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. Four important kinds of thermodynamic processes: Isothermal Isochoric Isobaric Adiabatic Why internal energy of ideal gas depends on T only. Difference between molar heat capacities at constant volume Cv & at constant pressure Cp.

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 is any collection of objects that may exchange energy with its surroundings. Popcorn in a pot can be a ‘thermodynamic system’. In the thermodynamic process shown here, heat is added to the system, Qin is positive! The system does work on its surroundings to lift pot lid. Wdone is positive! © 2016 Pearson Education Inc.

Thermodynamics systems In a thermodynamic process, changes occur in the state of the system. Be careful with signs! Q is positive when heat flows into a system. W is work done BY system, positive for expansion. Both can appear NEGATIVE © 2016 Pearson Education Inc.

Thermodynamics systems Q + heat flows into a system. +500J of heat added Q - heat flows out from a system. Cool gas with ice – 500 J W + done BY system as it expands. Piston expands +10 cm Gas does + 500 J of work W – done ON system as it is compressed. 500 J work done on gas

Work done during volume changes Consider a single molecule in a gas. Collides with a surface moving to the right, which moves, so volume occupied by gas increases, Molecule does positive work on the piston. © 2016 Pearson Education Inc.

Work done during volume changes If piston moves left, so volume occupied by gas decreases; positive work is done ON the gas molecule 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 is dW = pA dx In finite change of volume from V1 to V2: © 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.

System undergoing an expansion with constant pressure. Work on a pV-diagram Work done equals area under curve on pV-diagram If Pressure is constant, easiest! W = pDV = N/m2 x m3 = Nm W = p(V2 – V1) System undergoing an expansion with constant pressure. © 2016 Pearson Education Inc.

System undergoing an expansion with varying pressure. Work on a pV-diagram Work done equals area under curve on pV-diagram If pressure varies, easy!?? W =  p dV Integral depends on the pressure function p(Volume) System undergoing an expansion with varying pressure. © 2016 Pearson Education Inc.

Work on a pV-diagram System undergoing a compression with varying pressure. Work is done ON the gas, not BY the gas. Work is negative © 2016 Pearson Education Inc.

Work depends on path chosen: Slide 1 of 4 Consider three different paths on a pV-diagram for getting from state 1 to state 2. © 2016 Pearson Education Inc.

Work depends on path chosen: Slide 2 of 4 System does a LARGE amount of work on path WHY?

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: Slide 3 of 4 System does a small amount of work on path WHY??

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: Slide 4 of 4 Along smooth curve from 1 to 2, work done is different from that for other paths. © 2016 Pearson Education Inc.

First law of thermodynamics Change in internal energy U of system equals heat added minus work done by system! 1st Law of Thermodynamics just a generalization of principle 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 DU = Qin - Wout Both Q (heat in) & W (work done by) depend on path chosen between states, but DU is independent of the path. WHY??? If changes are infinitesimal, write 1st law as dU = dQ – dW. © 2016 Pearson Education Inc.

Internal energy The internal energy of a cup of coffee depends on just its thermodynamic state. how much water & ground coffee it contains… and… its temperature. It does not depend on the history of how the coffee was prepared—that is, the thermodynamic path that led to its current state. © 2016 Pearson Education Inc.

First law of thermodynamics In a thermodynamic process, internal energy U of a system may increase. Example: More heat is added to system than system does work. Internal energy of system increases. © 2016 Pearson Education Inc.

First law of thermodynamics In a thermodynamic process, internal energy U of a system may decrease. Example: More heat flows out of the system than work is done ON the system (note sign of work!!) Internal energy of the system decreases. © 2016 Pearson Education Inc.

First law of thermodynamics In a thermodynamic process, internal energy U of a system may remain the same. Example: Heat added to system equals work done BY system. Internal energy of system is unchanged. © 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. When you do a push-up, your body does work, so W > 0. Your body also warms up during exercise; but by perspiration & other means the body rids itself of this heat, so Q < 0. © 2016 Pearson Education Inc.

First law of exercise thermodynamics Since Q is negative & W is positive, U = Q − W < 0 and your body’s internal energy decreases. That’s why exercise helps you lose weight! It uses up some of internal energy stored in your body in the form of fat. Yay! Studying Physics helps you maintain your shape! © 2016 Pearson Education Inc.

First law of exercise thermodynamics Since Q is negative & W is positive, U = Q − W < 0 and your body’s internal energy decreases. The same holds for studying! You do work lifting your book and calculator and pencil! Yay! Thermodynamics proves that studying Physics helps you maintain your shape! © 2016 Pearson Education Inc.

Thermodynamic process for a human day © 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 four different processes for constant amount of ideal gas, all starting at state a. © 2016 Pearson Education Inc.

The four processes on a pV-diagram Paths on pV-diagram for all four different processes for constant amount of ideal gas, all starting at state a. Adiabatic – no heat in or out © 2016 Pearson Education Inc.

The four processes on a pV-diagram Paths on pV-diagram for all four different processes for constant amount of ideal gas, all starting at state a. Isochoric – no work done © 2016 Pearson Education Inc.

The four processes on a pV-diagram Paths on pV-diagram for all four different processes for constant amount of ideal gas, all starting at state a. Isobaric – constant pressure © 2016 Pearson Education Inc.

The four processes on a pV-diagram Paths on pV-diagram for all four different processes for constant amount of ideal gas, all starting at state a. Isothermal – constant Temp © 2016 Pearson Education Inc.

Four kinds of thermodynamic processes Four specific kinds of thermodynamic processes that occur often in practical situations: Adiabatic: No heat is transferred into or out of the system, so Qin = 0. Therefore: DU = U2 – U1 = Qin –Wout = -W Since no heat allowed in or out, change in internal energy is opposite of work. If work is done BY system, internal energy decreases. If work is done ON system, internal energy increases. © 2016 Pearson Education Inc.

Adiabatic process When cork is popped on a bottle of champagne, pressurized gases inside the bottle expand rapidly & do positive work on outside air. VERY little time for gases to exchange heat with their surroundings, so expansion is nearly adiabatic. Hence the internal energy of expanding gases decreases (ΔU = –W ) and temperature drops. This makes water vapor condense & form a miniature cloud. © 2016 Pearson Education Inc.

Exhaling adiabatically Put your hand a few centimeters in front of your mouth, open your mouth wide, and exhale. Your breath will feel warm on your hand, because the exhaled gases emerge at roughly the temperature of your body’s interior. Now purse your lips as though you were going to whistle, and again blow on your hand. The exhaled gases will feel much cooler. In this case the gases undergo a rapid, essentially adiabatic expansion as they emerge from between your lips, so the temperature of the exhaled gases decreases. © 2016 Pearson Education Inc.

Four kinds of thermodynamic processes Four specific kinds of thermodynamic processes that occur often in practical situations: Adiabatic Isochoric: The volume remains constant, so W = 0. Therefore: DU = U2 – U1 = Qin – 0 = Qin Since no work is done by, nor on the system, change in internal energy is just how much heat energy enters or leaves. If the system is cooled, internal energy decreases. If heat is added, internal energy increases. © 2016 Pearson Education Inc.

Four kinds of thermodynamic processes Four specific kinds of thermodynamic processes that occur often in practical situations: Adiabatic Isochoric Isobaric: The pressure remains constant, so W = p(V2 – V1). Therefore DU = U2 – U1 = Qin – pDV If dV is negative (compress the system!), internal energy could still decrease if you pull enough heat out! Qin< 0 © 2016 Pearson Education Inc.

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

Four kinds of thermodynamic processes Four specific kinds of thermodynamic processes that occur often in practical situations: Adiabatic: Isochoric: Isobaric: Isothermal: The temperature remains constant. How?? PV = nRT means PV is constant! Use this in W =  p dV © 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.

Adiabatic vs Isothermal processes In adiabatic process, no heat is transferred in or out of gas, so Q = 0. As gas expands, it gas does positive work W on its environment, so its internal energy decreases. Gas temperature drops! Note that adiabatic curve at any point is always steeper than an isotherm at that point. © 2016 Pearson Education Inc.

Internal energy of an ideal gas The internal energy of an ideal gas depends only on its temperature, not on its pressure or volume. Test this with “free expansion” (Joule expansion) © 2016 Pearson Education Inc.

Free Expansion A “free expansion” is when gas does no work expanding into a larger volume. Wout = 0 Assume insulated so Qin = 0 By first law, DU = Qin – Wout = 0 Since U is a function of T only, T won’t change in this process! © 2016 Pearson Education Inc.

Free Expansion In reality, most gases DO cool down as they expand into a vacuum. Change kinetic energy into internal intermolecular potential energy He gas above 40K and H gas above 200 K both get *warmer*. Change intermolecular PE into KE! The real key: Free expansion is an IRREVERSIBLE process! Hints at “ENTROPY” © 2016 Pearson Education Inc.

Heat capacities of an ideal gas Consider a constant volume process. Add heat & raise temperature of ideal gas in rigid container with constant volume (ignoring container thermal expansion). © 2016 Pearson Education Inc.

Heat capacities of an ideal gas For a Constant volume process DU = Qin - Wout Wout = 0 for Cv process DU = Qin = nCvDT CV = molar heat capacity at Constant Volume. © 2016 Pearson Education Inc.

Heat capacities of an ideal gas CV = molar heat capacity at Constant Volume. From the First Law of Thermodynamics DU = Qin = nCvDT So between ANY two temperatures in an ideal gas, no matter what the process is, you know change in U: DU = nCvDT Change in internal energy of ideal gas no matter what process! © 2016 Pearson Education Inc.

Heat capacities of an ideal gas Now consider a constant pressure process to same T… Add heat, let gas expand just enough to keep pressure constant as temperature rises, to the same final T. © 2016 Pearson Education Inc.

Heat capacities of an ideal gas First Law: DU = Qin - Wout How much heat is added? Define Cp = molar heat capacity at Constant Pressure Qin = nCpDT © 2016 Pearson Education Inc.

Heat capacities of an ideal gas at Constant P First Law: DU = Qin - Wout But now, work IS done! DU = nCpDT – pDV From ideal gas law pDV = nRDT DU = nCpDT – nRDT But wait…. There’s more!! Didn’t we just say we know DU no matter what process is involved?? © 2016 Pearson Education Inc.

Relating Cp and CV for an ideal gas DU = nCvDT Change in internal energy of ideal gas no matter what process! If is same (ending at same T!) Then DU = nCpDT – nRDT becomes n CvDT = n CpDT – nRDT or… n CvDT + nRDT= n CpDT © 2016 Pearson Education Inc.

Relating Cp and CV for an ideal gas n CvDT + nRDT = n CpDT So… Cp = Cv + R Cp > CV To produce same temperature change, more heat is required at Constant Pressure than at Constant Volume © 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 since is the same in both cases. Cp > CV Cp = CV + R R is the gas constant R = 8.314 J/mol ∙ K. © 2016 Pearson Education Inc.

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