up a long black pipe and out of the room and draws fresh air into the stove.  When the smoke is in contact with inner the surface of the pipe, it heats the pipe by conduction.  Conduction also carries the thermal energy from the inner surfaces of the stove and the pipe to the outer surfaces, and heats the air close to the surfaces.  The hot air then begins to rise by convection.  Cooler air rushes in to replace the rising air, and a convection current begins to flow in a convection cell.  This distributes the warm air throughout the room.  The hot, black, outer surface of the stove is also a good emitter of infrared thermal radiation.  This thermal radiation is absorbed by the surfaces of different objects in the room.

Regulating Heat Flow

Heat always flows from a region of higher temperature to a region of lower temperature.  If flows by conduction, convection, and radiation.  Often we are interested in regulating the rate at which the thermal energy is transferred.  We may want to keep an object at a temperature different from that of its surrounding for a long time by slowing down heat flow.  Or we may want an object to cool down rapidly by increasing the rate at which thermal energy is transferred.  In developing a method to do this effectively, we always have to consider the importance of the three different ways by which heat flows.

If we surround an object at temperature T2 with a layer of material, to insulate it from its surrounding at temperature T1, then the thermal conductivity of the surrounding material determines, how fast heat can flow through it.

conductivity.gif (2379 bytes)

The thermal conductivity k is defined through the equation

.

For a slab of infinitesimal thickness we rewrite the equation as

.

This equation is called the law of heat conduction.  dQ/dt is the rate at which heat flows across the area A, in Joules per second or Watts.  dT/dx is the change in the temperature over the distance dx in degrees Kelvin or Celsius per meter.  It is the temperature gradient.  The thermal conductivity k is a property of the material.

Thermal conductivities: (kcal/sec)/(oCm)

Aluminum 4.9 ´ 10-2
Copper 9.2 ´ 10-2
Steel 1.1 ´ 10-2
Air 5.7 ´ 10-6
Ice 4 ´ 10-4
Wood 2 ´ 10-5
Glass 2 ´ 10-4
Asbestos 2 ´ 10-5

1 kcal = 4186 J

To minimize heat flow through a layer of material by conduction, choose the right material, make the layer as thick as possible, and make the surface area as small as possible.

Problems:

A bar of gold is in thermal contact with a bar of silver of the same length and area.  One end of the compound bar is maintained at 80 oC and the opposite end is at 30 oC.  When the heat flow reaches steady state, find the temperature at the junction.

Solution:
When a steady state is reached, then the same amount of heat crosses any cross sectional area perpendicular to the bar per second.  (Otherwise the energy would increase in certain regions, the temperature would increase there, the temperature gradient would change, and we would not have a steady state.)  If the temperature at the junction is T, then in the gold we have
DQ/Dt = (314 W/(moC))´A´(80 oC - T)/(L/2)
and in the silver we have
DQ/Dt = (427 W/(moC))´A´(T - 30 oC)/(L/2).
We can therefore write
(314 W/(moC))´A´(80 oC - T)/(L/2) = (427 W/(moC))´A´(T - 30oC)/(L/2).
(314 W/(moC))´(80 oC - T) = (427W/(moC))´(T - 30 oC).
(80 oC - T) = 1.36´(T - 30 oC).
120.8 oC = 2.36 T.
T = 51.2 oC.

A Thermopane window of area 6 m2 is constructed of two layers of glass, each 4 mm thick, separated by an air space of 5 mm.  If the inside is at 20 oC and the outside is at -30 oC, What is the rate of heat loss through the window?
Solution:
When a steady state is reached, then the same amount of heat crosses any cross sectional area per second.  Let the temperature of the inner glass-air boundary be T1 and the temperature of the outer glass-air boundary be T2. Then for the inner piece of glass we have
DQ/Dt = (0.8 W/(moC))´6 m2´(20 oC - T1)/0.004 m.
For the air layer we have
DQ/Dt = (0.0234 W/(moC))´6 m2´(T1 - T2)/0.005 m.
For the outer piece of glass we have
DQ/Dt = (0.8 W/(moC))´6 m2´(T2 + 30 oC)/0.004 m.
The first and third equation yield (20 oC- T1) = (T2 + 30 oC), T2 = -10oC - T1.
Inserting this expression for T2 into the second equation we have
DQ/Dt = (0.0234 W/(moC))´6 m2´(2 T1+10 oC)/0.005 m.
Combining this equation with the first equation then yields
(0.0234 W/(moC))´(2 T1+10 oC)/0.005 m = (0.8 W/(moC))´(20o C - T1)/0.004 m.
0.0234´(2 T1+10 oC) = 20 oC - T1.
1.0468 T1 = 19.776 oC. T1 = 18.882 oC.
The first equation now yields
DQ/Dt = (0.8 W/(moC))´6 m2´(1.118 oC)/0.004 m = 1.34 kW.

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