Work done in Isothermal and adiabatic Process
- 1. Work Done in Isothermal AndWork Done in Isothermal And Adiabatic ProcessAdiabatic Process From: DEEPANSHU CHOWDHARYFrom: DEEPANSHU CHOWDHARY Roll no: 05Roll no: 05 Class: 11Class: 11thth AA
- 2. Isothermal processIsothermal process • P,V may change but temperature isP,V may change but temperature is constant.constant. • The cylinder must have conducting wallsThe cylinder must have conducting walls • It must happen very slowly so that heatIt must happen very slowly so that heat produced during compression isproduced during compression is absorbed by surroundings and heat lostabsorbed by surroundings and heat lost during compression is supplied byduring compression is supplied by surroundings.surroundings.
- 3. Adiabatic processAdiabatic process • In an adiabatic process, the system isIn an adiabatic process, the system is insulated from the surroundings and heatinsulated from the surroundings and heat absorbed or released is zero. Since there isabsorbed or released is zero. Since there is no heat exchange with the surroundings,no heat exchange with the surroundings, • When expansion happens temperature fallsWhen expansion happens temperature falls • When gas is compressed, temperatureWhen gas is compressed, temperature rises.rises.
- 4. Kinds of ProcessesKinds of Processes Often, something is held constant. Examples: dV = 0 isochoric or isovolumic process dQ = 0 adiabatic process dP = 0 isobaric process dT = 0 isothermal process
- 5. • Work done when PV = nRT = constantWork done when PV = nRT = constant P = nRT / VP = nRT / V Isothermal processesIsothermal processes ∫ −=−= final initial )curveunderarea(dVpW ∫∫ −=−= f i f i V V V V /nRT/nRT VdVVdVW )/VV(nRT ifnW −= p V 3 T1 T2 T3T4
- 6. AdiabaticAdiabatic ProcessesProcesses An adiabatic process is process in which there is no thermal energy transfer to or from a system (Q = 0) A reversible adiabatic process involves a “worked” expansion in which we can return all of the energy transferred. In this case PVγ = const. All real processes are not. p V 2 1 3 4 T1 T2 T3T4
- 7. 8 Adiabatic ProcessAdiabatic Process •For an ideal gas, and most real gasses,For an ideal gas, and most real gasses, •đQ = dU + PdVđQ = dU + PdV •đQ = CđQ = CVVdT + PdV,dT + PdV,.. •Then, whenThen, when đQđQ = 0,= 0, VC PdV dT −=
- 8. 9 Adiabatic ProcessAdiabatic Process nR VdPPdV C PdV nR VdPPdV dT nR PV T V + =− + == Then, and, For an ideal gas, PV=nRT, so
- 9. 10 Adiabatic ProcessAdiabatic Process nR VdP nRC CnR PdV nR VdP nRC PdV nR VdPPdV C PdV nR VdPPdV dT nR pV T V V VV + + = + += + += + == 0 11 0 Then, and,
- 10. 11 Adiabatic ProcessAdiabatic Process V p V P PV V V C C VdPPdVVdP C C PdV CCnR VdP C CnR PdV = +=+ = =+ + + = γ γ where, 0 0
- 11. 12 Adiabatic ProcessAdiabatic Process ( ) constant constantlnlnln constantlnln ,integratedbecanwhich,0 0 = ==+ =+ =+ =+ γ γγ γ γ γ PV PVPV PV P dP V dV VdPPdV
- 12. 13 Adiabatic ProcessAdiabatic Process constant constant as,expressedbealsocanthis ofhelpWith the constant 1 1 = = = = − − γ γ γ γ P T TV nRTPV PV
- 13. 14 γγ for “Ideal Gasses”for “Ideal Gasses” 33.1 6 2 1:polyatomic 40.1 5 2 1:diatomic 67.1 3 2 1:monatomic 2 1 =+= =+= =+= += γ γ γ ν γ
- 14. Combinations of Isothermal &Combinations of Isothermal & AdiabaticAdiabatic ProcessesProcesses All engines employ a thermodynamic cycle W = ± (area under each pV curve) Wcycle = area shaded in turquoise Watch sign of the work!
- 15. ISOTHERMAL PROCESS:ISOTHERMAL PROCESS: CONST. TEMPERATURE,CONST. TEMPERATURE, ∆∆T = 0,T = 0, ∆∆U =U = 00 NET HEAT INPUT = WORK OUTPUTNET HEAT INPUT = WORK OUTPUT ∆∆Q =Q = ∆∆U +U + ∆∆W ANDW AND ∆∆Q =Q = ∆∆WW ∆U = 0 ∆U = 0 QQOUTOUT WorkWork InIn Work OutWork Out QQININ WORK INPUT = NET HEAT OUTWORK INPUT = NET HEAT OUT
- 16. ISOTHERMAL EXAMPLEISOTHERMAL EXAMPLE (Constant T):(Constant T): PAVA = PBVB Slow compression at constant temperature: ----- No change in UNo change in U. ∆∆U =U = ∆∆TT = 0= 0 B A PA V2 V1 PB
- 17. ISOTHERMAL EXPANSION (ISOTHERMAL EXPANSION (ConstantConstant T)T):: 400 J of energy is absorbed by gas as 400 J of work is done on gas. ∆T = ∆U = 0 ∆U = ∆T = 0 BB AA PA VA VB PB PAVA = PBVB TA = TB ln B A V W nRT V = Isothermal Work
- 18. ∆∆Q =Q = ∆∆U +U + ∆∆W ;W ; ∆∆W = -W = -∆∆U orU or ∆∆U = -U = -∆∆WW ADIABATIC PROCESS:ADIABATIC PROCESS: NO HEAT EXCHANGE,NO HEAT EXCHANGE, ∆∆Q = 0Q = 0 Work done at EXPENSE of internal energy INPUT Work INCREASES internal energy Work Out Work In−∆U +∆U ∆Q = 0 ∆W = -∆U ∆U = -∆W
- 19. ADIABATIC EXAMPLE:ADIABATIC EXAMPLE: Insulated Walls: ∆Q = 0 B A PPAA VV11 VV22 PPBB Expanding gas doesExpanding gas does work with zero heatwork with zero heat loss.loss. Work = -Work = -∆∆UU
- 20. ADIABATIC EXPANSION:ADIABATIC EXPANSION: 400 J of WORK is done, DECREASING the internal energy by 400 J: Net heat exchange is ZERO. ∆∆Q =Q = 00 ∆Q = 0 B A PPAA VVAA VVBB PPBB PPAAVVAA PPBBVVBB TTAA TT BB = A A B BP V P Vγ γ =
- 21. ADIABATIC EXAMPLE: ∆Q = 0 AA BB PPBB VVBB VVAA PPAA PAVA PBVB TTAA TT BB = PPAAVVAA = P= PBBVVBB γ γ Example 2: A diatomic gas at 300 K and 1 atm is compressed adiabatically, decreasing its volume by 1/12. (VA = 12VB). What is the new pressure and temperature? (γ = 1.4)
- 22. ADIABATIC (Cont.): FIND PADIABATIC (Cont.): FIND PBB ∆Q = 0 PB = 32.4 atm or 3284 kPa 1.4 12 B B A B V P P V = 1.4 (1 atm)(12)BP = PPAAVVAA = P= PBBVVBB γ γ AA BB PPBB VVBB 12VVBB 1 atm1 atm 300 K Solve for PSolve for PBB:: A B A B V P P V γ = ÷
- 23. ADIABATIC (Cont.): FIND TADIABATIC (Cont.): FIND TBB ∆Q = 0 TB = 810 K (1 atm)(12V(1 atm)(12VBB)) (32.4 atm)(1 V(32.4 atm)(1 VBB)) (300 K)(300 K) TT BB == AA BB 32.4 atm32.4 atm VVBB 1212VVBB 1 atm1 atm 300 K Solve for TSolve for TBB TTBB=?=? A A B B A B P V P V T T =
- 24. ADIABATIC (Cont.):ADIABATIC (Cont.): If VIf VAA= 96 cm= 96 cm33 and Vand VAA= 8 cm= 8 cm33 , FIND, FIND ∆∆WW ∆Q = 0 ∆∆W = -W = - ∆∆U = - nCU = - nCVV ∆∆TT && CCVV== 21.1 j/mol K21.1 j/mol K AA B 32.4 atm32.4 atm 1 atm1 atm 300 K 810 K SinceSince ∆∆Q =Q = 0,0, ∆∆W = -W = - ∆∆UU 8 cm8 cm33 96 cm96 cm33 Find n fromFind n from point Apoint A PV = nRTPV = nRT PVPV RTRT n =n =
- 25. ADIABATIC (Cont.):ADIABATIC (Cont.): If VIf VAA= 96 cm= 96 cm33 and Vand VAA= 8 cm= 8 cm33 , FIND, FIND ∆∆WW AA BB 32.4 atm32.4 atm 1 atm 300 K 810 K 8 cm8 cm33 96 cm96 cm33 PVPV RTRT n =n = == (101,300 Pa)(8 x10(101,300 Pa)(8 x10-6-6 mm33 )) (8.314 J/mol K)(300 K)(8.314 J/mol K)(300 K) nn = 0.000325 mol= 0.000325 mol && CCVV= 21.1 j/mol K= 21.1 j/mol K ∆∆TT = 810 - 300 = 510 K= 810 - 300 = 510 K ∆∆W = -W = - ∆∆U = - nCU = - nCVV ∆∆TT ∆W = - 3.50 J