PHYSICS 231 Lecture 28 Engines and fridges PHY 231 1 Thermodynamics Piston is moved downward slowly so that the gas remains in thermal equilibrium The temperature is the same at all times in the gas vin vout PHY 231 Vout Vin Work is done on the gas 2 Non isobaric compression In general the pressure can change when lowering the piston The work done on the gas when going from an initial state i to a final state t t f iis th the area under d th the P P V V diagram The work done by the gas is the opposite of the work done on the gas PV nRT PHY 231 3 Work done on gas signs Work Area under the curve on P V diagram V Vf Vi If the arrow goes from right to left V 0 positive work is done on the gas If the arrow goes from left to right V 0 negative work is done on the gas the gas has done positive work on the piston PHY 231 4 P A P Quiz B v v P C v I which In hi h case iis th the work kd done on th the gas largest l t Case B C B area under d curve is i larger l and arrow goes from right to left positive work is done on the gas g PHY 231 5 First Law of Thermodynamics By performing work on an object the internal energy can be changed Think about deformation pressure Think about heat transfer By transferring heat to an object the internal energy can be changed The change in internal energy depends on the work done on the object and the amount of heat transferred to the object object Remember the internal energy is the energy associated with translational translational rotational rotational vibrational motion of atoms and potential energy 6 PHY 231 First Law of thermodynamics U Uf Ui Q W U change in internal energy Q energy Q gy transfer through g heat if heat is transferred to the system W energy transfer through work if work is done on the system This law is a general rule for conservation of energy that includes heat PHY 231 7 Types of processes A Isovolumetric B Adiabatic C Isothermal D Isobaric D I b i V 0 Q 0 T 0 P 0 P 0 First law of thermo thermoDynamics U Q W PV nRT PV T constant PV T constant PHY 231 8 Isovolumetric processes line A p V 0 W 0 area under d the h curve is zero U Q Use U W Q with W 0 In case of f m monatomic m ideal g gas U 3 2nR T if P then T PV T constant so U negative U negative and Q negative Heat is extracted from the gas if P then T PV T constant v so U positive and d Q positive Heat is added to the gas Cv 3 2 R so Q Cvn T Cv molar specific heat at const vol PHY 231 9 isobaric process p P 0 v Use U W Q In case of ideal gas W P V U 3 2nR T if V then T PV T constant W positive work done on gas U negative Q negative heat extracted Q if V then T PV T constant W negative work done by gas U positive Q positive heat added Q U W 3 n Q U W 3 2nR T PV V 3 3 2nR T n nR T 5 2nR T n 5 n used ideal gas law Q nCp T with Cp 5 2 R Cp Cv10 R Cp molar heat capacity at constant pressure PHY 231 isothermal processes p T 0 The temperature is not changed Q W Use U W Q with U 0 v Vf W nRT ln Vi if if V W positive and Q negative Work is done on the g gas and heat extracted if if V W negative and Q positive Work is done by the gas and d heat h t added dd d PHY 231 11 p Adiabatic process line B Q 0 v No heat is added extracted from the system U W Use U W Q with Q 0 In case of ideal gas U 3 2nR T if T U U negative ti and d W negative W ti The gas has done work if T U positive and W positive Work is done on the gas Sudden expansions or compressions are adiabatic PHY 231 12 Cyclic processes An Engine The system returns to its original state state Therefore Therefore the internal energy must be the same after completion of f the th cycle l U 0 U 0 U Uf Ui Q W U change in internal energy Q heat positive if heat is transferred to the system W work positive if work is done on the system PHY 231 13 Cyclic Process step by step 1 Process A B P A B Negative work is done on the gas the g gas is doing g positive p work W Area under P V diagram 50 10 10 3 1 0 0 0 105 50 10 10 3 5 0 1 0 105 W 12000 J work done on gas U 3 2nR T 3 2 PBVB P PAVA 1 5 1E 5 50E 03 5E 5 10E 03 0 The internal energy has not changed U Q W so Q U W 12000 J Heat that was added to the 14 system was used to do thePHY work 231 Cyclic process step by step 2 Process B C W Area under P V diagram 3 1 0 0 0 10 5 10 50 10 W 4000 J Work was done on the gas U 3 2nR T 3 2 PCVC PBVB 1 5 1E 5 10E 3 1E 5 50E 3 6000 J The internal energy has decreased by 6000 J U Q W so Q U W 6000 4000 J 10000 J 10000 J of energy has been transferred out of the system PHY 231 15 Cyclic process step by step 3 Process C A W Area under P V diagram W 0J W 0 No work was done on the gas U 3 2nR T 3 2 PAVA PCVC 1 5 5E 5 10E 3 1E 5 10E 3 6000 J The internal energy has increased by 6000 J U Q W so Q U W 6000 0 J 6000 J 6000 J of energy has been transferred into the system PHY 231 16 Summary of the process AREA A B Quantity Work W Heat Q Process U A B A B 12000 12000 J 12000 J 0 B C 4000 J 10000 J 6000 C A 0J 6000 J 6000 SUM 8000 J 8000 J 0 B C C A PHY 231 17 What did we do The gas performed net work 8000J Area while heat was supplied 8000J We have built an engine Convert heat to mechanical work What if the p process was done in the reverse …
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