Work in Thermodynamic Processes We are now ready to combine the energy transfer mechanism of heat with another mechanism: work We will focus our discussion on work done on a gas Important in discussion of how system gets from one state to another (described by state variables P, T, V, n, U) Will provide connection between microscopic and macroscopic energy transfer mechanisms In Chap. 5 we spoke about work done by forces on objects undergoing a displacement Now we will talk about the work done on a gas by the environment W (or by gas on environment Wenv) Work done on gas (W) can be positive, negative, or zero W = 0 when no “mechanical action” taking place W = –Wenv Work influences internal energy

Work in Thermodynamic Processes Work done by a gas on a piston in a cylinder: Wenv = FDy = PADy = PDV = P(Vf – Vi) When the piston moves inward, Vf < Vi and Wenv < 0 When the piston moves outward, Vf > Vi and Wenv > 0 When piston doesn’t move, Wenv = 0 Work done by the piston on the gas = W = –Wenv When the piston moves inward, Vf < Vi and W > 0 When the piston moves outward, Vf > Vi and W < 0 When piston doesn’t move, W = 0 (from University Physics, 11th Ed.)

Work in Thermodynamic Processes This expression can only be used if the pressure remains constant during the expansion or compression This is called an isobaric (constant pressure) process (same Greek root as “barometer”) If the pressure changes, the average pressure may be used to estimate the work done If pressure and volume are known at each step of the process, a PV diagram helps visualize process with curves (paths) connecting initial and final states The area under the curve on a PV diagram is equal in magnitude to the work done on a gas (W) True whether or not P remains constant W is positive (negative) when volume decreases (increases)

PV Diagrams For a gas being compressed at constant pressure: For a gas being compressed with varying pressure: The work done depends on the particular path (shaded area represents work done on the gas; here W > 0) (shaded area, the area under the curve, represents work done on the gas) (same initial and final states, but work done is different in each case)

Other Thermodynamic Processes Isovolumetric (or isochoric) Volume stays constant For example, maintain constant piston position Vertical line on the PV diagram No work done on gas Isothermal Temperature stays constant Heat flows between the system and a reservoir to keep the system’s temperature constant For example, keep oven temperature constant Adiabatic No energy exchanged with the surroundings via heat For example, provide excellent insulation for system Or, process occurs so quickly that there is no time for heat to flow in or out of system PV Processes Interactive

Example Problem #12.5 A gas expands from I to F along the three paths indicated in the figure. Calculate the work done on the gas along paths IAF IF, and IBF. Solution (details given in class): – 810 J – 507 J – 203 J

First Law of Thermodynamics Both work and heat can change the internal energy of a system Work can be done on a rubber ball by squeezing it, stretching it, or throwing it onto a wall Energy can flow to the ball via heat by leaving it out in the sun or putting it into a hot oven These 2 methods of increasing the internal energy of a system lead to the first law of thermodynamics: The change in internal energy of a system is equal to the heat flow into the system plus the work done on the system DU = internal energy change of system Q = energy transferred to system by heat from the outside W = work done on the system (really a statement of energy conservation!)

Sign Conventions for First Law Quantity Definition Meaning of + sign Meaning of – sign Q Energy transferred by heat flow Heat flow into the system Heat flow out of the system W Work Work done on the system Work done by the system DU Internal energy change Internal energy increase Internal energy decrease In–class example – Fill in the boxes with +, –, or 0: Situation System Q W DU (a) Rapidly pumping up bicycle tire Air in the pump (b) Pan of room-temperature water resting on a hot stove Water in the pan (c) Air leaking quickly out of a balloon Air rushing out of balloon

Applications of the First Law Isolated system: does not interact with its surroundings No energy transfer takes place and no work is done Internal energy of the isolated system remains constant Q = W = DU = 0 Cyclic process: originates and ends at the same state DU = 0 and Q = –W The net work done per cycle by the gas is equal to the area enclosed by the path representing the process on a PV diagram (important for describing heat engines)  PV diagram for an ideal monatomic gas confined in a cylinder by a movable piston undergoing a cyclic process

Applications of the First Law Isothermal process: constant- temperature process Consider ideal monatomic gas contained in cylinder with movable piston The cylinder and gas are in thermal contact with a large source of energy (reservoir) Allow energy to transfer into the gas by heat The gas expands (volume increases) and pressure falls while maintaining a constant temperature Since , DU = 0 Therefore DU = 0 = Q + W So W = – Q < 0 (system supplies work to outside world)

Applications of the First Law Adiabatic process: no energy transferred by heat The work done is equal to the change in the internal energy of the system (W = DU) One way to accomplish a process with no heat exchange is to have it happen very quickly (mostly true in an internal combustion engine) In an adiabatic expansion, the work done is negative and the internal energy decreases For an ideal monatomic gas (see Chap. 10): Isovolumetric process: no change in volume, so W = 0 The energy added to the system goes into increasing the internal energy of the system (DU = Q) Temperature will increase Isobaric process: no change in pressure W = – PDV, so DU = Q – PDV

The First Law and Human Metabolism The first law can be applied to living organisms The internal energy stored in humans goes into other forms needed by the organs and into work and heat When we eat, we replenish our supply of internal energy The metabolic rate (DU / Dt) is directly proportional to the rate of oxygen consumption by volume Basal metabolic rate (to maintain and run organs, etc.) is about 80 W (other rates shown in Table 12.4) The efficiency of the body is the ratio of the mechanical power output to the metabolic rate What you get out divided by what you put in Efficiencies range from about 20% (cycling) to 3% (shoveling)

Example Problem #12.16 A gas is taken through the cyclic process described by the figure. Find the net energy transferred to the system by heat during one complete cycle. If the cycle is reversed  that is, the process follows the path ACBA  what is the net energy transferred by heat per cycle? Solution (details given in class): 12 kJ –12 kJ

Interactive Example Problem: Triangular Cyclic Process on a PV Diagram Simulation and solution details given in class. (ActivPhysics Online Exercise #8.13, copyright Addison Wesley publishing)

Heat Engines A heat engine is a device that converts internal energy to other useful forms, such as electrical or mechanical energy Electrical power plants Internal combustion engine in cars A heat engine carries some “working substance” through a cyclical process Energy is transferred from a source at a high temperature (Qh) Work is done by the engine (Weng) Energy is expelled to a source at a lower temperature (Qc) Example of piston steam engine (in-class movie): Animation

Heat Engine Since it is a cyclical process, its initial and final internal energies are the same So, DU = 0 = Q + W and Q = Qnet = – W = Weng The work done by the engine equals the net energy absorbed by the engine Therefore, The work is equal to the area enclosed by the curve of the PV diagram (if working substance is a gas) Thermal efficiency is defined as the ratio of the work done by the engine to the energy absorbed at the higher temperature (what you get divided by what you put in):

Second Law of Thermodynamics The efficiency of heat engines is addressed in the second law of thermodynamics: No engine can have 100% efficiency Other statements and implications of the second law: A system cannot convert heat solely into mechanical work Heat never flows spontaneously from a colder body to a hotter body It is impossible for any process to have as its sole result the transfer of heat from a cooler to a hotter body There is no such thing as a free lunch! Summary of 1st and 2nd laws: 1st law: We cannot get more energy out than we put in 2nd law: We cannot break even

Internal Combustion Engines An idealized model of the thermodynamic processes of a gas engine is called the Otto cycle Intake stroke: Gas-air mixture drawn into cylinder (a  b) Compression stroke: Intake valve closes, mixture compressed (adiabatic compression; b  c) Ignition: Spark plug ignites mixture (heating at constant volume; c  d) Power stroke: Hot burned mixture pushes mixture down, doing work (adiabatic expansion; d  e) Cooling: Energy released via heat at constant volume (e  b) Exhaust stroke: Exhaust valve opens and burned mixture is pushed out of cylinder (b  a) (from College Physics, Giambattista) (from University Physics, 11th Ed.)

Heat Pumps: Refrigerators, Air conditioners Energy is extracted from the cold reservoir (food cabin) and transferred to hot reservoir (kitchen) Air conditioner works on the same principle Both are examples of a heat pump (reverse heat engines) (from University Physics, 11th Ed.) (from University Physics, 11th Ed.) Heat Pumps Interactive

Carnot Engine The most efficient heat engine possible is the Carnot engine A theoretical heat engine operating in an ideal, reversible cycle (a cycle in which system can be returned to its initial state along the same path) called the Carnot cycle Note that a truly reversible process is an idealization, and real processes are irreversible (although some are close to being reversible) PV diagram of the cycle: Consists of 2 adiabatic and 2 isothermal processes, all reversible Net work done W is net energy received in one cycle: W = Qh – Qc

Carnot Cycle Assume the substance that changes temperature is an ideal gas Gas is contained in cylinder with movable piston at one end Steps of the cycle: A  B: Gas expands isothermally while in contact with reservoir at Th B  C: Gas expands adiabatically (Q = 0) C  D: Gas compressed isothermally while in contact with reservoir at Tc < Th D  A: Gas compressed adiabatically Note that upward (downward) red arrows on piston indicate removal (addition) of sand

Carnot Engine Carnot showed that the efficiency of the engine depends on the temperatures of the reservoirs: Temperatures must be in Kelvin All Carnot engines operating between the same two temperatures will have the same efficiency The efficiency increases as Tc is lowered and as Th is raised Efficiency is 0 if Th = Tc Efficiency is 100% only if Tc = 0 K (not possible from third law of thermodynamics) In most practical cases Tc is near room temperature (300 K) so generally Th is raised to increase efficiency All real engines are less efficient than the Carnot engine (friction, cycle completed in brief time period)

Example Problem #12.27 One of the most efficient engines ever built is a coal-fired steam turbine engine in the Ohio River valley, driving an electric generator as it operates between 1870°C and 430°C. What is its maximum theoretical efficiency? Its actual efficiency is 42.0%. How much mechanical power does the engine deliver if it absorbs 1.40  105 J of energy each second from the hot reservoir? Solution (details given in class): 0.672 (or 67.2%) 58.8 kW