 # Thermodynamics April 27, 2015April 27, 2015April 27, 2015.

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Thermodynamics April 27, 2015April 27, 2015April 27, 2015

Thermodynamics Thermodynamics is the study off heat and thermal energy. Thermal properties (heat and temperature) are based on the motion off individual molecules, so thermodynamics is a lot like chemistry.

Heat vs. Temperature Temperature is a measure of the average kinetic energy of the molecules of a substance. – –Think of it as a measure of how fast the molecules are moving. The unit is C ° or K. Temperature is NOT heat! – –Heat is the internal energy that is transferred between bodies in contact. The unit is Joules (J). – –A difference in temperature 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.

Ideal 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 is obtained from temperature in C ° by adding 273

Problem Suppose an ideal gas occupies 4.0 litters at 23 C ° and 2.3 atm. What will be the volume off the gas if the temperature is lowered to 0 C ° and the pressure is increased to 3.1 atm.

Ideal Gas Law P V = n R T – –P: pressure (in Pa) – –V: volume (in m 3 ) – –n: number of moles – –R: gas law constant – –8.31 J/(mol K) – –T: temperature (in K) P V = N k B T –N: number of molecules –k B : Boltzman’s constant –1.38 x 10 -23 J/K

Problem Determine the number of moles of an ideal gas that occupy 10.0 m 3 at atmospheric pressure and 25 C °.

Kinetic Theory April 27, 2015April 27, 2015April 27, 2015

Problem Suppose an ideal gas occupies 4.0 liters at 23 C °. and 2.3 atm. What will be the volume off the gas if the temperature is lowered to 0 C ° and the pressure is increased to 3.1 atm?

Problem Determine the number of moles of an ideal gas that occupy 10.0 m 3 at atmospheric pressure and 25 C °.

Ideal Gas Law PV = n R T (using moles) P V = N k B T (using molecules) – –P: pressure (Pa) – –V: volume (m3) – –N: number of molecules – –k B : Boltzman’s constant 1.38 x 10 -23 J/K – –T: temperature (K)

Problem Suppose a near vacuum contains 25,000 molecules off helium in one cubic meter at 0 C °. What is the pressure?

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

Average Kinetic Energy K ave = 3/2 k B T – –K ave : average kinetic energy (J) – –k B : Boltzmann’s Constant (1.38 x 10 -23 J/K) – –T: Temperature (K) The molecules have a range off kinetic energies; K ave is just the average off that range.

Problem What is the average kinetic energy and the average speed off oxygen molecules in a gas sample at 0 C ° ?

First Law of Thermodynamics April 27, 2015April 27, 2015April 27, 2015

System Boundary The system boundary controls how the environment affects the system. If the boundary is “closed to mass”, that means that mass can’t get in or out.. If the boundary is “closed to energy”,, that means energy can’t get in or out..

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

Internal Energy U is the sum off the kinetic energies off all molecules in a system (or gas).. U = N K ave U = N (3/2 k B T) U = n (3/2 RT) Since k B = R/N A

First Law of Thermodynamics ΔU = Q + W – –ΔU:: change in internal energy off 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”..

Problem A system absorbs 200 J of heat energy from the environment and does 100 J of work on the environment. What is its change in internal energy?

Problem How much work does the environment do on a system if its internal energy changes from 40,000 J to 45,000 J without the addition of heat?

Gas Processes April 27, 2015April 27, 2015April 27, 2015

Gas Process The thermodynamic state off 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

Isothermal Process (constant temperature) T1T1 T2T2 T3T3 P V PV = nRT ΔT = 0 Isotherms

Isobaric Process (constant pressure) T1T1 T2T2 T3T3 P V PV = nRT ΔP = 0 Isobaric Compression Isobaric Expansion

Isometric Process (constant volume) T1T1 T2T2 T3T3 P V PV = nRT ΔV = 0

Adiabatic Process (insolated) P V PV = nRT ΔQ = 0 Pressure, Volume and Temperature all change in an adiabatic process

Work Calculation off 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.

Problem Calculate the work done by a gas that expands from 0.020 m 3 to 0.80 m 3 at constant atmospheric pressure. How much work is done by the environment when the gas expands this much?

Problem What is the change in volume of a cylinder operating at atmospheric pressure if its internal energy decreases by 230 J when 120 J of heat are removed from it?

Work (isobaric) P1P1 P2P2 V1V1 V2V2 AB CD W AB = P 2 ΔV W CD = P 1 ΔV W AB > W CD

Work (isobaric) P1P1 P2P2 V1V1 V2V2 AB CD W ABD > W ACD

Problem One mole of a gas goes from state A (200 kPa and 0.5 m 3 ) to state B (150 kPa and 1.5 m 3 ). What is the change in temperature of the gas during this process?

2 nd Law of Thermodynamics April 27, 2015April 27, 2015April 27, 2015

Work Done in a Cycle When a gas undergoes a complete cycle, it starts and ends in the same state. The gas is identical before and after the cycle, so there is no identifiable change in the gas. ΔU = 0 for a complete cycle. The environment, however, has been changed.

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.

Problem Calculate the heat necessary to change the temperature of one mole off an ideal gas from 300K to 500K A) at constant volume. B) at constant pressure (assume 1 atm).

Heat Engines April 27, 2015April 27, 2015April 27, 2015

Heat Engines Heat engines can convert heat into useful work. According to the 2nd Law of Thermodynamics 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.

Heat Transfer Hot Reservoir Q H Cold Reservoir Q C HEAT Heat flows until Q H = Q C

Heat Engine Hot Reservoir Q H Cold Reservoir Q C Q H = W + Q C Engine HEAT WORK

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.

Heat Engine Hot Reservoir Q H Impossible Engine There is no reason for heat to flow to the engine. 100% efficiency is impossible. “There is no free lunch” Engine HEAT Work

Physics Challenge An ideal gas is initially in a state that corresponds to point 1 on the graph above, where it has pressure P 1, volume V 1, and temperature T 1. The gas undergoes an isothermal process represented by the curve shown, which takes it to a final state 3 at temperature T 3. If T 2 and T 4 are the temperatures the gas would have at points 2 and 4, respectively, which of the following relationships is true? (A)T 1 < T 3 (B)T 1 < T 2 (C)T 1 < T 4 (D)T 1 = T 2 (E)T 1 = T 4

Adiabatic vs. Isothermal Initial State Isothermal Expansion Adiabatic Expansion In an 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. P V

Carnot Cycle P V Isothermal Expansion Heat In Heat Out Adiabatic Expansion Adiabatic Compression Isothermal Compression WORK

Work & Heat Engines Q H = W + Q C – –Q H : Heat that is put into the system and comes from the hot reservoir in the environment. – –W: Work that is done by the system on the environment. – –Q C : Waste heat that is dumped into the cold reservoir in the environment.

Problem A piston absorbs 3600 J of heat and dumps 1500 J of heat during a complete cycle. How much work does it do during the cycle?

Efficiency In general, efficiency is related what fraction of the energy put into a system is converted 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 that flows in is converted to work. The rest is waste heat that is dumped into the cold reservoir.

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

Problem A certain coal-fired steam plant is operating with 33% thermodynamic efficiency. If this is a 120 MW plant, at what rate is heat energy used?

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

Problem Calculate the Carnot efficiency of a heat engine operating between the temperatures of 60 and 1500 C °. How much work is produced when 15 kJ of waste heat is generated?

Entropy April 27, 2015April 27, 2015April 27, 2015

Entropy Entropy is disorder, or randomness. The entropy of the universe is increasing.

Entropy ΔS = Q/T – –ΔS: 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.