1 Thermal Physics 13 - Temperature & Kinetic Energy 15 - Laws of Thermodynamics

2 Assignments Read 13.2,4,6-10 Read 14 sections 1-4 for review Read 15 sections1-9 Ch 13 Questions 3,5,7 Problems: 33,35,37,43 Ch 15 Questions 1-3 Problems 7,17,19,21,23,24,32

3 Introduction to Thermodynamics Temperature and Heat Gas Law Review

4 Thermodynamics This is the study of heat and thermal energy. Thermal properties (heat and temperature) are based on the motion of molecules. This is one part of our studies that relates to Chemistry.

5 Total Energy E = U + K + E int U: potential energy K: kinetic energy E int : internal or thermal energy Potential and kinetic energies are specifically for large objects and represent mechanical energy Thermal energy relates to the kinetic energy of the molecules of a substance.

6 Temperature and Heat Temperature is a measure of the average kinetic energy of the molecules of a substance - a measure of how fast the molecules are moving. Unit: o C or K Temperature is NOT heat! Heat is the internal energy that is transferred between bodies in contact. Unit: joules (J) or calories (cal) A difference in temperatures will cause heat energy to be exchanged between bodies in contact When two bodies are at the same temperature, no heat is transferred; this is called Thermal Equilibrium.

7 Ideal Gas Law (combined Gas Law) P 1 V 1 /T 1 = P 2 V 2 /T 2 P 1,P 2 : initial and final pressure (any unit) V 1,V 2 : initial and final volume (any unit) T 1,T 2 : initial and final temperature (in Kelvin) Temperature in K = o C

8 Sample Problem An ideal gas occupies 4.0 L at 23 o C and 2.3 atm. What will be the volume of the gas if the temperature is lowered to 0 o C and the pressure is 3.1 atm?

9 Ideal Gas Equation PV = nRT (using moles) PV = N k B T (using molecules) P: pressure (Pa) V: volume (m 3 ) n: number of moles N: number of molecules R: Universal Gas constant (8.31 J/mol K) k B : Boltzman’s constant (1.38 x J/K) T: temperature (in K)

10 Sample Problem Determine the number of moles of an ideal gas that occupies 10.0 m 3 at atmospheric pressure and 25 o C. PV = nRT

11 Sample Problem Suppose a near vacuum contains 25,000 molecules of helium in one cubic meter at 0 o C. What is the pressure? PV = nRT n = N molecules N A Avogadro’s # PV = Nk B T

12 The relationship between the Universal Gas Law Constant, R (8.31 J/mol K) and Boltzman’s constant k B (1.38 x J/K) is… Na k B = R 6.02 x (1.38 x J/K) = 8.32 J/molK mole

13 Kinetic Theory of Gases 1. Gases consist of a large number of molecules that make elastic collisions with each other and the walls of the container. 2. Molecules are separated, on average, by large distances and exert no forces on each other except when they collide. 3. There is no preferred position for a molecule in the container, and no preferred direction for the velocity.

14 Average Kinetic Energy of a Gas K ave = 3/2 k B T K ave : average kinetic energy (J) k B : Boltzmann’s constant (1.38 x J/K) T: Temperature (K) The molecules have a range of kinetic energies; is just the average of that range.

15 Sample Problem What is the average kinetic energy and the average speed of oxygen molecules in a gas sample at O o C?

16 Sample Problem Suppose nitrogen and oxygen are in a sample of gas at 100 o C. A. What is the ratio of the average kinetic energies for the two molecules? B. What is the ratio of their average speeds?

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18 Notation Warning! U is potential energy in mechanics. However, U is E int (thermal energy) in thermodynamics This means that when we are in thermo, U is thermal energy which is related to temperature. When in mechanics, it is potential energy (related to configuration or position).

19 More about U U is the sum of the kinetic energies of all molecules in a system (or gas). U = N K ave U = N (3/2 k B T) U = n (3/2 R T) –Since k B = R/N A Heat added is + Heat lost is - Work on system is - Work by system is +

20 First Law of Thermodynamics   U = Q - W   U : change in internal energy of system (J)  Q: heat added to the system (J). This heat exchange is driven by temperature difference.  W: work done on the system (J). Work will be related to the change in the system’s volume.  This law is sometimes paraphrased as “you can’t win”

21 Problem #8  U = Q - W  U = 200 J J  U = 100 J Heat added is + Heat lost is - Work on system is - Work by system is +

22 Problem #9  U = Q - W  J  J = 0 - W W = - 5,000 J

23 Gas Processes The thermodynamic state of a gas is defined by pressure, volume, and temperature. A “gas process” describes how gas gets from one state to another state. Processes depend on the behavior of the boundary and the environment more than they depend on the behavior of the gas.

24  U = 0. Q = W

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26 Isometric/isochoric V is constant W = 0 Isobaric P is constant W = P  V

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28 Work Calculation of work done on a system (or by a system) is an important part of thermodynamic calculations. Work depends upon volume change Work also depends upon the pressure at which the volume change occurs.

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31 Problem #10 W = P  V = 1 atm (.8-.02) m 3 W = 1.013x10 5 N/m 2 (.78m 3 ) W = + work done by a gas W = - work done on a gas

32 Problem #11 U = Q - W 230 = W W = -350J W = P  V   V =W/P = -350 J / x 10 5 J/m 3  V =

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34 Work - isobaric

35 Work is path dependent

36 Problem #12

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38 Work done by a Gas When a gas undergoes a complete cycle, it starts and ends in the same state. The gas is identical before and after the cycle. There is no identifiable change in the gas.  U = 0 for a complete cycle. However, the environment, has been changed.

39 Work done by the Gas

40 #13 Problem Consider the cycle ABCDA, where –State A: 200 kPa, 1.0 m 3 –State B: 200 kPa, 1.5 m 3 –State C: 100 kPa, 1.5 m 3 –State D: 100 kPa, 1.0 m 3  A. Sketch the cycle.  B. Graphically estimate the work done by the gas in one cycle.  C. Estimate the work done by the environment in one cycle.

41 A: 200 kPa, 1.0 m 3 B: 200 kPa, 1.5 m 3 C: 100 kPa, 1.5 m 3 D: 100 kPa, 1.0 m 3

42 10.(II) Consider the following two-step process. Heat is allowed to flow out of an ideal gas at constant volume so that its pressure drops from 2.2 atm to 1.4 atm. Then the gas expands at constant pressure, from a volume of 6.8 L to 9.3 L, where the temperature reaches its original value. See Fig. 15–22. Calculate (a) the total work done by the gas in the process, (b) the change in internal energy of the gas in the process, and (c) the total heat flow into or out of the gas.

43 Problem 15/10 Fig

44 (a)No work is done during the first step, since the volume is constant. The work in the second step is given by (b) Since there is no overall change in temperature, (c) The heat flow can be found from the first law of thermodynamics.

45 2 nd Law of Thermodynamics No process is possible whose sole result is the complete conversion of heat from a hot reservoir into mechanical work. (Kelvin-Planck statement.) No process is possible whose sole result is the transfer of heat from a colder to a hotter body. (Clausius statement.)

46 Heat Engines The Carnot Cycle Efficiency

47 Heat Engines Heat engines can convert heat into useful work. According to the 2 nd Law of T-D, heat engines always produce some waste heat. Efficiency can be used to tell how much heat is needed to produce a given amount of work. NOTE: A heat engine is not something that produces heat. A heat engine transfers heat from hot to cold, and does mechanical work in the process.

48 Q H = W + Q L Operating Temps: Q H and Q L New Sign Convention: Q H, Q L and W are all positive

49 Efficiency of Heat Engine In general, efficiency is related by what fraction of the energy put into a system is convereted to useful work. In the case of a heat engine, the energy that is put in is the heat that flows into the system from the hot reservoir. Only some of the heat flowing in is converted to work. The rest if waste heat that is dumped into the cold reservoir.

50 Efficiency of Heat Engine Efficiency = W/Q H = (Q H - Q C )/Q H * W: Work done by engine on enviroment Q H : Heat absorbed from hot reservoir Q C : Waste heat dumped to cold reservoir Efficiency is often given as percent efficiency.

51 Adiabatic vs Isothermal Expansion In adiabatic expansion, no heat energy can enter the gas to replace energy being lost as it does work on the environment. The temperature drops, and so does the pressure.

52 Carnot Cycle Go to page 419 in your book.

53 Efficiency of Carnot Cycle For a Carnot engine, the efficiency can be calculated from the temperatures of the hot and cold reservoirs. Carnot Efficiency: e = 1 - T C T H T C :Temperature of cold reservoir (K) T H :Temperature of hot reservoir (K)

54 Sample Problem 23.(II) A Carnot engine performs work at the rate of 440 kW while using 680 kcal of heat per second. If the temperature of the heat source is 570°C, at what temperature is the waste heat exhausted? T L

55 Sample Problem 21.(II) A nuclear power plant operates at 75% of its maximum theoretical (Carnot) efficiency between temperatures of 625°C and 350°C. If the plant produces electric energy at the rate of 1.3 GW, how much exhaust heat is discharged per hour?

56 Entropy Disorder or randomness The entropy of the universe is increasing. This will lead to what philosophers have often discussed as “Heat Death of the Universe. Does the entropy of your room tend to increase or decrease?

57 Entropy   S = Q/T  S: Change in entropy (J/K) Q: Heat going into system (J) T: Kelvin temperature (K) If change in entropy is positive, randomness or disorder has increased. Spontaneous changes involve an increase in entropy. Generally, entropy can go down only when energy is put into the system.

58 Problem 15/42 42.(II) 1.0 kg of water at 30°C is mixed with 1.0 kg of water at 60°C in a well-insulated container. Estimate the net change in entropy of the system.