lecture39 - Fast enough to have Q=0 Slow enough to have...

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Heat capacitance of an ideal gas dT dQ 1 / V n C V at constant volume / P dT dQ 1 n C P at constant pressure PdV dU dQ dW dU dQ dW dQ dU dT dU 1 dT dQ 1 / V n n C V R C n n n C V P dT PdV 1 dT dU 1 dT dQ 1 / P For an ideal gas : dT dU dT dU dT dU / / P V R C C V P V P C C R f C V 2 1 R C V 1 R C P 40 . 1 , C , C 5 f : re temperatu room at gas diatomic 67 . 1 , C , C 3 f : gas monoatomic 5 7 2 7 P 2 5 v 3 5 2 5 P 2 3 v R R R R nRT PV
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Relating heat capacities at constant volume and pressure T U T U ~ ~
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Adiabatic processes for an ideal gas For an adiabatic process: dQ = 0 dU = dQ -PdV For an ideal gas: PV=nRT dT n dU dT dU 1 V V C n C dV C V V nRT - dT n ) 1 ( R V V P V C C C C  V dV T 1 - - dT     const TV const V T const V T 1 1 ln ln ln ln 1 ln   const TV 1     1 2 2 1 1 1 V T V T const PV 2 2 1 1 V P V P nRT PV
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Adiabatic processes for an ideal gas (2)  const TV 1     1 2 2 1 1 1 V T V T const PV 2 2 1 1 V P V P dT n dU V C dW ) P (P 1 1 ) P (P ) T (T n 2 2 1 1 2 2 1 1 2 1 V V V V R C W C W V V Adiabatic process:
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Unformatted text preview: Fast enough to have Q=0 Slow enough to have equilibrium Example: ? 2 60 1 2 2 2 1 2 1 1 1 P Q b T T a l V C T l V atm P V V 1 P V 2 1 2 (a) 2 1 2 2 2 1 1 1 T T T V P T V P 2 2 1 1 V P V P 2 1 1 2 V V P P atm l l atm P 1 2 1 2 2 (b) 2 2 1 1 2 2 2 1 1 1 V P V P T V P T V P 2 1 1 2 V V P P atm P 2 1 2 2 1 2 1 1 2 V V T T 2 1 60 2 C T...
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This note was uploaded on 10/21/2010 for the course PHYS 221 taught by Professor Herrera-siklody during the Fall '08 term at Iowa State.

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lecture39 - Fast enough to have Q=0 Slow enough to have...

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