Thermodynamics Day I: Effects of heat on matter, Ideal Gases, Heat Engines, Laws of Thermodynamics I. Basic definitions: *HEAT (Q): energy transferred from one substance/body to another by virtue of a difference in temperature. Note that it is not correct to speak of a substance or object as containing heat. An object may contain what is referred to as internal energy (U). That energy is only called heat while it is in the process of being transferred from one body or substance to another. Heat is a name given to a temporary form of energy. Q

A bU U→

There are 3 types of processes by which heat can be transferred from one object to another, called CONDUCTION, CONVECTION, and RADIATION. You have perhaps heard of these before. There are different mathematical relationships governing the rates of heat transfer by each of the three methods. Of the 3, we will be focusing on the simplest, conduction, in which heat is transferred from one body to another by direct contact. *TEMPERATURE (T): Temperature is somewhat harder to define than heat but is best thought of as a reflection of the average kinetic energy per atom/ molecule of a substance. It is worth noting that it is not really possible to directly measure the temperature of anything: all methods of measure the temperature of an object are based upon a measurement of some other physical property such as * length of a column of liquid (traditional thermometer) * electrical resistance (electrical thermometer) * gas volume (gas thermometer) * density of water (Galileo thermometer) * bending of a bimetallic strip (thermostat) We will see later that in an ideal gas, the average kinetic energy of the atoms/ molecules

is directly proportional to the Kelvin Temperature: 32avg bKE k T= . This implies then

that 3 brms

k Tvm

= , where m is the mass in kg of a single atom/ molecule of the gas and kb

is the “Boltzmann constant”, which equals 1.38 X 10-23 J/K THE KELVIN TEMPERATURE SCALE IS OUR MAIN SCALE IN THIS COURSE. (TK = TC + 273). EQUATIONS SUCH AS THE ONES ABOVE WILL NOT WORK IN CELSIUS. THE ONLY TIME YOU CAN GET AWAY WITH USING CELSIUS IS IF THE EQUATION REFERS TO A CHANGE IN TEMPERATURE. THE CHANGE IN CELSIUS TEMPERATURE OF AN OBJECT OR SUBSTANCE IS THE SAME AS THE CHANGE IN KELVIN TEMPERATURE ANYWAY. II. Effects of Heat on matter:

A. temperature increase or phase change: Q = mc∆T / Q = mLf or mLv B. thermal expansion of solids: ∆l = αlo∆T (for liquids and gases, ∆V = βVo∆T)

C. rate of heat conduction: Heat Current: 2 1T TQH kAt L

−Δ= =Δ

III. Intro to heat engines: A. basic definition: A heat engine is a device for the purpose of transferring thermal energy into useful work. There are a many types of heat engines, 2 familiar ones being steam engines and internal combustion engines. In both cases, a fuel is burned to generate heat and some of this heat is successfully converted into useful work. Engines run in “cycles”, which consist of repeating steps. One example is shown below. We will be considering somewhat more idealized engine cycles than this. We consider the enclosed gas as the “working substance” in a heat engine and all of the mathematics that we will do will relate to the changing temperature, volume, pressure, and internal energy of the gas (get ready to review the ideal gas law!).

B. sign convention: the enclosed gas inside of a heat engine expands and contracts during different steps of the engine cycle. When it is expanding, it is doing positive work on the outside world. Alternately, we could say that negative work is being done on the gas. When the gas is contracting, positive work is being done on the gas and negative work is being done by the gas. Choosing a sign convention for work is thus a little tricky. The textbook uses the convention that “W” in thermodynamic equations means work done on the gas. This is somewhat counterintuitive, as it means that the work term is negative while the gas is doing useful work and positive while the gas is being compressed. More on this later. C. engine efficiency: The efficiency of an engine is defined as the useful work

outputted by the engine divided by the heat inputted to the engine: out

in

WeQ

= This should

seem like common sense, as we always define efficiency as output/input. A useful way of thinking of the energy transfer in an engine is the following overview:

Heat is inputted at the “hot” side of the engine (Qin). Some of this energy is transferred into useful work (W) while the rest is exhausted (Qout). Some textbooks use QH and QC instead of Qin and Qout. H and C basically stand for the hot and cold sides of the engine. By conservation of energy, Qin = Wout + Qout. This may also be written as

[ ]H CQ W Q= + . The Absolute value is used because remember that by convention, in all the textbook formulas, W is the work done on the gas, not by the gas. Thus the

efficiency formula may be rewritten as [ ]H

We

Q= or H C

H

Q QeQ−

=

D. Carnot Efficiency: Sadi Carnot was a French theoretical physicist in the early 1800s who studied engine efficiency. Strictly through theoretical work, Carnot determined that the ideal engine would consist of a cycle of 4 reversible processes: I. an isothermal expansion. “Isothermal” means at constant temperature. In order to expand or (contract) isothermally, the process must take place slowly and the gas enclosed gas must be in contact with an outside “heat sink”, which will keep the gas’ temperature constant. so that the temperature of the enclosed gas can be kept constant Otherwise, gases cool as they expand and get hotter as they are compressed. A perfectly isothermal process is probably impossible but a process can be nearly isothermal. II. an adiabatic expansion. “Adiabatic” means no heat exchanged with the outside environment. In order to expand or contract adiabatically, the process must take place very quickly, so that there is no time for heat to flow into or out of the enclosed gas. III. an isothermal compression IV. an adiabatic compression.

This idealized process represented on the “P-V” diagram (you’ll be seeing a lot more of those) became known as a “Carnot Cycle”. You are not really responsible for knowing the steps of the Carnot Cycle. An extremely important result of Carnot’s work, however, was his finding that in an ideal engine, the heats QH and QC must be proportional to the operating temperatures TH and TC.

Since [ ]H

We

Q= or H C

H

Q QeQ−

= , this means that H Cideal

H

T TeT−

= or 1 C

H

TT

− .

On the textbook, the equation is written as H Cc

H

T TeT−

= (C for Carnot).

Carnot’s equation is considered incredibly important because it places a theoretical limit on the efficiency of an engine. For example, a gasoline engine which burns fuel at 800oC and exhausts at 300o C can achieve, at most, an efficiency of 46.5% (1 – 573/1073). Real engines can never achieve Carnot ideal efficiency because of friction and such. Well designed engines may achieve efficiencies as high as 60-80% of the theoretical ideal (Carnot) value.

Thermodynamics Day II and III: IV. Ideal Gases Gases are probably the best understood phase of matter. You probably learned a good deal about the properties of gases in Chemistry. We will essentially review this material, approaching it from a slightly different perspective than you did in Chemistry. A. Gas Quantities: Gases are quantified according to 4 basic quantities: V (volume, usually in m3) P (pressure, usually in Pa) T (temperature, always in K) and N (number of atoms/molecules) or n (number of moles) (N = n*Na, Avogadro’s number, 6.022X1023 particles/mole) B. Early Gas Laws: The gas laws are mathematical relationships which were discovered empirically by European scientists during the 17th and 18th centuries. In order of discovery, they are PI. Boyle’s Law (1660): P1V1 = P2V2 or PV = constant (*assumes that n, T are fixed)

II. Charles’ Law (late 1700s): 1 2

1 2

V VT T

=

or .V constT=

(*assumes n, P are fixed) III. Gay-Lussac’s Law (early 1800s):

1

1 2

P PT T

= 2 or .P constT=

(*assumes n, V fixed) IV. Avogadro’s’ Principle (1811):

1

1 2

V Vn n= 2 or .V const

n=

(*assumes P, T fixed)

V V

T (K) P

T(K) V

n

C: *Ideal Gas Law: Boyle’s, Charles’, Gay-Lussac’s, and Avogadro’s laws can

be combined to give a “combined gas law”: 1 1 2 2

1 1 2 2

PV PVn T n T

= or .PV constnT

=

What is remarkable is that this constant is very nearly the same for all gases! The constant, as you may remember, is usually called “R”, the universal gas constant, and the

relationship is written as PV RnT

= , or more familiarly,

PV = nRT, where R = 8.31 J/mol*K The equation may also be written in terms of number of molecules instead of number of moles. In this case, the equation becomes PV = NkbT , where kb = R/Na = 1.38 X 10-23 J/K, called “Boltzmann’s constant” D. Kinetic Molecular Theory of an ideal gas Real gases are not ideal gases. Ideal gases follow all of the gas laws perfectly, including the ideal gas law. They also never condense into liquids or solids. Real gases follow the ideal gas law very closely under conditions of very high Temperature and low Pressure but poorly under conditions of lower Temperature and higher Pressure. Real gases, of course, will eventually condense into liquids and/or solids which don’t even come close to obeying any of the gas laws. Why? Because the ideal gas law is actually based upon 5 fundamental assumptions that do not apply 100% to any real gas. They are 1. The particles in an ideal gas move in rapid motion with a bell-curve type distribution of speeds, in random directions and obey the laws of classical mechanics when they collide with each other or with the walls of the container that they are in. (random motion, obey Newton’s laws, etc.) 2. The space between the particles in an ideal gas are >> than the size of the particles themselves. The size of the particles themselves is completely negligible (Vparticles ≈ 0) 3. The particles in an ideal gas exert no attractive or repulsive forces on each other except when they collide. (Fparticle-particle≈0 except during collisions) 4. All collisions between particles and between particles and container walls are perfectly elastic. (all collisions perfectly elastic) 5. The average kinetic energy of an ideal gas particle is proportional to the Kelvin temperature of the gas. For monoatomic gases, more specifically KEavg. = 3/2 kbT

E. DERIVATION OF THE IDEAL GAS LAW Starting with the 5 assumptions of the kinetic molecular theory of an ideal gas, the ideal gas law can actually be proven using some clever physics/mathematics. This is sometimes called the “particle in a box derivation”. Picture a single particle in a rectangular box of length L and cross sectional area A. The particle has a mass m and begins with some speed v, which has components vx, vy, and vz.

Each time the particle collides with the wall, the collision is elastic. Since the wall is essentially infinitely massive compared to the particle, this means that the particle will bounce off the wall with a speed equal and a direction opposite to the way in which it was traveling before it hit the wall. Actually, only one of the vector components of the velocity will be reversed, in this case the vx. The change in momentum ∆p of the particle each time it hits the left wall is given by ∆p = m∆v = m(vx-(-vx)) = 2mvx. After colliding with the left wall, the particle will bounce back and hit the right wall, eventually coming back and hitting the left wall again. The time it will take to return to the left wall is given by t = ∆x/vx = 2L/vx The average force the left wall exerts on this particle is given by

Favg = ∆p/∆t = 22

2x x

x

mv mvL L

v

=

According to Newton’s 3rd law, the magnitude of the average force exerted by the particle on the wall is the same, Favg = mvx

2/L. Assuming that the particle moves in random direction, then the average force exerted by the particle on any wall should have this magnitude. Since Pressure = Force/Area, the pressure by this single particle would be given by

2

2x

x

mvmvF LP

A A AL= = =

Since AL = the volume of the container, 2 2

x xmv mvFPA AL V

= = =

The last expression on the previous page can be rearranged to yield PV = mvx2. So far,

we have assumed only that the particle travels in straight lines, the collisions with the wall are elastic, and that the mass of the particle is negligible compared to the mass of the wall. If we further assume that the vx, vy, and vz of the particle will, on average, be the same, we can take this a little further. By Pythagorean theorem, the overall speed of the particle “v” is given by 2 2 2

x y zv v v v= + + Therefore, 2 2 2 2

x y zv v v v= + + Assuming that vx, vy, and vz contribute (on the average), equally to the speed of the particle, then

2 2 2x y zv v v= = and 2 3 2

xv v= Rearranged, 2

2

3xvv = .

Substituting this into the result show in bold at the top of the page yields 2

3mvPV =

Since this result is for just a single particle, if our box contains “N” particles, then each

particle would contribute to the pressure and the result would become 2

3NmvPV =

By definition, K.E. = ½ mv2. Thus our equation above can be rewritten

22 2 ( . . )3 2 3 avg

NmvPV N K E= = . Finally, using assumption #5 that 32 bKE k T= , we

achieve the result 2 3(3 2 bPV N k T= ) or PV = NkbT !!!

(Since N = nNA and kb = R/N, this can also be written as the more familiar PV = nRT)

F. More definitions: INTERNAL ENERGY (U): Internal energy is a catchall for the total energy stored within the atoms or molecules of a substance. This includes both the kinetic energy of the atoms/molecules and the potential energy associated with the interatomic/ intermolecular bonding/structure/state. Because of the potential energy part, we can never strictly define the absolute internal energy of any real substance. We can speak only of changes in internal energy (∆U). These are of paramount importance in thermodynamics. Since the particles of an ideal gas do not attract or repel each other at all, an ideal gas has no potential energy term and U depends only on the kinetic energy of the atoms, which

for a monoatomic gas is related to the temperature by 3. .2 bK E k T= . For an ideal

monoatomic gas consisting of N atoms, 32 bU Nk= T . For an ideal gas consisting of n

moles, 32

U nR= T , so 32

U nRΔ = ΔT

WORK (W): You know one definition of work already: W = Fdcosθ. In thermodynamics, work is also of tremendous importance, particularly for the subject of heat engines. Heat engines are essentially devices for converting heat energy into work. Like heat, work is a name given to a temporary form of energy. Work is a transfer of energy from one body or substance to another not based on a difference in temperature between the two bodies. This latter part of the definition is to distinguish work from heat. When a gas expands, it does positive work on its environment and negative work is done on it. When a gas is compressed, it does negative work on its environment while positive work is done on it. It will be shown below that the amount of work done is given by Won = -Pavg∆V (alternatively, Wby = +Pavg∆V )

By definition, P = F/A, so the force exerted on the gas is given by F = P*A. When a piston containing an enclosed gas is compressed an amount d, the work done on the gas is given by W = Fd, which can be rewritten as W = PAd. The change in volume of the gas ∆V = -Ad. The work done on the gas can therefore be rewritten as -P∆V.

G. Laws of Thermodynamics The laws of thermodynamics were pieced together by a number of scientists during the 19th century. They have been stated in various different ways by different scientists and are thus not usually accredited to particular authors. The different ways in which they can be stated often seem like completely different laws to a beginning student. There is a “zeroth” law of thermodynamics, called the “zeroth” law because it the most basic but was overlooked until years after the other 2 laws had already been written. Zeroth law: If object A is in thermal equilibrium with object B and object B is in thermal equilibrium with object C, then object A is in thermal equilibrium with object C. Don’t worry about it. **1st law: The total energy of an isolated system is always conserved. In thermodynamics, the 1st law is usually written as ∆U = Q + W All of these quantities are taken with respect to the enclosed gas: ∆U refers to the change in internal energy of the gas. Q refers to the amount of heat that flows into the gas from the environment. W refers to the amount of work done on the gas*. If heat flows into the gas and/or positive work is done on the gas, the internal energy of the gas will increase, which implies that the Temperature will as well, since ∆U = 3/2 nR∆T (at least for a monoatomic gas). Heat flowing out of a gas and/or positive work done by the gas will result in a decrease in internal energy and in temperature. *in the past, the convention was used that W referred to the amount of work done by the gas. Using that convention, the law was written as ∆U = Q – W 2nd law: The entropy of a closed system may never decrease. It may stay the same (for reversible processes) or increase (for irreversible processes). Since no real process is perfectly reversible and the only really closed system is the universe as a whole, the 2nd law may also be stated as “for any real process, the entropy of the universe always increases”. Entropy, as you may recall, roughly relates to“disorder”. A more precise definition of entropy relates to the number of “microstates” of a system: S = kblnW, where W is the “multiplicity” of the system (essentially, the number of possible microstates corresponding to a given macrostate). To do half-decent calculations with entropy requires calculus so don’t worry about it. Implications of the 2nd law include that heat will always flow from an object at higher Temperature into an object at lower temperature, never the other way around. Also, there is no such thing as a perfectly efficient engine or a perfectly efficient refrigerator. Also, gases will spontaneously expand into empty space but never spontaneously contract to create empty space. Doubtless, there are many others.

H: Types of Thermodynamic Processes When a gas is heated/cooled and/or expanded/compressed, various changes occur in the temperature (T), pressure (P), volume (V), and internal energy (U) of the enclosed gas. In addition, heat (Q) may be exchanged with the environment and work (W) may be done. You are expected to be familiar with several idealized types of thermodynamic processes, called isochoric, isobaric, isothermal, and adiabatic. I. Isochoric. P

V II. Isobaric P

V III. Isothermal P

V IV. Adiabatic P

V (adiabatic curve is steeper than isothermal)

Isochoric means constant volume (∆V = 0). That means that as the gas is heated/cooled, the pressure increases/decreases proportionally. Since ∆V = 0, no work is done. Thus, during an isochoric process, ∆U = Q + W = Q + 0 ∆U = Q Isobaric means constant pressure. That means that as the gas is heated/cooled, the volume increases/decreases proportionally. Since the pressure is constant, we can easily calculate the work done on the gas using W = -P∆V Isothermal means that the temperature stays constant. This means that as the gas expands/contracts, the pressure decreases/increases inversely to the volume change. Since the internal energy of an ideal gas depends only on temperature, ∆U = 0 for an isothermal process. Thus the 1st law becomes ∆U = 0 = Q + W Q = -W Adiabatic means that no heat is exchanged with the environment (Q = 0). An adiabatic expansion is sometimes also referred to as a free expansion. During an adiabatic expansion, the temperature of the gas drops as internal energy is used to do work on the environment. During an adiabatic compression, the temperature increases. The pressure, volume, and temperature all change during an adiabatic process and calculus is needed to calculate them. The 1st law, though, becomes ∆U = Q + W = 0 + W ∆U = W

I. Engine cycles Probably the most common type of long problem involving thermodynamics has been a problem involving a P-V diagram for an engine cycle. Gases are assumed to be ideal, systems are assumed to be closed, and processes are assumed to be reversible. The gas is assumed to return to its initial conditions at the end of the cycle. Students are asked to calculate QH, QC, W, ∆U, etc. for various steps in the cycle and/or for the cycle as a whole. The ideal gas law may be needed to calculate Temperatures at various points in the cycle. It is important to realize that for any closed cycle, the change in temperature and thus the change in internal energy of the gas must be zero, since the gas returns to its original conditions. ∆Ucycle = 0 Qcycle = -Wcycle Here is an example problem:

a) calculate the work done on the gas during processes AB, BC, and CA and the net work for one complete cycle. AB is an isobaric process, so the work done is given by WAB = -P∆V WAB = -P(VB-VA) = - 20 Pa (3.0 m3 – 1.0 m3) = -40 J (the work done by the gas is +40 J) BC is an isochoric process. ∆V = 0 so W = 0. CA is a process which occurs at varying pressure but the pressure varies linearly with the volume. The work done can thus be easily calculated either by using the average pressure during the process or by calculating the area under line CA

3 340 20( )(1.0 3.0 ) 602CA

Pa PaW m+= − − = +m J (the work done by the gas is -60 J)

The work done on the gas for a complete cycle is -40 J + 0 + 60 J = 20 J. This means that the work done by the gas for a complete cycle is -20 J. This is not a heat engine, this is a refrigerator. In a heat engine, the gas does net positive work. In a refrigerator, positive work is done on the gas b) If the temperature at point A is 250 K, calculate the temperatures at points B and C: PV/nT is constant. Assuming the system is closed, PV/T is constant. Plugging in the numbers, TB = 750 K and TC = 1500 K c) fill in the signs of Q, W, and ∆U for each part of the cycle and for the cycle as a whole.