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Physics SF 016 Chapter 15 Remarks :15.1 Learning Outcome Distinguish between thermodynamic work done on the system and work done by the system. State and use first law of thermodynamics, Keypoint : Q U W
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Physics SF 016 Chapter 15 15.1.1 Signs for heat, Q and work, W (a) SurroundingsSign convention for heat, Q : (environment)Q = positive value Heat flow into the system System 3 Q 0 W 0Q = negative value (b) Surroundings Heat flow out of the system (environment) System 3 W 0
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Physics SF 016 Chapter 15Sign convention for work, W: Surroundings (environment)W = positive value Work done by the system System Q 0 W 0 SurroundingsW = negative value (environment) Work done on the system System W 0 4
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Physics SF 016 Chapter 15 Surroundings (environment) Q = positive value W = positive value System Q 0 W 0 Surroundings (environment) Q = negative value W = negative value System 5 W 0
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Physics SF 016 Chapter 15 Work done by gas (Expansion) Motion of piston When the air is expanded, the molecule loses kinetic energy and does positive work on piston. Air Expansion Work done on gas(Compression) Air Motion of Initially piston When the air is compressed, the molecule gains kinetic energy and does negative work on piston. Air 6 Compression
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Physics SF 016 Chapter 15 15.1.2 Work done in the thermodynamics system Consider the infinitesimal work A done by the gas (system) during Initial Gas the small expansion, dx in a cylinder with a movable piston as shown in Figure 15.3. dx F A Suppose that the cylinder has a Final cross sectional area, A and the pressure exerted by the gas (system) at the piston face is P. Figure 15.3 7
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Physics SF 016 Chapter 15 15.1.3 First law of thermodynamics It states that : “The heat (Q) supplied to a system is equal to the increase in the internal energy ( U) of the system plus the work done (W) by the system on its surroundings.” Q U W (15.2) and U U 2 U1 where Q : quantity of heat supplied W : work done U : change in the internal energy U1 : initial internal energy U 2 : final internal energy For infinitesimal change in the energy, dQ dU dW
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Physics SF 016 Chapter 15 •The first law of thermodynamics is P a generalization of the principle of 1 conservation of energy to include P1 3 energy transfer through heat as well as mechanical work. P2 2 •The change in the internal energy 4 ( U) of a system during any 0 V1 V2 V thermodynamic process is Figure 15.4 independent of path. For example a thermodynamics system goes from state 1 to state 2 as shown in Figure U1 2 U1 4 2 U1 3 2 16.5.
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Physics SF 016 Chapter 15 P(kPa )Example 1 : D C 300 150 B Figure 15.6 A 0 2.0 4.0 V ( 10 2 m3 ) A vessel contains an ideal gas in condition A, as shown in Figure 16.6. When the condition of the gas changes from A to that of B, the gas system undergoes a heat transfer of 10.5 kJ. When the gas in condition B changes to condition C, there is a heat transfer of 3.2 kJ. Calculate a. the work done in the process ABC, b. the change in the internal energy of the gas in the process ABC, c. the work done in the process ADC, d. the total amount of heat transferred in the process ADC. 10
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Physics SF 016 Chapter 15 a. The work done in the process ABC is given by : Calculate: WABC WAB WBC a. the work done in the process ABC, but WBC 0 b. the change in the internal energy of the gas in the process ABC, WABC PA VB VA c. the work done in the process ADC WABC 150 103 4.0 10 2 2.0 10 2 , d. the total amount of heat WABC 3000 J transferred in the process ADC.
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Physics SF 016 Chapter 15 b. By applying the 1st law of thermodynamics for ABC, thus Calculate: QABC U ABC WABC a. the work done in the process ABC, b. the change in the internal energy U ABC QAB QBC WABC of the gas in the process ABC, U ABC 10.5 103 3.2 103 3000 c. the work done in the process ADC , 4 d. the total amount of heat U ABC 1.07 10 J transferred in the process ADC.
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Physics SF 016 Chapter 15 c. The work done in the process ADC is given by Calculate: WADC WAD WDC a. the work done in the process ABC, but WAD 0 b. the change in the internal energy of the gas in the process ABC, WADC PD VC VD c. the work done in the process ADC , WADC 300 103 4.0 10 2 2.0 10 2 d. the total amount of heat transferred in the process ADC. WADC 6000 J
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Physics SF 016 Chapter 15 d. By applying the 1st law of thermodynamics for ADC, thus Calculate: QADC U ADC WADC a. the work done in the process ABC, and U ADC U ABC b. the change in the internal energy of the gas in the process ABC, QADC U ABC WADC c. the work done in the process ADC , d. the total amount of heat QADC 1.07 104 6000 transferred in the process ADC. 4 QADC 1.67 10 J
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Physics SF 016 Chapter 15 Remarks :15.2 Learning OutcomeThermodynamics processes (1 hours) Define the following thermodynamics processes: i) Isothermal, ΔU= 0 ii) Isovolumetric, W = 0 iii) Isobaric, ΔP = 0 iv) Adiabatic, Q = 0 Keypoint : Sketch P V graph to distinguish between isothermal process and adiabatic process. 15
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Physics SF 016 Chapter 15 There are four specific kinds of thermodynamic processes. It is : Isothermal process Isovolumetric @ Isochoric process Isobaric process Adiabatic process
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Physics SF 016 Chapter 15 15.2.1 Isothermal process is defined as a process that occurs at Isothermal changes constant temperature. When a gas expands or compresses isothermally (constant temperature) thus U 0 PV constant (16.3) Equation (16.3) can be expressed as Thus, PV1 1 P2V2 Q U W Q W If the gas expand isothermally, thus V2>V1 W = positive If the gas compress isothermally, thus V2<V1 W = negative
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Physics SF 016 Chapter 15 15.2.2 Adiabatic Process is defined as a process that occurs Notes : without heat transfer into or out of a system i.e. For Adiabatic expansion (V2>V1), W = positive value Q 0 but U =negative value hence the internal energy of the thus Q U W system decreases. U U 2 U1 W For Adiabatic compression (V2<V1), W = negative value but U =positive value hence For example, the compression stroke in an internal combustion engine is an the internal energy of the approximately adiabatic process. system increases.
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Physics SF 016 Chapter 15 15.2.3 Isovolumetric @ 15.2.4 Isobaric Isochoric is defined as a process that occurs atis defined as a process that occurs at constant pressure i.e.constant volume i.e. W 0 thus P 0 and W P V Q U W thus Q U W Q U U 2 U1In an isochoric process, all the energy Q U P Vadded as heat remains in the systemas an increase in the internal energy thusthe temperature of the system increases. For example, boiling water at constant pressure is an isobaric process.For example, heating a gas in a closedconstant volume container is an isochoricprocess.
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Physics SF 016 Chapter 1515.2.4 Pressure-Volume diagram (graph) for thermodynamic processesFigure 15.5 shows a P V diagram for eachthermodynamic process for a constant T4 T3 T2 T1amount of an ideal gas. Path A B Isothermal process (TB=TA) P Path A C Adiabatic process (TC<TA) A E P1 Path A D T4 Isochoric process (TD<TA) P2 B T3 D C T2 Path A E P3 T1 Isobaric process (TE>TA) 0 V1 V2 V3 V Figure 16.8
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Physics SF 016 Chapter 15Figure 15.5 shows a P V diagram for each From the Figure 15.5,thermodynamic process for a constant For comparison between theamount of an ideal gas. isothermal (A B) and adiabatic expansions (A C): The temperature fall (TC<TB) which P accompanies the adiabatic expansion results in a lower final pressure than that produced by the A E isothermal expansion (PC<PB). P1 The area under the isothermal is T4 greater than that under the P2 B T3 D adiabatic, i.e. more work is done by P3 C T2 the isothermal expansion than by T1 the adiabatic expansion. 0 V The adiabatic through any point is steeper than the isothermal through that point.
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Physics SF 016 Chapter 15 Example 3 : Air is contained in a cylinder by a frictionless gas-tight piston. a. Calculate the work done by the air as it expands from a volume of 0.015 m3 to a volume of 0.027 m3 at a constant pressure of 2.0 105 Pa. b. Determine the final pressure of the air if it starts from the same initial conditions as in (a) and expanding by the same amount, the change occurs isothermally.
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Physics SF 016 Chapter 15Example 3 : Solution : a. GivenAir is contained in a cylinder by africtionless gas-tight piston. V1 0.015 m3 ; V2 0.027 m3 ; P1 2.0 105 Paa. Calculate the work done by the air asit expands from a volume of 0.015 m3 to The work done by the air is:a volume of 0.027 m3 at a constantpressure of 2.0 105 Pa. W PV V 1 2 1 5b. Determine the final pressure of the air W 2.0 10 0.027 0.015if it starts from the same initialconditions as in (a) and expanding by the W 2400 Jsame amount, the change occursisothermally
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Physics SF 016 Chapter 15Example 3 : b. The final pressure for the isothermalAir is contained in a cylinder by a process isfrictionless gas-tight piston. PV1 1 P2V2a. Calculate the work done by the air asit expands from a volume of 0.015 m3 to 2.0 105 0.015 P2 0.027a volume of 0.027 m3 at a constantpressure of 2.0 105 Pa. P2 1.11 105 Pab. Determine the final pressure of the airif it starts from the same initialconditions as in (a) and expanding by thesame amount, the change occursisothermally
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Physics SF 016 Chapter 15 Remarks :15.3 Learning Outcome Thermodynamics work (4 hour)• Derive expression for work, W = pdV• Determine work from the area under p-V graph.• Derive the equation of work done in isothermal, isovolumetric and isobaric processes.• Calculate work done in :- isothermal process and use Keypoint : V2 P1 W nRT ln nRT ln V1 P2 isobaric process, use W PdV P V2 V1 isovolumetric process, use 25 W PdV 0
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Physics SF 016 Chapter 15 15.3.1 Work done in the thermodynamics systemThe work, dW done by the gas isgiven by dW Fdx cos A Initial Gas where 0 andF PA dW PAdx and Adx dV dx dW PdV AIn a finite change of volume from Final V1 to V2, F V2 Figure 15.6 dW PdV where V1 V2 W : work done W PdV (15.3) P : gas pressure V1 V1 : initial volume of the gas 26 V2 : final volume of the gas
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Physics SF 016 Chapter 15 15.3.2 Work done in the thermodynamics system P For a change in volume at constant P1 1 2 pressure, P W P V Work done at W P1 V2 V1 0 W P V2 V1 constant pressure 0 V P For any process in the system which the P2 2 volume is constant (no change in volume), the work done is W 0 Work done at constant W 0 volume P1 1 0 V 27 V1
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Physics SF 016 Chapter 15When a gas is expanded from V1 to V2 P Area under graph = work done by gas Work done by gas, V2 W pdV P1 1 V1 V2 1 W nRT dV V1 V V P2 2 nRT ln 2 W 0 V1 0 V1 V2 VWhen a gas is compressed from V1=> V2 Expansion PWork done on gas, V2 W pdV 2 V1 P2 Area under graph V2 1 = work done on gas W nRT dV V1 V V2 P 1 nRT ln 1 V1 W 0 Since V2< V1 the value of work done is (-) 0 V2Compression V1 V
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Physics SF 016 Chapter 15 15.3.3 Work done in Isothermal ProcessFrom the equation of state for an idealgas, nRT V2 PV nRT then P W nRT ln V V1 VTherefore the work done in theisothermal process which change of W nRT ln V2 ln V1volume from V1 to V2, is given V2 V2 W PdV W nRT ln (15.9) V1 V1 V2nRT W dV V1 V V2 1 W nRT dV V1 V
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Physics SF 016 Chapter 15 15.3.3 Work done in Isothermal ProcessFor isothermal process, the temperature By applying the 1st law ofof the system remains unchanged, thus Thermodynamics,thus V2 P1 Q U W and U 0 PV1 P2V2 1 V1 P2 Q WThe equation (16.9) can be expressed as V2 P1 Q nRT ln nRT ln P V1 P2 W nRT ln 1 (15.10) P2
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Physics SF 016 Chapter 1515.3.3 Work done in isobaric process 15.3.3 Work done in isovolumetric processThe work done during the isobaricprocess which change of volume from V1 Since the volume of the system into V2 is given by V2 isovolumetric process remains W PdV unchanged, thus V1 and P constant dV 0 V2 W P dV Therefore the work done in the isovolumetric process is V1 W P V2 V1 OR W PdV 0 (15.11) W P V (15.10)
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Physics SF 016 Chapter 15 Example 4 : a. Process 3 is a process at constant A quantity of ideal gas whose ratio of volume known as isovolumetric molar heat capacities is 5/3 has a (isochoric). temperature of 300 K, volume of 64 10 3 m3 and pressure of 243 kPa. It is b. The graph of gas pressure (P) against made to undergo the following three gas volume (V) for the changes changes in order: described is shown in Figure 15.7. 1 : adiabatic compression to a volume P( 104 Pa) 27 10 3 m3, Figure 15.7 2 : isothermal expansion to 64 10 3 m3 , 102 2 3 : a return to its original state. Process 2a. Describe the process 3.b. Sketch and label a graph of pressure P3 3 Process 1 Process 3 against volume for the changes described. 24.3 533 K 1 300 K 0 27 64 V ( 10 3 m3 )
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Physics SF 016 Chapter 15 Example 5 : A vessel of volume 8.00 10 3 m3 contains an ideal gas at a pressure of 1.14 105 Pa. A stopcock in the vessel is opened and the gas expands adiabatically, expelling some of its original mass until its pressure is equal to that outside the vessel (1.01 105 Pa). The stopcock is then closed and the vessel is allowed to stand until the temperature returns to its original value. In this equilibrium state, the pressure is 1.06 105 Pa. Explain why there was a temperature change as a result of the adiabatic expansion?Solution : Initial Final P V1 1 P2 V2 P3 V3 V2 T1 T2 T3 T1 Adiabatic Isochoric expansion process
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Physics SF 016 Chapter 15 Example 5 : A vessel of volume 8.00 10 3 m3 contains an ideal gas at a pressure of 1.14 105 Pa. A stopcock in the vessel is opened and the gas expands adiabatically, expelling some of its original mass until its pressure is equal to that outside the vessel (1.01 105 Pa). The stopcock is then closed and the vessel is allowed to stand until the temperature returns to its original value. In this equilibrium state, the pressure is 1.06 105 Pa. Explain why there was a temperature change as a result of the adiabatic expansion?Solution : When the gas expands adiabatically, it does positive work. Thus Q U W and Q 0 U W The internal energy of the gas is reduced to provide the necessary energy to do work. Since the internal energy is proportional to the absolute temperature hence the temperature decreases and resulting a temperature change.
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Physics SF 016 Chapter 15 Example 6 : a. Write an expression representing i. the 1st law of thermodynamics and state the meaning of all the symbols. ii. the work done by an ideal gas at variable pressure. [3 marks] b. Sketch a graph of pressure P versus volume V of 1 mole of ideal gas. Label and show clearly the four thermodynamics process. [5 marks] c. A monatomic ideal gas at pressure P and volume V is compressed isothermally until its new pressure is 3P. The gas is then allowed to expand adiabatically until its new volume is 9V. If P, V and for the gas is 1.2 105 Pa,1.0 10 2 m3 and 5/3 respectively, calculate i. the final pressure of the gas. ii. the work done on the gas during isothermal compression. (Examination Question Intake 2003/2004) [7 marks]
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Physics SF 016 Chapter 15 Example 6 : a. Write an expression representing i. the 1st law of thermodynamics and state the meaning of all the symbols. ii. the work done by an ideal gas at variable pressure. [3 marks]Solution : a. i. 1st law of thermodynamics: Q U W where U : change in internal energy Q : quantity of heat trans ferred W : work done ii. Work done at variable pressure: V2 V2 W PdV OR W nRT ln V1 V1
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Physics SF 016 Chapter 15 Example 6 : b. Sketch a graph of pressure P versus volume V of 1 mole of ideal gas. Label and show clearly the four thermodynamics process. [5 marks]Solution : b. PV diagram below represents four thermodynamic processes: P Isobaric process A E Isothermal process PA Isochoric process T4 D B T3 adiabatic process C T2 T1 0 VA V
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Physics SF 016 Chapter 15 Example 6 :c. A monatomic ideal gas at pressure P and volume V is compressed isothermally until its new pressure is 3P. The gas is then allowed to expand adiabatically until its new volume is 9V. If P, V and for the gas is 1.2 105 Pa,1.0 10 2 m3 and 5/3 respectively, calculate i. the work done on the gas during isothermal compression. [7 marks]Solution : i. The work done during the isothermal compression is V1 W nRT ln and nRT PV V V 1 W PV ln 3 W 5 1.2 10 1.0 10 2 ln V 3 W 1.32 103 J
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Physics SF 016 Chapter 15 THE END… Good luck For Second Semester Examination PDT STUDENTS, DO YOUR BEST, BEAT THE REST PHYSICS ‘A’, INSYAALLAH ...
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