Thermodynamics Heat & Work The First Law of Thermodynamics

Transcription

1 Thermodynamics Heat & Work The First Law o Thermodynamics Lana Sheridan De Anza College April 26, 2018

2 Last time more about phase changes work, heat, and the irst law o thermodynamics

3 Overview P-V diagrams applying the irst law in various cases heat transer

4 e gas be Work done on a gas er variservation a process transer r leaving Here, the volume decreases, so the work done on the gas is positive. a change e system, s section, rk. Work vestigate cylinder volume V the pisas on the y the piscompress in essenn o the ownward 0.4b), the ause A dy e gas as (20.8) P A a V dy Figure 20.4 Work is done on a gas contained in a cylinder at a pressure P as the piston is b Work done on a gas V W = P dv The work done Von a gas i equals the negative o the area under the PV curve. The area The work is negative donehere is the because area under the the P-V curve volume (with is decreasing, the appropriate resulting sign). in positive work. P P i P V V i i V To e ing I dep at e imp o d cur N PV d T i d F par illu

5 The work done on a gas equals the negative o the area under the PV curve. The area P-V diagrams are very is negative useul in here thermodynamics. because volume is decreasing, resulting Example o a P-V diagram: in positive work. Aside: P-V Diagrams P P i P depends on the vo at each step o the important graphic o diagram allows curve on a PV diag Notice that the PV diagram. There The work done initial state to a diagram, evalua For the process V particular path tak V V i illustrate this imp They are graphs ofigure pressure 20.5 vs volume A gas is compressed or a ixed quantity (Fig. 20.6). In the (n moles) o an ideal quasi-statically gas. (slowly) rom state reduced rom V i t i to state. An outside agent must increases rom P i t They contain a lotdo opositive inormation. work on the gas to along this path is 2 1 compress it. Figure orm Serway & Jewett, 9th ed, page 602. rom P to P at con i

6 Aside: P-V Diagrams Example o a P-V diagram: Pressure i Volume These diagrams represent the (thermodynamic) equilibrium states o the ideal gas sample, each point is a state, imply the temperature o a known sample, and show the internal energy o the gas. (E int T ) PV = nrt 1 Figure (modiied) rom Halliday, Resnick, and Walker, page 537.

7 Aside: under P-V the PV curve. Diagrams The area P P i equals the negative o the area is negative here because the volume is decreasing, resulting Example in positive o work. a P-V diagram: P i depends on the volume and temper at each step o the process, the state important graphical representation c o diagram allows us to visualize a pr curve on a PV diagram is called the p When Notice these that diagrams the integral include a in path, Equatio they PV diagram. Thereore, we can ident show reversible processes, The eg. compression work done o on gas, a gas in a qua initial show all state theto intermediate a inal state is the n diagram, thermal states evaluated passedbetween through, the in show the work done on the gas, For the process o compressing a g particular indicate path the heat taken transerred between to V the in V V i illustrate the gas. this important point, consi Figure 20.5 A gas is compressed (Fig. 20.6). In the process depicted i quasi-statically (slowly) rom state E int = reduced W + Q rom V i to V at constant p i to state. An outside agent must increases rom P i to P by heating at c do positive work on the gas to along this path is 2P i (V 2 V i ). In Fig compress it. rom P i to P at constant volume V i an

8 same initial volume, temperature, and pressure, and is assumed to be ideal. In Figure 20.7a, the gas is thermally insulated rom its surroundings except at the bottom o Work done depends on the process the gas-illed region, where it is in thermal contact with an energy reservoir. An energy reservoir Dierent is a source paths o energy or processes that is considered to go rom to (V be i, so P great i ) to (V that, P a inite ) require transer o energy to or rom the reservoir does not change its temperature. The piston is held dierent amounts o work. A constant-pressure compression ollowed by a constant-volume process A constant-volume process ollowed by a constantpressure compression An arbitrary compression P P P P P P P i V V i i V P i a b c V In all o the processes shown above, there are temperature changes during the process. 1 Figure orm Serway & Jewett. V i i V P i V V i i V

9 Heat transer also depends on the process This process (shown) happens at constant temperature: 20.5 The First Law o The gas is initially at temperature T i. The hand reduces its downward orce, allowing the piston to move up slowly. The energy reservoir keeps the gas at temperature T i. The g initial tempe T i an conta by a th memb with v above Energy reservoir at T i Energy reservoir at T i a b c Heat Q is transerred to the gas, and the gas does positive work on its surroundings. (Equivalently, negative work is done on the gas.) Figure 20.7 Gas in a cylinder. (a) The gas is in contact with an energy reservoir. The walls o the cylin base in contact with the reservoir is conducting. (b) The gas expands slowly to a larger volume. (c) The hal o a volume, with vacuum in the other hal. The entire cylinder is perectly insulating. (d) The gas

10 Heat transer also depends on the process This process also happens at constant temperature, and has the same start and 20.5 end The First points, Law o (VThermodynamics i, P i ) to, P ): 603 The hand reduces its downward orce, allowing the piston to move up slowly. The energy reservoir keeps the gas at temperature T i. The gas is initially at temperature T i and contained by a thin membrane, with vacuum above. The membrane is broken, and the gas expands reely into the evacuated region. ir at T i No heat is transerred to the gas, and the gas does no work. c (We cannot represent the path or this process on a P-V diagram.) tact with an energy reservoir. The walls o the cylinder are perectly insulating, but the d

11 First Law o Thermodynamics Reminder: Internal energy, E int or U The energy that a system has as a result o its temperature and all other molecular motions, eects, and conigurations, when viewed rom a reerence rame at rest with respect to the center o mass o the system. 1st Law The change in the internal energy o a system is equal to the sum o the heat added to the system and the work done on the system. E int = W + Q This is just the conservation o energy assuming only the internal energy changes.

12 Applying the 1st Law i W > 0 Volume Some special cases o interest: i Process Pressure W > 0 or an isolated system, (b) 0 W = Q = 0 E int = 0 Volume a (c) Pressure 0 Pressure i iw > 0 Volume e can control how uch work it does. h Moving rom to i, it does negative work. Cycling clockwise yields Volume a positive net work. or a process (V i, P i ) to (V i, P i ), a cycle, E int = 0 i Q = W Pressure Pressure i W net > 0 Volume d (e) 0 W < 0 Volume () 0 Volume

13 a system is always the negative o the work done by the Eq in terms o the work W on done on the system, on. This tells us the ollowing:the internal energy o a e i heat Theis igure absorbed shows by our the system paths on or i a p-v positive diagram work along is which a gas nversely, canthe be taken internal rom energy statetends i to to state decrease. Rank i heat the paths, is greatest to egative least, work according is done ontothe system. (a) the change E int in the internal energy o the gas, Applying the 1st Law & P-V diagrams questions p ur paths on a p-v diagram taken rom state i to state. g to (a) the change E int in e gas, (b) the work W done gnitude o the energy transn the gas and its environi A 1, 2, 3, 4 B 4, 3, 2, 1 int,are not true dierentials;that is,there are no such unctions as end only Con 1, the 4, state 3, 2o the system.the quantities dq and dw are are usually represented by the symbols d Q and d W. For our imply as Dininitesimally the same small energy transers. 1 Based on question rom Halliday, Resnick, and Walker, page 491. V

14 a system is always the negative o the work done by the Eq in terms o the work W on done on the system, on. This tells us the ollowing:the internal energy o a e i heat Theis igure absorbed shows by our the system paths on or i a p-v positive diagram work along is which a gas nversely, canthe be taken internal rom energy statetends i to to state decrease. Rank i heat the paths, is greatest to egative least, work according is done ontothe system. (b) the work W done on the gas, Applying the 1st Law & P-V diagrams questions p ur paths on a p-v diagram taken rom state i to state. g to (a) the change E int in e gas, (b) the work W done gnitude o the energy transn the gas and its environi A 1, 2, 3, 4 B 4, 3, 2, 1 int,are not true dierentials;that is,there are no such unctions as end only Con 1, the 4, state 3, 2o the system.the quantities dq and dw are are usually represented by the symbols d Q and d W. For our imply as Dininitesimally the same small energy transers. 1 Based on question rom Halliday, Resnick, and Walker, page 491. V

15 a system is always the negative o the work done by the Eq in terms o the work W on done on the system, on. This tells us the ollowing:the internal energy o a e i heat Theis igure absorbed shows by our the system paths on or i a p-v positive diagram work along is which a gas nversely, canthe be taken internal rom energy statetends i to to state decrease. Rank i heat the paths, is greatest to egative least, work according is done ontothe system. (c) the energy transerred to the gas as heat Q. Applying the 1st Law & P-V diagrams questions p ur paths on a p-v diagram taken rom state i to state. g to (a) the change E int in e gas, (b) the work W done gnitude o the energy transn the gas and its environi A 1, 2, 3, 4 B 4, 3, 2, 1 int,are not true dierentials;that is,there are no such unctions as end only Con 1, the 4, state 3, 2o the system.the quantities dq and dw are are usually represented by the symbols d Q and d W. For our imply as Dininitesimally the same small energy transers. 1 Based on question rom Halliday, Resnick, and Walker, page 491. V

16 Applying the 1st Law: new Vocabulary There are ininitely many paths (V i, P i ) to (V, P ) that we might consider. Ones that are o particular interest or modeling systems in engines, etc. are processes that keep one o the variables constant. There is a technical name or each kind o process that keeps a variable constant.

17 Applying the 1st Law: new Vocabulary Adiabatic process is a process where no heat is transerred into or out o the system. Q = 0 This type o transormation occurs when the gas is in a thermally insulated container, or the process is very rapid, so there is no time or heat transer. Since Q = 0: E int = W

18 Applying the 1st Law: new Vocabulary Isobaric process is a process that occurs at constant pressure. P i = P = P This type o transormation occurs when the gas is ree to expand or contract by coming into orce equilibrium with a constant external environmental pressure. In this case, the expression or work simpliies: W = V V i P dv = P(V V i )

19 Applying the 1st Law: new Vocabulary Isovolumetric process is a process that occurs at constant volume. V i = V = V In an isovolumetric process the work done is zero, since the volume never changes. W = 0 E int = Q (An isovolumetric process can also be called an isochoric process.)

20 Applying the 1st Law: new Vocabulary Isothermal process is a process that occurs at constant temperature. T i = T = T This kind o transormation is achieved by putting the system in thermal contact with a large constant-temperature reservoir. In an isothermal process, assuming no change o phase (staying an ideal gas!): E int = 0

21 Isothermal Expansion o an Ideal Gas Since T = 0, PV = nrt reduces to: PV = a where a is a constant. We could also write this as: 606 P = a VChapter 20 The First Law o Thermody Plotting this unction: P i P P i Isotherm PV = constant The curve is a hyperbola. Isothermal Exp Suppose an ideal g This process is des hyperbola (see App stant indicates that Let s calculate th The work done on t process is quasi-stat V i V V

22 Isothermal Expansion o an Ideal Gas Since nrt is constant, the work done on the gas in an isothermal expansion is: V W = V i V P dv nrt = V i V dv ( ) V = nrt ln V i ( ) Vi W = nrt ln V

23 Questions Which path is isobaric? inal state. Figure 20.9 The PV diagram or an isothermal expansion o an ideal gas rom an initial state to a P Because T is c n and R: To evaluate th result at the in D A B C T 1 T 2 T 3 T 4 Figure (Quick Quiz 20.4) Identiy the nature o paths A, B, 1 Based on Quick C, Quiz and 20.4, D. Serway & Jewett, page 606. V Numerically, t shown in Figu done on the g the work done Q uick Quiz 2 ric, isotherm

24 Questions Which path is isothermal? inal state. Figure 20.9 The PV diagram or an isothermal expansion o an ideal gas rom an initial state to a P Because T is c n and R: To evaluate th result at the in D A B C T 1 T 2 T 3 T 4 Figure (Quick Quiz 20.4) Identiy the nature o paths A, B, 1 Based on Quick C, Quiz and 20.4, D. Serway & Jewett, page 606. V Numerically, t shown in Figu done on the g the work done Q uick Quiz 2 ric, isotherm

25 Questions Which path is isovolumetric? inal state. Figure 20.9 The PV diagram or an isothermal expansion o an ideal gas rom an initial state to a P Because T is c n and R: To evaluate th result at the in D A B C T 1 T 2 T 3 T 4 Figure (Quick Quiz 20.4) Identiy the nature o paths A, B, 1 Based on Quick C, Quiz and 20.4, D. Serway & Jewett, page 606. V Numerically, t shown in Figu done on the g the work done Q uick Quiz 2 ric, isotherm

26 Questions Which path is adiabatic? inal state. Figure 20.9 The PV diagram or an isothermal expansion o an ideal gas rom an initial state to a P Because T is c n and R: To evaluate th result at the in D A B C T 1 T 2 T 3 T 4 Figure (Quick Quiz 20.4) Identiy the nature o paths A, B, 1 Based on Quick C, Quiz and 20.4, D. Serway & Jewett, page 606. V Numerically, t shown in Figu done on the g the work done Q uick Quiz 2 ric, isotherm

27 Example 20.6: Boiling water This example illustrates how we can apply these ideas to liquids and solids also, and even around phase changes, as long as we are careul. Suppose 1.00 g o water vaporizes isobarically at atmospheric pressure ( Pa). Its volume in the liquid state is V i = V liq = 1.00 cm 3, and its volume in the vapor state is V = V vap = 1671 cm 3. Find the work done in the expansion and the change in internal energy o the system. Ignore any mixing o the steam and the surrounding air; imagine that the steam simply pushes the surrounding air out o the way.

28 Example 20.6: Boiling water V i = V liq = 1.00 cm 3 V = V vap = 1671 cm 3 L v = J/kg Work done,

29 Example 20.6: Boiling water V i = V liq = 1.00 cm 3 V = V vap = 1671 cm 3 L v = J/kg Work done, W = P(V V i ) = ( Pa)( ) 10 6 m 3 = 169 J Internal energy, E int?

30 Example 20.6: Boiling water V i = V liq = 1.00 cm 3 V = V vap = 1671 cm 3 L v = J/kg Work done, W = P(V V i ) = ( Pa)( ) 10 6 m 3 = 169 J Internal energy, E int? Know W, must ind Q: Q = L v m = ( J/kg)(1 3 kg) = 2260 J So, E int = W + Q = 2.09 kj

31 Summary P-V diagrams applying the irst law in various cases Homework Serway & Jewett: Read chapter 20 and look at the examples. prev: Ch 20, OQs: 1, 13; Probs: 25, 27, 29, 31, 35, 39, 41, 59, 71, 77