An Introductory Course in Waterloo Collegiate Institute

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An Introductory Course in Waterloo Collegiate Institute
SPH3UW Thermal Physics An Introductory Course in Thermodynamics Waterloo Collegiate Institute

Welcome to Thermodynamics
Heat, Temperature, and Power Thermal Expansion Ideal Gas Law Kinetic Theory of Gases First Law of Thermodynamics Work and PV Diagrams Isothermal Processes Adiabatic Processes Isobaric Processes Isochoric Process Heat Engines and the Second Law of Thermodynamics

Thermal Physics electrons and holes in semiconductors
converting energy into work magnetism thin films and surface chemistry thermal radiation (global warming) and much more…

Some Definitions Absolute Zero: the lowest possible temperature, at which all molecular motion would cease and a gas would have zero volume. Calorie: the amount of heat required to raise the temperature of one gram of water by one Celsius degree. Calorimeter: device which isolates objects to measure temperature changes due to heat flow. Celsius (C): temperature scale in which the freezing point of water is 0 and the boiling point of water is 100 Convection: heat transfer by the movement of a heated substance, due to the differences in density. Conduction: heat transfer from molecule to molecule in substances due to differences in temperature.

Heat Heat, represented by the variable Q, is a type of energy that can be transferred from one body to another Heat is measured in Joules Energy must be transferred in order to be called heat. (So heat may be gained or lost, but not possessed. It is incorrect to say, “a gas has 4000 J of heat” Internal energy: The sum of the energies of all of the molecules in a substance. Represented by the variable U (for example: the total of the Kinetic and Potential energy at the molecular level is called the Internal Energy of the system. U=KE+PE) Temperature: Related to the average kinetic energy per molecule of a substance.

What is Heat? Up to mid-1800’s heat was considered a substance -- a “caloric fluid” that could be stored in an object and transferred between objects. After 1850, kinetic theory. A more recent and still common misconception is that heat is the quantity of thermal energy in an object. The term Heat (Q) is properly used to describe energy in transit, thermal energy transferred into or out of a system from a thermal reservoir … (like cash transfers into and out of your bank account) Q U Sign of Q : Q > 0 system gains thermal energy Q < 0 system loses thermal energy W > 0 work done on system W < 0 work done by the systems So we give Q + W a name: The Internal Energy

Still More Heat We will be discussing three states of matter (solid, liquid, and gas). The molecules of a solid are fixed in a rigid structure. The molecules of a liquid are loosely bound and may mix with one another freely. (While a liquid has a definite volume, it still takes the shape of its container. The molecules of a gas interact with each other slightly, but usually move at higher speeds than that of solid of liquid. In all three states of matter the molecule are moving and therefore have Kinetic Energy. But, they also have Potential Energy due to the bonds between them. The sum of the potential and kinetic energies of the molecule of the substance is also known as its Internal Energy. When a warmer substance comes in contact with a cooler substance, some of the kinetic energy of the molecules in the warmer substance is transferred to the cooler substance. The energy representing this kinetic energy of the molecule that is transferred from the warmer to the cooler substance is called heat energy.

What is temperature? A mercury thermometer The mercury rises up
the tube as it expands. This is movement. An object (say, a cup of hot chocolate) The mercury is gaining (internal) energy from the hot chocolate. This transfer of energy is what we call heat. When the transfer stops, the objects are in thermal equilibrium.

Temperature Scales Celsius (0C)
Zero defined by an ice-water bath at 1 atm. Unit defined by water-steam (100ºC) at 1 atm. Kelvin (absolute K) Zero defined by absolute zero, but we cannot reach that temperature experimentally K defined by the triple-point of water (0.01ºC at 4.58 mm of mercury, water can exist in all three states of matter) Unit is the same as the Celsius scale Fahrenheit (0F) Zero and unit based on salt-water, water freezes at 32 0F and boils at 212 0F )

Converting Between the Scales
From Celsius to Kelvin: From Fahrenheit to Celsius:

Example You place a small piece of melting ice in your mouth. Eventually, the water all converts from the ice at T1= F to body temperature, T2 = F. Express these temperatures as 0C and K. Plan: We convert Fahrenheit to Celsius temperature, then from Celsius to Kelvin

Mechanical Equivalent of Heat
James Joule showed that mechanical energy could be converted to heat and arrived at the conclusion that heat was another form of energy. He showed that 1 calorie (c) of heat was equivalent to J of work. (that is 1 calorie is defined as the heat needed to raise the temperature of 1 gram of water 1 0C) 1 cal = J Kilocalorie(C) – the amount of heat needed to raise the temperature of 1,000 grams of water by 1 °C. (Used with food, Food calories (C) are determined by burning the food and measuring the amount of energy that is released.) British Thermal Units (BTUs) are the amount of heat to raise one pound of water by 1 °F.

Heat Transfer and Temperature Change
The change in temperature that a substance experiences depends upon two things: its identity (specific heat) and the amount of material (mass). The equation that connects the amount of heat, Q, and the resulting temperature change , T in 0C, is: Where Q is the quantity of heat (calories) m is the mass of the sample in grams and c is the intrinsic property called specific heat capacity in 1 cal/g°C. Note: that positive Q is interpreted as heat coming in (T is positive, so T increases), while negative Q corresponds to heat going out (T is negative, so T decreases).

Heat Transfer Example During a bout with the flu an 80. kg man ran a fever of C instead of the normal body temperature of C. Assuming that the human body is mostly water (c=1cal/g 0C), how much heat, in calories and Joules, is required to raise his temperature by that that amount? We could also use:

Heat Transfer and Phase Changes
Consider an ice cube. Since water freezes at 0 0C, the temperature of the ice cube is 0 0C. If we add heat to the ice, its temperature does not rise. Instead the thermal energy absorbed by the ice goes into loosening the intermolecular bonds of the ice, thereby transforming it into a liquid. The temperature remains at 0 0C. In each phase change (solid to liquid, liquid to gas), absorbed heat causes no temperature change so Q=mcT does not apply. The equation we use is: Where L is the latent heat of transformation (solid to liquid, or vice versa L is latent heat of fusion. From liquid to gas, L is called latent heat of vaporization). This equation tells us how much heat must be transferred in order to cause a sample of mass m to undergo a phase change.

Example on Temperature and Phase
You want to cool 0.25kg of water, initially at 25 0C, by adding ice, initially at -20 0C. How much ice should you add so that the final temperature will be 0 0C with all the ice melted [cwater = 4190 J/kg K, cice=2100 J/kg K, L=334000J/kg]? The ice and water are the objects that exchange heat. The water undergoes a temperature change only, while the ice undergoes both a temperature and phase change. We require the mass of the ice. Let’s first determine the negative heat added to the water. For Ice, first we determine the heat needed to warm the ice.

Example on Temperature and Phase
You want to cool 0.25kg of water, initially at 25 0C, by adding ice, initially at -20 0C. How much ice should you add so that the final temperature will be 0 0C with all the ice melted [cwater = 4190 J/kg K, L=334000J/kg]? For Ice, now we need the heat to phase shift it from solid to liquid. The sum of these quantities must be zero

Heat Transfer and Thermal Expansion
When a substance undergoes a temperature change, it changes in size. Changes in length due to temperature is governed by: Where  is the coefficient of linear expansion of the material Similarly, there are formulas for changes in Area and volumes Water contracts as it is cooled until it reaches 4 °C. From then on, cooling causes expansion. This is because freezing water has a crystalline structure.

Understanding You need to slide an aluminum ring onto a rod. At room temperature (20oC), the internal diameter of the ring is 50.0 mm and the diameter of the rod is 50.1 mm. Should you heat or cool the ring to make it fit onto the rod? If the coefficient of linear expansion for aluminum is 2.5 x 10-5 oC-1, at what temperature will the ring just barely fit onto the rod? a) Heat, even the hole will expand b) Therefore the final temperature is 20oC+80oC or 100oC

The Kinetic Theory of Gases
Unlike the condensed phases of matter (solid or liquid), molecules of a gas move freely and rapidly. A confined gas exerts a force on the walls of its container. The moving molecules strike the walls and rebound. The magnitude of the force per unit area is called pressure, and is denoted by P: The SI unit for pressure is the N/m2, the pascal (Pa). We also need a way to talk about the vast number of molecules in a given sample of gas. One mole of atoms or molecules (or teachers)contains of these elementary quantities. NA is known as Avogadro’s Number.

Molecular Picture of Gas
Gas is made up of many individual molecules Number density is Number of molecules/Volume: N/V = r/m r is the mass density m is the mass for one molecule Number of moles: n = N / NA NA = Avogadro’s Number = 6.022x1023 mole-1 Mass of 1 mole of “stuff” in grams = molecular mass in u e.g., 1 mole of N2 has mass of 2x14=28 grams Atomic mass unit

The Ideal Gas Law Three physical properties – Pressure (P), volume (V), and temperature (T) describe a gas. At low densities, all gasses approach ideal behaviour; This means that these three variables are related by the equation: Where n is the number of moles of gas and R is the constant (8.31 J/mol K), called the universal gas constant. This equation is known as the Ideal Gas Law.

The Ideal Gas Law Therefore Boltzmann’s constant
An important consequence of this equation is that, for a fixed volume of gas, an increase in P gives a proportional increase in T. By applying Newton’s Second Law we can find that the pressure – exerted by N molecules of gas in a container of volume V is related to the average kinetic energy of the molecules by the equation: Boltzmann’s constant Therefore m=molecular mass (kg) M=molar mass (kg/mole) root-mean-square

The Ideal Gas Law We recall that the sum of the Kinetic and Potential energy (at the molecular level) is called the Internal energy of the system. With the average kinetic energy of a molecule related to temperature by: Atoms of a single element Then for an ideal monatomic gas (No potential energy) containing N molecules, the Internal energy will just be N time this equation n is the number of Moles and R the Ideal gas constant.

The Ideal Gas Law Calculate the internal energy of the air in a typical room with a volume of 50 m3. treat the air as if it were a monatomic ideal gas at 1 atm = 1.01 x 105 PA Hint:

P V = N kB T P V = n R T The Ideal Gas Law
P = pressure in N/m2 (or Pascals) V = volume in m3 N = number of molecules T = absolute temperature in K k B = Boltzmann’s constant = 1.38 x J/K Note: P V has units of N-m or J (energy!) P V = n R T n = number of moles R = ideal gas constant = NAkB = 8.31 J/(mol*K) 20

Ideal Gas Law ACT II PV = nRT
You inflate the tires of your car so the pressure is 30 psi, when the air inside the tires is at 20 degrees C. After driving on the highway for a while, the air inside the tires heats up to 38 C. Which number is closest to the new air pressure? 1) 16 psi 2) 32 psi 3) 57 psi Careful, you need to use the temperature in K P = P0 (38+273)/(20+273) 23

Boyle’s Law The quantities P, V, n, and T aren’t independent but are related by an equation of state. You can perform various experiments where two of these quantities are held fixed and a relation between the other two is determined. These mini-laws have names like Boyle’s Law and Charles’s Law. PV=constant n and T held fixed

Charles’s Law The quantities P, V, n, and T aren’t independent but are related by an equation of state. You can perform various experiments where two of these quantities are held fixed and a relation between the other two is determined. These mini-laws have names like Boyle’s Law and Charles’s Law. V/T=constant n and P held fixed

Ideal Gas Law: ACT 1 PV = nRT
A piston has volume 20 ml, and pressure of 30 psi. If the volume is decreased to 10 ml, what is the new pressure? (Assume T is constant.) 1) ) ) 15 When n and T are constant, we have PV = constant (Boyle’s Law) Change this to the pressure change with temperature, but give the temperature in C. V=20 P=30 V=10 P=??

Balloon is still open to atmospheric pressure, so it stays at 1 atm
Balloon ACT 1 What happens to the pressure of the air inside a hot-air balloon when the air is heated? (Assume V is constant) 1) Increases 2) Same 3) Decreases Balloon is still open to atmospheric pressure, so it stays at 1 atm

FB = r V g, r is density of outside air!
Balloon ACT 2 What happens to the buoyant force on the balloon when the air is heated? (Assume V remains constant) 1) Increases 2) Same 3) Decreases FB = r V g, r is density of outside air! 32

P and V are constant. If T increases N decreases.
Balloon ACT 3 What happens to the number of air molecules inside the balloon when the air is heated? (Assume V remains constant) 1) Increases 2) Same 3) Decreases PV = NkT P and V are constant. If T increases N decreases. 34

Balloon Summary In terms of the ideal gas law, explain briefly how a hot air balloon works. Hot air has less mole density than cool air. So less hot air is required in order to achieve the same pressure as cool air. This makes the density of hot air less allowing it to float. When temperature increases the volume of the gas increases, thus reducing the density of the gas making it lighter that then surrounding air, which causes the balloon to rise. Note! this is not a pressure effect, it is a density effect. As T increases, the density decreases the balloon then floats due to Archimedes principle. The pressure remains constant!

The Laws of Thermodynamics
We have learned about two ways in which energy may be transferred between a system and its environment. One is work (force act over a distance), the other is heat (energy transferred due to a difference in temperature). The study of energy transfers involving work and heat, and the resulting changes in internal energy, temperature, volume, and pressure is called thermodynamics. The order of discovery: (A hint at the logic of Thermodynamics) The 2nd Law was discovered First The 1st Law was discovered Second The 0th Law was the third to be discovered The 4th Law is called the 3rd Law

The Zeroth Law of Thermodynamics
When two objects are brought into contact, heat will flow from the warmer object to the cooler one until they reach thermal equilibrium. If Objects 1 and 2 are each in thermal equilibrium with Object 3, then Objects 1 and 2 are in thermal equilibrium with each other.

The First Law of Thermodynamics
Simply put, the first Law of Thermodynamics is a statement of the Conservation of energy that includes heat. So what does the First Law say? In words, the internal energy of a body (such as a gas) can be increased by heating it or by doing mechanical work on it James Joule showed that mechanical energy could be converted to heat and arrived at the conclusion that heat was another form of energy. He showed that 1 calorie of heat was equivalent to J of work. 1 cal = J

Changing the Internal Energy
U is a “state” function --- depends uniquely on the state of the system in terms of p, V, T etc. (e.g. For a classical ideal gas, U = NkT ) There are two ways to change the internal energy of a system: WORK done by the system on the environment Wby = -Won HEAT is the transfer of thermal energy into the system from the surroundings Q Thermal reservoir Work and Heat are process energies, not state functions.

The First Law of Thermodynamics
The first law deals with changes in the internal energy of the system. It is important to note that changes in the internal energy of a system can only occur if the system is not isolated. This tells us that the system is imbedded in its surroundings in a way that there can be energy transfer. There are two types of energy transfer (Work and Heat), and the difference between these two is determined only by what occurs in the surroundings of the system. Work: is an energy transfer between a system and its surroundings that is the result of an organized motion in the surroundings. For example: You can increase the internal energy of a piece of plastic by vigorously rubbing it. You can decrease the internal energy of a gas by letting it expand against some external pressure (such as a piston). Piston examples are commonly used to illustrate many concepts of Thermodynamics, so Let’s look into this more deeply.

The First Law of Thermodynamics
Consider an example system of a piston and cylinder with an enclosed dilute gas characterized by P,V,T & n.

The First Law of Thermodynamics
What happens to the gas if the piston is moved inwards?

The First Law of Thermodynamics
If the container is insulated the temperature will rise, the atoms move faster and the pressure rises. Is there more internal energy in the gas? Yes!

The First Law of Thermodynamics
External agent did work in pushing the piston inward. W = Fd =(PA)Dx W =PDV Dx

The First Law of Thermodynamics
Work done on the gas equals the change in the gases internal energy, W = DU Dx

The First Law of Thermodynamics
Direction of arrow (left to right is very important We can represent the force the gas exerts on the piston or the external force that compresses the piston by a PV diagram. V P 1 2 V V2 (+) area (+) work done by gas (-) work done on gas This tells us that the work done by the gas during the expansion (from 1 to 2) is just the area under the curve. By convention, this area is positive for positive work done by the gas. This also corresponds to negative work done on the gas. We must be careful to distinguish the work done on the gas and the work done by the gas. V P 1 2 V V2 (-) area (-) work done by gas (+) work done on gas

The First Law of Thermodynamics
Example: One mole of monatomic ideal gas is enclosed under a frictionless piston. A series of processes occur, and finally the state of this ideal gas returns to its initial state (see PV diagram). Answer the following questions in terms of P0, V0, and R. Determine the temperature at each vertex. Determine the change in internal energy (U) for each process. Determine the work done by the gas for each process. (a) Use V P 1 2 3 V V V0 P0 1/2P0

The First Law of Thermodynamics
Example: One mole of monatomic ideal gas is enclosed under a frictionless piston. A series of processes occur, and finally the state of this ideal gas returns to its initial state (see PV diagram). Answer the following questions in terms of P0, V0, and R. Determine the temperature at each vertex. Determine the change in internal energy (U) for each process. Determine the work done by the gas for each process. (a) Use V P 1 2 3 V V V0 P0 1/2P0

The First Law of Thermodynamics
Example: One mole of monatomic ideal gas is enclosed under a frictionless piston. A series of processes occur, and finally the state of this ideal gas returns to its initial state (see PV diagram). Answer the following questions in terms of P0, V0, and R. Determine the temperature at each vertex. Determine the change in internal energy (U) for each process. Determine the work done by the gas for each process. (a) Use area under graph for each segment V P 1 2 3 V V V0 P0 1/2P0 Note: No work for constant volume. Total work 3/4P0V0 is same as area enclosed by triangle.

The First Law of Thermodynamics
Summary: T U Wby Won 12 3(PoVo)/R 4.5PoVo 3PoVo -3PoVo 23 -2(PoVo)/R 31 -1(PoVo)/R -1.5PoVo -2.25PoVo +2.25PoVo V P 1 2 3 V V V0 P0 1/2P0

First Law of Thermodynamics
Let’s change the situation: Keep the piston fixed at its original location. Place the cylinder on a hot plate. What happens to gas?

The First Law of Thermodynamics
Heat flows into the gas. Atoms move faster, internal energy increases. Q = heat in Joules DU = change in internal energy in Joules. Q = DU

The First Law of Thermodynamics
Heat: Heat is an energy transfer between a system and its surroundings that is the result of random motion in the surroundings. (Note: there is a difference between a work process [organized motion in surroundings] and a heat process [random motion in surroundings]). Heat will always flow spontaneously from the system at higher temperature to the system of lower temperature, but heat can be made to flow in the opposite direction as well if work is done in the process (a refrigerator).

The First Law of Thermodynamics
What if we added heat and pushed the piston in at the same time?

The First Law of Thermodynamics
Work is done on the gas, heat is added to the gas and the internal energy of the gas increases! DU=Q+W F

First Law of Thermodynamics
Energy Conservation :The change in internal energy of a system (U) is equal to the heat flow into the system (Q) minus the work done by the system (W) DU = Q - W Work done by system Increase in internal energy of system V P 1 2 3 V V2 P1 P3 ideal gas Heat flow into system Equivalent way of writing 1st Law: DU = Q + W Work done on system Increase in internal energy of system Heat flow into system 07

First Law of Thermodynamics
If we re-examine our PV diagram and now calculate the heat transfer (Q) for each process. Since the work done by the gas and the change in internal energy has already been calculated, it is very easy to calculate the heat transfer from the First Law: DU = Qinto + Won V P 1 2 3 V V V0 P0 1/2P0

First Law of Thermodynamics
If we re-examine our PV diagram and now calculate the heat transfer (Q) for each process. Since the work done by the gas and the change in internal energy has already been calculated, it is very easy to calculate the heat transfer from the First Law: DU = Qinto + Won V P 1 2 3 V V V0 P0 1/2P0

First Law of Thermodynamics
If we re-examine our PV diagram and now calculate the heat transfer (Q) for each process. Since the work done by the gas and the change in internal energy has already been calculated, it is very easy to calculate the heat transfer from the First Law: DU = Qinto + Won V P 1 2 3 V V V0 P0 1/2P0

First Law of Thermodynamics
U Wby Qin 12 3(PoVo)/R 4.5PoVo 3PoVo 7.5PoVo 23 -2(PoVo)/R -3PoVo 31 -1(PoVo)/R -1.5PoVo -2.25PoVo -3.75PoVo V P 1 2 3 V V V0 P0 1/2P0

(1) Positive, heat flows into soup (2) Zero (is close enough)
Signs Example You are heating some soup in a pan on the stove. To keep it from burning, you also stir the soup. Apply the 1st law of thermodynamics to the soup. What is the sign of : 1) Q 2) W 3) DU (1) Positive, heat flows into soup (2) Zero (is close enough) (3) Positive, Soup gets warmer

Signs Energy is also used to mean the potential to do work; the greater the ability of something to change the things around it, the more energy it has. Mathematically, the first law is expressed: U= Q+W. Where Q is the change in the quantity of heat added to the system, U is the change in the internal energy and W is the work done on or by the system. The sign convention here is that if U is positive the amount of internal energy increases. This means that Q stands for the heat energy put into the system and W for the work done on the system. This is known as the ‘physicists’ convention’. Work W>0: Work is done on the system by the surroundings W<0: Work is done by the system on the surroundings Heat Q>0: Heat is added to the system from the surroundings Q<0: Heat is released by the system to the surroundings

First Law of Thermodynamics
For each process in this cycle, indicate in the table below whether the quantities W, Q, and U are positive (+), negative (-), or zero (0). W is the work done BY the GAS. Process W Q U 1 -> 2 2 -> 3 3 -> 1 + + + - - - - - V P 1 2 3 V V V0 P0 1/2P0

W = P V :For constant Pressure
Work Done by a Gas ACT M M y The work done by a gas as it contracts is A) Positive B) Zero C) Negative W = F d cosq < 0 = P A d = P A Dy = P DV Note that equation only works for constant pressure! W = P V :For constant Pressure W > 0 if V > 0 expanding system does positive work W < 0 if V < 0 contracting system does negative work W = 0 if V = 0 system with constant volume does no work

Work Done by a Gas ACT A gas is kept in a cylinder that can be compressed by pushing down on a piston. You add 2500 J of heat into the system, and then you push the piston 1.0 m down with a constant force of 1800 N. What is the change in the gas’s internal energy. We define the variables in the First Law, being careful of the signs: Heat added : Q=+2500 J We did work on gas, so W is positive: Therefore:

Pressure as a Function of Volume
Work is the area under the curve of a PV-diagram. Work depends on the path taken in “PV space.” When a process is depicted on a PV diagram, directional arrows are used on the graph. In a process that moves to the right, the gas will do positive work on the surroundings; in a process that moves to the left, the gas will do negative work on the surroundings (ie the surroundings do work on the gas) The precise path serves to describe the kind of process that took place.

Work Done by a Gas Work done by gas equals area inside graph

P-V Diagrams When you are describing a process that involves a gas expanding or compressing, you will need to use specific nomenclature. A process is called isobaric if the pressure remains constant throughout. It is called isothermal if the temperature remains constant. If no heat energy enters or leaves the gas, it is called adiabatic.

Different Thermodynamic Paths
The work done depends on the initial and final states and the path taken between these states.

Thermodynamic Systems and P-V Diagrams
ideal gas law: PV = nRT (nR = NkB) for n fixed, P and V determine “state” of monatomic ideal gas system T = PV/nR U = (3/2)nRT = (3/2)PV Examples: which point has highest T? B which point has lowest U? C to change the system from C to B, energy must be added to system V P A B C V V2 P1 P3

First Law of Thermodynamics Isobaric Example
2 moles of monatomic ideal gas is taken from state 1 to state 2 at constant pressure P=1000 Pa, where V1 =2m3 and V2 =3m3 (R=8.31 J/k mole) Find: 1) T1 2) T2 3)DU 4) W 5) Q V P 1 2 V V2 1. PV1 = nRT1  T1 = PV1/nR = 120K 2. PV2 = nRT2  T2 = PV2/nR = 180K 3. DU = (3/2) nR DT = 1500 J DU = (3/2) P DV = 1500 J (has to be the same) 4. W = P DV = J 5. Q = DU + W = = 2500 J 21

First Law of Thermodynamics Isochoric Example
2 moles of monatomic ideal gas is taken from state 1 to state 2 at constant volume V=2m3, where T1=120K and T2 =180K. Find Q. P P2 P1 2 Q = DU + W DU = (3/2) nR DT = 1500 J W = P DV = 0 J Q = DU + W = = 1500 J 1 V V Requires less heat to raise T at const. volume than at const. pressure (no energy used for work)

Heat Transfer and Temperature Change for Gases
The value of the heat transfer in a system, Q, is also path-dependent. We recall for a solid or liquid Q=mcT. For gases, however, the system is a little more complicated, because the value of the proportionality constant between Q and T depends on whether the volume or pressure is kept constant. If the volume remains constant (isochoric) during heat transfer, then CV is the molar heat capacity at constant volume If the pressure remains constant (isobaric) during the heat transfer CP is the molar heat capacity at constant pressure

Total Work Done V P Wtot = ?? V P W = PDV (>0) DV > 0 V P
1 2 3 4 Wtot = ?? V P W = PDV (>0) 1 2 3 4 DV > 0 V P W = PDV = 0 1 2 3 4 DV = 0 V P 1 2 3 4 Wtot > 0 V P W = PDV (<0) 1 2 3 4 DV < 0 V W = PDV = 0 1 2 3 4 P DV = 0 General rule: work done is area under P-V curve (even if not horizontal).

Question 25 L of gas is enclosed in a cylinder/piston apparatus at 2 atm of pressure and 300 K. If 100 kg of mass is placed on the piston causing the gas to compress to 20 L at constant pressure. This is done by allowing heat to flow out of the gas. What is the work done on the gas? What is the change in internal energy of the gas? How much heat flowed out of the gas? Po = 202,600 Pa, Vo = m3, To = 300 K, Pf = 202,600 Pa, Vf=0.020 m3, W =-PDV = -202,600 Pa (0.020 – 0.025)m3 =1013 J energy added to the gas.

Question 25 L of monatomic gas is enclosed in a cylinder/piston apparatus at 2 atm of pressure and 300 K. If 100 kg of mass is placed on the piston causing the gas to compress to 20 L at constant pressure. This is done by allowing heat to flow out of the gas. What is the work done on the gas? What is the change in internal energy of the gas? How much heat flowed out of the gas?

Question 25 L of gas is enclosed in a cylinder/piston apparatus at 2 atm of pressure and 300 K. If 100 kg of mass is placed on the piston causing the gas to compress to 20 L at constant pressure. This is done by allowing heat to flow out of the gas. What is the work done on the gas? What is the change in internal energy of the gas? How much heat flowed out of the gas? DU = Won - Qout Qout = Won - DU = 1013J – (-1518J) = 2531 J heat out

Second Law of Thermodynamics
Entropy, S, is a measure of the disorder, or randomness of a system. The greater the disorder of a system, the greater the entropy. If a system is highly ordered (like particles is a solid) we say that the entropy is low. The second law of thermodynamics states that all spontaneous processes proceeding in an isolated system lead to an increase in entropy. The increase or decrease in entropy can be found by Where Q is heat flow into or out of a system and T is the average Kelvin temperature

Thermal Efficiency of a Heat Engine
The thermal efficiency, e, of the heat engine is equal to the ratio of the heat we get out to the heat we put in. Where Qin is heat absorbed, and Qout is heat discharged Unless Qout=0, the engines efficiency is always less than 1.

Efficiency If we re-examine our PV diagram and now calculate the efficiency of the system. The efficiency of a cycle is defined as the ratio of the work done by the gas to the heat Qin that flows into the system. Any heat that is expelled into the surroundings is not included in the calculation of Qin. From the point of view of efficiency, this expelled heat is lost and its energy is not used by the system: Wby Qin 12 3PoVo 7.5PoVo 23 -3PoVo 31 -2.25PoVo -3.75PoVo Total 0.75 PoVo V P 1 2 3 V V V0 P0 1/2P0

Heat Engine: Efficiency
The objective: turn heat from hot reservoir into work The cost: “waste heat” 1st Law: Qin -Qout = W efficiency e  W/Qin =W/Qin = (Qin-Qout)/Qin = 1-Qout/Qin TH TC Qin Qout W HEAT ENGINE 13

A hot (98 C) slab of metal is placed in a cool (5C) bucket of water.
ACT A hot (98 C) slab of metal is placed in a cool (5C) bucket of water. What happens to the entropy of the metal? A) Increase B) Same C) Decreases What happens to the entropy of the water? DS = Q/T Heat leaves metal: Q<0 Heat enters water: Q>0

Description of Disorder
Isolated systems tend toward greater disorder, and entropy is a measure of that disorder S = kB ln (W) kB is Boltzmann’s constant W is a number proportional to the probability that the system has a particular configuration This version is a statement of what is most probable rather than what must be The Second Law also defines the direction of time of all events as the direction in which the entropy of the universe increases

Second Law of Thermodynamics
The entropy change (Q/T) of the system+environment  0 never < 0 order to disorder Consequences A “disordered” state cannot spontaneously transform into an “ordered” state No engine operating between two reservoirs can be more efficient than one that produces 0 change in entropy. This is called a “Carnot engine” 31

Carnot Cycle

Carnot Now suppose you want to make an engine that runs on energy from a source of high temperature heat (at temperature TH) such as a burner. And, furthermore, suppose you want your heat engine to work in the most efficient way possible. The first thing you have to do is to get energy into your working fluid. Obviously, the best process to do this will be a constant temperature (isothermal) heat transfer, which will result in an expansion of the working gas at TH (shown on the graph as A-->B). Any other expansion process that results in a reduction of the temperature of the working gas below TH, will also result in a reduction of efficiency. [An analogy can be made here between temperature and pressure. Imagine a water turbine running from a dam -- ideally you want to keep the water level at the maximum so that you have the maximum pressure to turn the turbine. If you draw off water too fast then the water level (and therefore pressure) will drop and the power output of the turbine will be reduced.]

Carnot The isothermal expansion (A-->B) produces a work output -- yay! -- but at the end of the process (point B) you are faced with a dilemma. You want to get your piston back to it's original position, so that you can repeat the expansion/work output process again. But if you simply compress the working gas, then you have to do exactly the same amount of work to get back to the original state as you produced during the expansion process (and, of course, you end up rejecting your thermal energy back to the high temperature heat source). This would mean that you would get no net work output from your heat engine, which would be very unsatisfactory.

Carnot The answer, of course, is to perform another expansion process with the high temperature heat source disconnected from the system (shown on the graph as B-->C). The temperature (and pressure) of the working gas will fall, but because the system is not exchanging heat then you are not losing any of the 'potential' of the high temperature heat from the heat source. This then allows you to compress the working gas at low temperature and pressure (shown on the graph as C-->D), which requires much less work to return the piston almost to its original position. Even though you have to put work back into the engine to do this, it is much less than the work output during the isothermal expansion process (A-->B) -- and so, overall, the engine produces a NET work output. This compression process must, of course, reject heat -- which it does to a low temperature heat sink at temperature TC (usually the ambient environment). Again, an isothermal process is the best way to do this, in order to keep the temperature difference between the heat source and sink always at the maximum value.

Carnot Of course, the engine is not quite back to its original state, as the temperature of the working gas is still at TC. To return the temperature to TH, we disconnect the system from the low temperature heat sink, and perform another compression process (shown on the graph as D-->A). This requires a work input, which will be of exactly the same value as the work output in B-->C. But, of course -- to reiterate -- the engine will still have an overall net work output, since the isothermal work output A-->B is much greater than the isothermal work input C-->D. Thus the engine has produced a work output, and returned to its original state, ready to do the whole thing all over again. This is how the Carnot cycle works.

Carnot Cycle Idealized Heat Engine No Friction DS = Q/T = 0 Reversible Process Process A is expansion Process B is compression A B T U Q W 1->2 Isothermal expansion QH absorbed T=TH QH>0 QH 2->3 Adiabatic expansion No heat exchange, Temp drops to TC TC - TH -W3 3->4 Isothermal compression Heat QC discarded T=TC QC<0 -QC 4->1 Adiabatic compression No heat exchange, Temp rises to TH TH - TC W4 Note: work from (2->3) and (4->1) balance, yet more work by during (1->2) than on during (3->4).

The Carnot Cycle in Slow Motion
Stage 1 In the first stage the piston moves upward while the chamber absorbs heat from a source and the gas begins to expand. The portion of the graphic from point 1 to point 2 represents this behaviour. Because the temperature of the gas does not change, all the heat drawn in from the source goes into work performed by the expansion of the gas. There is only heat QH flowing into the system, there is no change in Potential Energy, therefore since DU = Q + W, the work by the system is –Q (the gas does positive work).

The Carnot Cycle in Slow Motion
Stage 2 In the second stage the heat source is removed; and the piston continues to move upward and the gas is still expanding while cooling (lowering in temperature from TH to TC). The portion of the graphic from point 2 to point 3 represents this behaviour. This stage is adiabatic expansion the gas (no heat transfer). Because the system expands it does negative work (the gas does positive work), its internal energy and temperature decrease because it receives no influx of heat from the surroundings

The Carnot Cycle in Slow Motion
Stage 3 In the third stage the piston begins to move downward and the cool gas (TC) is placed in thermal contact with heat reservoir at temperature TC, the gas is isothermally compressed at this temperature TC , thus the gas expels heat QC to the reservoir (the engine gives energy to the environment). Because the system contracts (decrease in volume and increase in pressure) the system does positive work on the gas (gas does negative work) and, rather than increasing its internal energy, the gas discards heat to the low temperature reservoir.

The Carnot Cycle in Slow Motion
Stage 4 In the final stage the piston moves downward and the cool gas is compressed to its original state. Its temperature also rises to its orignal state (point 4 to point 1). No exchange of heat with the surroundings take place. Because the system is compressed, its internal energy and temperature increase (since no heat is discarded), and the work done on the gas by the environment is W42 (The gas does negative work)

Carnot Cycle Review During step one the gas does a positive amount of work.  Step two is adiabatic, with the gas doing a positive amount of work.   During step three there is a negative amount of work done by the gas. Finally in step four, which is adiabatic, the work done by the gas is negative.  Notice that the total work done (the remaining area) is positive because positive work is done at high temperatures and negative work is done at lower temperatures.  During step one heat is absorbed (Q > 0) and during step three heat is released (Q < 0).  More heat is absorbed than is released for the entire cycle.  This is the basis of how engines work: Heat (from the hot reservoir) is transferred into mechanical work (piston moving).

Engines and the 2nd Law TH TC QH QC W HEAT ENGINE The objective: turn heat from hot reservoir into work The cost: “waste heat” 1st Law: QH -QC = W efficiency e  W/QH =W/QH = 1-QC/QH S = QC/TC - QH/TH  0 S = 0 for Carnot Therefore, QC/QH  TC/ TH QC/QH = TC/ TH for Carnot Therefore e = 1 - QC/QH  1 - TC/ TH e = 1 - TC/ TH for Carnot e = 1 is forbidden! e largest if TC << TH 36

Example Consider a hypothetical refrigerator that takes 1000 J of heat from a cold reservoir at 100K and ejects 1200 J of heat to a hot reservoir at 300K. 1. How much work does the refrigerator do? 2. What happens to the entropy of the universe? 3. Does this violate the 2nd law of thermodynamics? TH TC QH QC W QC = 1000 J QH = 1200 J Since QC + W = QH, W = 200 J DSH = QH/TH = (1200 J) / (300 K) = 4 J/K DSC = -QC/TC = (-1000 J) / (100 K) = -10 J/K DSTOTAL = DSH + DSC = -6 J/K  decreases (violates 2nd law)

Review Which of the following is forbidden by the second law of thermodynamics? 1. Heat flows into a gas and the temperature falls 2. The temperature of a gas rises without any heat flowing into it 3. Heat flows spontaneously from a cold to a hot reservoir 4. All of the above Not 1, because volume could increase and lower temperature Not 2, because increase of pressure can increase temperature 43

Summary First Law of Thermodynamics Ideal Gas Law
Average KE of Gas Molecule Internal energy of ideal Gas Expansion Work Heat Engine Efficiency Carnot Efficiency

Summary II Isobaric Work: Isochoric Work: Isothermal Work:
Adiabatic Work: Isothermal  Internal Energy The internal energy of an ideal gas depends only on temperature

Summary III Cyclic process (originates and ends at same state) U=0
Wby=Qinto Isovolumetric (constant volume) Qinto=U Isobaric (constant pressure) Isothermal (constant temperature) Adiabatic (no heat exchange) Q=0 U=Won U=-Wby

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