The First Law of Thermodynamics Chapter 19 The First Law of Thermodynamics © 2016 Pearson Education Inc.

Learning Goals for Chapter 19 Calculate work done by a system when its volume changes. Interpret & use 1st law of thermodynamics. (4) important kinds of thermodynamic processes. Adiabatic Isothermal Isochoric Isobaric Why internal energy of ideal gas depends on temperature only. Difference between Cv & Cp molar heat capacities © 2016 Pearson Education Inc.

Key Ideas to Concentrate Upon! PV diagrams as a way to see cyclic processes Pressure (atm, kPA, barr, lb/in2) Volume (liters, m3, cm3)

Key Ideas to Concentrate Upon! PV diagrams as a way to see cyclic processes Pressure (atm, kPA, barr, lb/in2) Volume (liters, m3, cm3)

Key Ideas to Concentrate Upon! PV diagrams as a way to see cyclic processes Pressure (atm, kPA, barr, lb/in2) Isothermal isochoric Adiabatic Isobaric Volume (liters, m3, cm3)

Introduction Steam locomotive operates using laws of thermodynamics So car engines… And air conditioners… And refrigerators! Revisit conservation of energy in form of 1st law of thermodynamics. © 2016 Pearson Education Inc.

Thermodynamics systems A thermodynamic system = collection of objects that may exchange energy with its surroundings. Popcorn in a pot is a thermodynamic system. In thermodynamic process shown, heat is added to system….and System then does work on its surroundings to lift pot lid. © 2016 Pearson Education Inc.

Thermodynamics systems In Thermodynamic process, changes occur in state of system. Careful of signs! Q = positive when heat flows into a system. W = work done by system W = + for expansion W = - for work done ON a system . © 2016 Pearson Education Inc.

Work done during volume changes Consider a molecule in gas colliding with a moveable piston Volume of gas increases … Positive work done BY gas on piston. © 2016 Pearson Education Inc.

Work done during volume changes If piston moves left, so volume of gas decreasing, work is done ON the gas during collision. Gas molecules do negative work on piston. © 2016 Pearson Education Inc.

Work done during volume changes Infinitesimal work done by system during small expansion dx = dW = pA dx. For finite change of volume from V1 to V2: © 2016 Pearson Education Inc.

Work on a pV-diagram Work done = area under pV curve If gas DOES + work, volume expands! System undergoing an expansion with varying pressure. © 2016 Pearson Education Inc.

Work on a pV-diagram Work done = area under pV curve If gas DOES + work, volume expands! System undergoing an expansion with constant pressure. W = p(V2 – V1) © 2016 Pearson Education Inc.

Work on a pV-diagram Work done = area under pV curve If volume decreases, work is done ON the gas System undergoing an compression with varying pressure. © 2016 Pearson Education Inc.

Work depends on path chosen! Consider (3) different paths on pV-diagram for getting from state 1 to state 2. © 2016 Pearson Education Inc.

Work depends on path chosen! Remember PV = nRT; from specific point of P & V to another point, gas ends up @ same T no matter HOW © 2016 Pearson Education Inc.

Work depends on path chosen! Gas does LARGE amount of work under path © 2016 Pearson Education Inc.

Work depends on path chosen! Gas does LARGE amount of work under path Why? HIGH Pressure expansion (1 => 3) How? ADD heat! Going from 3 => 2 no DV No work done! © 2016 Pearson Education Inc.

Work depends on path chosen! Gas does small amount of work under path © 2016 Pearson Education Inc.

Work depends on path chosen! Gas does small amount of work under path Why? LOW Pressure expansion (4 => 2) How? ADD less heat! Going from 1 => 4 no DV No work done! © 2016 Pearson Education Inc.

Work depends on path chosen! Along smooth curve from 1 to 2, work done is different from that for either of other paths. © 2016 Pearson Education Inc.

First law of thermodynamics Change in internal energy U of system = heat added minus work done by system: First law of thermodynamics = general reflection of conservation of energy. © 2016 Pearson Education Inc.

The First Law of Thermodynamics DEinternal = +Q in – W out Types of +Qin: Chemical Energy (Gasoline, Diesel) through combustion Conduction of Heat from Hot reservoir (steam from a boiler or nuclear power plant) Absorption of heat from solar energy

The First Law of Thermodynamics DEinternal = +Q in – W out Types of –W out: Expansion of Pistons (Internal combustion & steam engines) Turning of a rotor (solar powered fan)

First law of thermodynamics Both Q & W depend on path chosen between states… But net change of internal E depends on change in T only… is independent of path! If changes are infinitesimal, first law = dU = dQ – dW. © 2016 Pearson Education Inc.

First law of thermodynamics Suppose more heat is added to system + Q in than system does work +W by gas by expanding. Internal energy of system increases (gas heats up!) © 2016 Pearson Education Inc.

First law of thermodynamics Suppose heat added TO system + Q in equals work done BY system +W by gas Internal energy of system is unchanged (no change in T) © 2016 Pearson Education Inc.

First law of thermodynamics Suppose more heat flows out of system - Q in than work is done ON the gas -W on gas Internal energy of system decreases (gas cools!) © 2016 Pearson Education Inc.

Example Using the first law. 2500 J of heat is added to a gas under a piston in a closed cylinder, and 1800 J of work is done by the system as its piston expands. What is the change in internal energy of the system? Solution: Both the heat added and the work done ON the system add to the system’s internal energy, which increases by 4300 J.

Example 19-7: Using the first law. 2500 J of heat is added to a system, and 1800 J of work is done by the system. What is the change in internal energy of the system? DE = +2500 J – (1800J) = +700 J Solution: Both the heat added and the work done ON the system add to the system’s internal energy, which increases by 4300 J. DEinternal = +Q in – W out

Example 19-7: Using the first law. 2500 J of heat is added to a gas under a piston in a closed cylinder, and 1800 J of work is done on the system as the piston is pushed back down. What is the change in internal energy of the system? Solution: Both the heat added and the work done ON the system add to the system’s internal energy, which increases by 4300 J.

Example 19-7: Using the first law. 2500 J of heat is added to a system, and 1800 J of work is done on the system. What is the change in internal energy of the system? DE = +2500 J – (-1800J) = +4300 J Solution: Both the heat added and the work done ON the system add to the system’s internal energy, which increases by 4300 J. DEinternal = +Q in – W out

First law of exercise thermodynamics Your body is a thermodynamic system. Do a push-up! Your body does work! Wby system > 0. Your body also warms up during exercise; by perspiration & radiation & conduction, body gives off this heat, so Qin < 0. © 2016 Pearson Education Inc.

First law of exercise thermodynamics Since Qin negative & Wby system positive, U = Q − W < 0 Your body’s internal energy decreases. Exercise helps you lose weight, using up some of internal energy stored in your body in form of fat. © 2016 Pearson Education Inc.

First law of thermodynamics Both Q & W depend on path chosen between states… But net change of internal E depends on change in T only… is independent of path! © 2016 Pearson Education Inc.

Internal energy Internal energy of coffee depends on just its thermodynamic state — how much water & ground coffee it contains, & its temperature. Internal energy does NOT depend on history of how coffee was prepared— (thermodynamic path leading to current state) © 2016 Pearson Education Inc.

Cyclic thermodynamic process © 2016 Pearson Education Inc.

Cyclic thermodynamic process: Qin + © 2016 Pearson Education Inc.

Cyclic thermodynamic process © 2016 Pearson Education Inc.

Four kinds of thermodynamic processes Isochoric: Volume remains constant Isobaric: Pressure remains constant Isothermal: Temperature remains constant Adiabatic: No heat is transferred in or out of system © 2016 Pearson Education Inc.

The four processes on a pV-diagram Paths on pV-diagram for all (4) different processes for a constant amount of an ideal gas, all starting at state a. Isobaric Isothermal Isochoric Adiabatic © 2016 Pearson Education Inc.

Calculating Work Done BY Gas The work done in moving a piston by an infinitesimal displacement is: Easy: if V doesn’t change or P constant over volume change! Harder: If P & V BOTH change Figure 19-10. The work done by a gas when its volume increases by dV = A dl is dW = P dV.

Four kinds of thermodynamic processes Isochoric: Volume remains constant, No volume change – no work done! W = 0 By First law DU = Qin System heats up if heat added System cools down if heat lost © 2016 Pearson Education Inc.

Four kinds of thermodynamic processes Isobaric: Pressure remains constant W = p(V2 – V1) DU = Q in - pDV © 2016 Pearson Education Inc.

Isobaric process Most cooking involves isobaric processes. Air pressure above saucepan, or inside microwave oven, remains essentially constant while food is being heated. © 2016 Pearson Education Inc.

KEY idea: Internal energy of an ideal gas Internal energy of ideal gas depends only on temperature, not on pressure or volume. Temperature does NOT change during a “free” expansion. © 2016 Pearson Education Inc.

Heat capacities of an ideal gas CV = molar heat capacity at Constant Volume. Measure CV: Add known Qin = nCvdT Measure increased temperature of gas in rigid container Constant volume! Ignore any thermal expansion of container © 2016 Pearson Education Inc.

Heat capacities of an ideal gas CV = molar heat capacity at Constant Volume. First Law: DU = Qin - Wout Wout = 0 DU = Qin = nCvdT © 2016 Pearson Education Inc.

Heat capacities of an ideal gas Cp = molar heat capacity at Constant Pressure. Measure Cp Add known Qin = nCPdT to gas Moveable piston Fixed weight = constant pressure! Gas expands just enough to keep pressure constant as temperature rises. © 2016 Pearson Education Inc.

Heat capacities of an ideal gas Cp = molar heat capacity at Constant Pressure. First Law: DU = Qin - Wout Wout = pDV DU = nCvdT – pDV And from ideal gas law pDV = nRDT © 2016 Pearson Education Inc.

Relating Cp and CV for an ideal gas To produce same temperature change, more heat is required at Constant Pressure than at Constant Volume is same! n CvDT = n CpDT – pDV So… Cp > CV pDV = nRDT So… Cp = Cv + R © 2016 Pearson Education Inc.

The ratio of heat capacities Ratio of heat capacities g is: For monatomic ideal gases, For diatomic ideal gases, © 2016 Pearson Education Inc.

Four kinds of thermodynamic processes Isothermal: Temperature remains constant; P & V both vary! But Internal Energy reflects Temperature ONLY U can’t change! DU = 0 Qin = Wout © 2016 Pearson Education Inc.

Calculating Work – Isothermal Processes An isothermal process is one in which the temperature does not change. Figure 19-6. PV diagram for an ideal gas undergoing isothermal processes at two different temperatures. Figure 19-7. An ideal gas in a cylinder fitted with a movable piston.

Calculating Work – Isothermal Processes An isothermal process is one in which the temperature does not change. For isothermal process to take place, assume system is in contact with a heat “reservoir”. Assume that system remains in equilibrium throughout all steps. Figure 19-6. PV diagram for an ideal gas undergoing isothermal processes at two different temperatures. Figure 19-7. An ideal gas in a cylinder fitted with a movable piston.

Isothermal processes are S L O W To keep T constant, SLOWLY add a bit of heat, let gas expand a bit, repeat.. Use a large reservoir of surrounding material to keep T constant. © 2016 Pearson Education Inc.

Calculating Work: Isothermal Process For isothermal process, P = nRT/V. Integrate to find work done in taking gas from A to B: Figure 19-11. Work done by an ideal gas in an isothermal process equals the area under the PV curve. Shaded area equals the work done by the gas when it expands from VA to VB.

Four kinds of thermodynamic processes Adiabatic: No heat is transferred in or out of system so Qin = 0 by 1st law: DU = U2 – U1 = 0 – W by system Internal Energy drops if system DOES work; T decreases Internal Energy rises if work is done ON system; T increases © 2016 Pearson Education Inc.

Adiabatic process are FAST Pop cork on bottle of champagne! Pressurized gases inside bottle expand rapidly & do positive work on outside air. Little time for gases to exchange heat with surroundings, so expansion is nearly adiabatic. (ΔQ = 0) Internal energy of expanding gases decreases (ΔU = –W ) Temperature drops. Water vapor condenses & forms a miniature cloud. © 2016 Pearson Education Inc.

Exhaling adiabatically Put your hand a few centimeters in front of your mouth, open your mouth wide, & exhale. Breath feels warm because exhaled gases emerge ~ temperature of your body’s interior. Now compress your lips to whistle, & blow on your hand. Exhaled gases feel much cooler. Exhaled gases undergo rapid, essentially adiabatic expansion as they emerge from between your lips! Temperature of exhaled gases decreases. © 2016 Pearson Education Inc.

Adiabatic processes for an ideal gas No heat transferred in or out of gas, so Qin = 0. As gas expands, it does positive work W on its environment & internal energy decreases Temperature drops. Note that an adiabatic curve at any point is always steeper than an isotherm at that point! WHY??? © 2016 Pearson Education Inc.

Work by Adiabatic Expansion of a Gas Incremental Work is dU = Qin – PdV For an adiabatic expansion dU = – PdV (no heat transfer!) But dU = nCV dT ALWAYS so nCVdT = - PdV

Work by Adiabatic Expansion of a Gas nCVdT = - PdV so nCVdT = -nRT dV V or (dT/T) + R (dV/V) = 0 Cv recall… R = Cp – Cv So R/Cv = (Cp/CV) – 1 = g - 1

Work by Adiabatic Expansion of a Gas So dt/T + (g – 1) dV/V = 0 Integrate! ln T + (g – 1)ln V = constant Or TV (g – 1) = constant The Adiabatic Gas Law relationship between volume and temperature!

Work by Adiabatic Expansion of a Gas So TV (g – 1) = constant Use Ideal Gas Law: PV = nRT => T = PV/nR (P/nR)V V (g – 1) = constant Equivalent to PVg = constant

Work by Adiabatic Expansion of a Gas First Law dU = Qin – PdV Work : PdV = - nCv DT Where DT = Tfinal – Tinitial = T2- T1 So Work done by gas = nCv (T1 – T2) If gas cools down, positive work done adiabatically! If work done ON gas, T2 > T1, gas heats up

Work by Adiabatic Expansion of a Gas So Work done by gas adiabatically nCv (T1 – T2) But T1 = P1V1/nR and T2 = P2V2/nR So Work done by gas adiabatically = Cv/R (P1V1 – P2V2) = 1 (P1V1 – P2V2) (g – 1)

Work Done by Processes! Isochoric Isobaric Isothermal Adiabatic Heat In Q in nCvDT n CpDT nCvDT + nRT ln(V2/V1) Work Out W out pDV nRT ln(V2/V1) nCv (T1 – T2) or 1 (P1V1 – P2V2) (g – 1) Change in Energy DU = Qin –Wout nCv (T2 – T1) © 2016 Pearson Education Inc.