Physics 150. Thermodynamics. Chapter 15

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1 Physics 150 Thermodynamics Chapter 15

2 The First Law of Thermodynamics Let s consider an ideal gas confined in a chamber with a moveable piston If we press the piston è the gas in the chamber compresses è piston is doing work è internal energy of the gas changes ΔU = Q + W gas-filled chamber moveable piston of mass m 2

3 The first law of thermodynamics ΔU = Q + W The change in internal energy of a system is equal to the heat flow into the system plus the work done on the system (conserva]on of energy). Q is the heat If HEAT IS ADDED to the system (gas) è Q > 0 If HEAT IS RELEASED by the system (gas) è Q < 0 W is the work If WORK IS DONE ON the system (gas) è gas compresses è W > 0 If WORK IS DONE BY the system (gas) è gas expands è W < 0 3

4 Conceptual ques]on The first law of thermodynamics Q1 An amount of heat equal to 2500 J 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? A J B. 700 J C J D J ΔU = Q + W If heat is added è Q > 0 è Q = 2500 J If the work is done on è W > 0 è W = 1800 J ΔU = 2500J +1800J = 4300J 30 4

5 Thermodynamic processes Gas is expanding è the work is done by the gas è the work on the gas is nega]ve W = Fd W = PAd W = PΔV If work is to be non-zero, the gas must expand or contract. When V f < V i è the gas has contracted and W > 0 When V f > V i è the gas has expanded and W < 0. 5

6 PV Diagrams Pressure Pressure P i P i P f V i V f Volume V i V f Volume The pressure is staying constant. The magnitude of the work done is the area underneath the curve. Both the pressure and the volume change The magnitude of the total work is s]ll the area under the curve. 6

PV Diagrams Pressure P i P f isobaric process isochoric process isothermal process Constant pressure (isobaric process) W = PΔV = P(V f V i ) Constant volume (isochoric

7 PV Diagrams Pressure P i P f isobaric process isochoric process isothermal process Constant pressure (isobaric process) W = PΔV = P(V f V i ) Constant volume (isochoric process) W = PΔV = 0 Constant temperature (isothermal process) V i V f adiaba]c process (steeper curve than for isothermal process) Volume No heat transfer (adiaba]c process) Q = 0 7

8 Thermodynamic processes for an ideal gas Constant volume (isochoric process) W = PΔV = 0 Q = nc V ΔT PV = nrt ΔU = Q +W = nc V ΔT Constant pressure (isobaric process) W = PΔV = nrδt Q = nc P ΔT ΔU = Q +W = nc P ΔT nrδt C V = C P R " (isothermal process) ΔU = 0 Q = W = nrt ln V f $ # Constant temperature V i % ' & 8

9 Exercise: Thermodynamic processes How much heat is required in an isochoric process to raise the temperature of 5 moles of an ideal, monatomic gas from -50 o C to 50 o C? " Q = nc V ΔT = n$ 3 # 2 R % '(T f T i ) & Q = (50 ( 50)) = 6236J 2 How much heat is needed to isothermally change the volume of 20 moles of an ideal gas from 1.0 m 3 to 0.4 m 3 if the temperature remains at o C?! Q = nrt ln V f # " V i $ & % First, convert the temperature to Kelvins: 273 +( 23) = 250 K. " Q = ln 0.4 % $ ' = 38095J # 1 & 9

10 Reversibility vs. Irreversibility A reversible process is a process that can happen equally well from state A to state B or state B to state A. Perfectly elas]c collision is a reversible process An Irreversible Process 10

11 The Second Law of Thermodynamics Any process that involves dissipa]on of energy is not reversible. Any process that involves heat transfer from a holer object to a colder object is not reversible. The second law of thermodynamics (Clausius Statement): Heat never flows spontaneously from a colder body to a holer body. 11

Heat engines A heat engine is a device designed

expel the remaining energy Q C into a cold

12 Heat engines A heat engine is a device designed to take heat and convert it to mechanical work. take a certain amount of energy from a hot reservoir convert some of that energy to work W expel the remaining energy Q C into a cold reservoir. C = W C = W +Q C No losses! 100% efficiency! 12

13 Efficiency of an engine The efficiency of a heat engine is equal to how much energy it converts to work from the energy is takes in from the hot reservoir. e = net work output heat input = W net e = Q C =1 Q C Efficiency (e) is a unitless quan]ty that goes from 0 (W=0) to 1 (Q C =0). 13

14 Refrigerators and heat pumps Heat pumps and refrigerators both operate under the principle of removing heat Q C from one environment and expelling heat into another environment by performing work W. Q C +W + = 0 W +Q C = Q C >0 because is being added to the system W>0 because is being added to the system <0 because is being removed from the system Performance of heat pumps is measured with the coefficient of performance K: Heat pumps K P = heat delivered net work input = W net Refrigerators K R = heat removed net work input = Q C W net 14

15 Exercise: Heat engine e = W net a) How much heat does an engine with efficiency of 33.3 % absorb in order to deliver 1.00 kj of work? e = W net = W net e = 1000J 0.33 = J b) How much heat is exhausted by the engine? = W +Q C Q C = W Q C = J 1000J = J 15

Reversible Engines and Heat Pumps Let s imagine a combina]on of a heat engine and a heat pump. The work generated by the heat engine, W H.E., can be used as the work we put into the heat pump, W H.P.. They operate between the same temperature extremes.

16 Reversible Engines and Heat Pumps Let s imagine a combina]on of a heat engine and a heat pump. The work generated by the heat engine, W H.E., can be used as the work we put into the heat pump, W H.P.. They operate between the same temperature extremes. No engine can have an efficiency greater than that of an ideal reversible engine that uses the same two reservoirs. The efficiency of this ideal reversible engine is e r = 1 T T C H. Hot Reservoir, T H,HE,HP W HE W HP Q C,HE Q C,HP Cold Reservoir, T C 16

17 Carnot cycle The ideal reversible engine with the highest efficiency. A heat engine is cyclic it returns to the same pressure, volume and temperature that it had aper the cycle has been completed 1. The engine sucks in heat from the hot reservoir but, by expanding, keeps the temperature constant 2. The gas con]nues expanding adiaba]cally, lowering the temperature 3. The gas exhausts heat and compresses, keeping the temperature constant 4. The gas con]nues compressing adiaba]cally and returns to its original configura]on 17

18 The Carnot cycle illustrated 18

19 Exercise: Engine efficiency e r =1 T C T H An engine operates between temperatures 650 K and 350 K at 65.0% of its maximum efficiency. a) What is the efficiency of this engine? The maximum possible efficiency is e r =1 T C T H e r =1 350 K 650K = The engine operates at e = 0.65e r = 0.30 or 30% efficiency. 19

20 Exercise: Engine efficiency e = Q C =1 Q C An engine operates between temperatures 650 K and 350 K at 65.0% of its maximum efficiency. b) If J is exhausted to the low temperature reservoir, how much work does the engine do? e = Q C =1 Q C ( 1 e) = Q C = Q C 1 e = 6300J = 9000J W net = Q C W net = = 2700J 20

21 Entropy Heat flows from objects of high temperature to objects at low temperature because this process increases the disorder of the system. Entropy is a measure of a system s disorder. If an amount of heat Q flows into a system at constant temperature, then the change in entropy is ΔS = Q T Every irreversible process increases the total entropy of the universe. Reversible processes do not increase the total entropy of the universe. 21

22 Entropy and the second law of thermodynamics A thermally isolated system will never see its entropy decrease, i.e., ΔS 0 for any process in a thermally isolated system. 22

23 The Third Law of Thermodynamics It is impossible to cool a system to absolute zero. All gases liquefy at low temperatures so no experimental data available for low temperatures Extrapolate! 23

The First Law of Thermodynamics

Thermodynamics The First Law of Thermodynamics Thermodynamic Processes (isobaric, isochoric, isothermal, adiabatic) Reversible and Irreversible Processes Heat Engines Refrigerators and Heat Pumps The Carnot