Presentation on theme: "Chapter 18: Heat,Work,the First Law of Thermodynamics"— Presentation transcript:
1Chapter 18: Heat,Work,the First Law of Thermodynamics Thermodynamic systemsThermodynamic systemThermodynamic system is any collection of objects that isconvenient to regard as a unit, and that may have the potentialto exchange energy with its surroundings.A process in which there are changes in the state of a thermo-dynamic system is called a thermodynamic process.Signs for heat and work in thermodynamicsQWQ is + if heat is added to systemQ is – if heat is lost by systemW is + if work is done by the system.W is – if work is done on the system.
2Work done during volume changes Piston in a cylinder moves by dldue to expansion of gas at pressure pForce on piston: F = p·AIncremental work done:dW = F·dl = p·A dl = p·dVp
3Work done during volume changes (cont’d) Work and pV diagrampWork done depends on the path taken.pApAProcess A B :If VB>VA, W > 0Process B A :pBIf VB>VA, W < 0Process A B :If VA=VB, W = 0
4Work done during volume changes (cont’d) Work and pV diagram (cont’d)pWork done depends on the path taken.DpAA C : W = 0C B : W = pB(VB-VA)B C : W = pB(VA-VB)pBCA C B : W = pB(VB-VA)A D : W = pA(VB-VA)D B : W = 0A D B : W = pA(VB-VA)
5Internal energy and the first law of thermodynamics Tentative definition:The internal energy U of a system is the sum of the kineticenergies of all of its constituent particles, plus the sum ofall the potential energies of interactions among theseparticles.The first law of thermodynamicsNow define the change in internal energy using the first lawof thermodynamics.It is known from experiments that while Q and W depend onthe path, DU does not depend on the path.
6Internal energy and the first law of thermodynamics A cyclic processA: is a process in which the initialand final states are the same.pApBBVAVBIf the directions of the arrows are reversed, then
7Internal energy and the first law of thermodynamics Infinitesimal changes in stateThe first law of thermodynamics:For the system we will discuss, the work dW is given by pdV
8Kinds of thermodynamics process Adiabatic processAn adiabatic process is defined as one with no heat transfer intoor out of a system.DU = -W, Q=0Isobaric processAn isobaric process is a constant-pressure process.W = p(V2-V1) , p = constantIsothermal processAn isothermal process is a constant-temperature process.For a process to be isothermal, any heat flow into or out of thesystem must occur slowly enough that thermal equilibrium ismaintained.T = constant
9Kinds of thermodynamics process (cont’d) Work done by a gas depends on the path taken in pV space!Isochoric processAn isochoric process is aconstant-volume process.
10Kinds of thermodynamics process (cont’d) Isothermal vs. adiabatic process
11Work done by expanding gases: path dependence VpIsotherm at T12V1V2p2p1Isothermal expansion (1 2 on pV diagram) for ideal gas:W = nRT ln (V2/V1)
12Work done by expanding gases: path dependence Expression for the work done in an isobaric expansion of an ideal gas (3 2 on pV diagram).pIsotherm at T12V1V2p2p1W = p2(V2 – V1)3V
13Internal energy of an ideal gas Free expansionthe same final statethe same initial stateControlled expansion(The temperature is keptconstant and the expansionprocess is done very slowly)Uncontrolled expansion= Free expansionWork doneNo work doneHeat exchangedNo heat exchangedDifferent paths
14Internal energy of an ideal gas Another property of an ideal gasFrom experiments it was found that the internal energy ofan ideal gas depends only on its temperature, not on itspressure or volume.However, for non-ideal gas some temperature change occur.The internal energy U is the sum of the kinetic and potentialenergies for all the particles that make up the system. Non-idealgas have attractive intermolecular forces. So if the internal energyis constant, the kinetic energies must decrease. Therefore as thetemperature is directly related to molecular kinetic energy, for anon-ideal gas a free expansion results in a drop in temperature.
15Heat capacities of an ideal gas Two kinds of heat capacitiesCV : molar heat capacity at constant volumeeasier to measureCp : molar heat capacity at constant pressureBut why there is a difference between these two capacities?For a given temperature increase, the internal energy changeDU of an ideal gas has the same value no matter what the process.constant-volume (no work)T1 and T2 are the sameconstant-pressureFor an ideal gas
16Heat capacities of an ideal gas Ratio of heat capacitiesExamplefor an ideal monatomic gasfor most diatomic gas
17Adiabatic processes of an ideal gas p, V, and g( by definition)( adiabatic process)( ideal gas)
18Adiabatic processes of an ideal gas p, V, and g (cont’d)( ideal gas)( adiabatic process)For adiabatic process, ideal gas( ideal gas)For adiabatic process, ideal gas
19Exercises Problem 1 p b c Solution pc (a) pa a d (b) V Va Vc (c) So assumingand
20Exercises Problem 2 p Three moles of an ideal gas are taken around the cycle acb shownin the figure. For this gas, Cp=29.1J/(mol K). Process ac is at constantpressure, process ba is at constantvolume, and process cb is adiabatic.Ta=300 K,Tc=492 K, and Tb=600 K.Calculate the total work W for the cycle.bpcpaacVVaVcSolutionPath ac is at constant pressure:Path cb is adiabatic (Q=0):Path ba is at constant volume:
21ExercisesProblem 3You are designing an engine that runs on compressed air. Air entersthe engine at a pressure of 1.60 x 106 Pa and leaves at a pressure of2.80 x 105 Pa. What must the temperature of the compressed air befor there to be no possibility of frost forming in the exhaust ports of theengine? Frost will form if the moist air is cooled below 0oC.SolutionUsing g=7/5 for air,
22ExercisesProblem 4An air pump has a cylinder m long with a movable piston. The pump isused to compress air from the atmosphere at absolute pressure 1.01 x 105 Painto a very large tank at 4.20 x 105 Pa gauge pressure. For air, CV=20.8J/(mol K). (a) The piston begins the compression stroke at the open end of thecylinder. How far down the length of the cylinder has the piston moved whenair first begins to flow from the cylinder into tank? Assume that the compression is adiabatic. (b) If the air is taken into the pump at 27.0oC, whatis the temperature of the compressed air? (c) How much work does the pumpdo in putting 20.0 mol of air into the tank?SolutionFor constant cross-section area, the volume is proportional to the length,and Eq becomes and the distance the piston hasmoved is:Eq.19.24
23Exercises Problem 5 Solution (b) Raising both side of Eq.1 T1V1g-1=T2V2g-1 to the power g and both sides ofEq.2 p1V1g = p2V2g to power g-1, dividing to eliminate the term and , andsolving for the ratio of the temperatures:Using the result for part (a) to find L2 and the using Eq.1 gives thesame result.(c) Of the many possible ways to find the work done, the most straightforwarded is to use the result of part (b) in Eq.3 W=nCV(T1-T2) :where an extra figure was kept for the temperature difference.Eq.1Eq.2Eq.3