PracticeEx1_Soln - R V P R V P T T T T V V V V A A A A c B...

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Solutions for the PracticeEx 1 1. (a) You only need two variables out of P,V and T to define a state of a gas: True (b) The energy of ideal gas remains constant : False (c) w rev > – w irrev : True (d) Entropy is a state function: False (e) dq = 0 so dS = 0 : True 2. (a) (P 1 ,V 1 ,T) (P 2 ,V 2 ,T) and irreversible isothermal process 1 1 V RT mg P and 2 2 2 V RT mg P RT mg RT mg RT V P RT V V P w 2 ) ( 1 2 1 2 2 RT w w U q (b) 2 ln 2 ln ln 2 1 1 2 2 1 RT mg RT mg RT RT V V RT dV V RT PdV w V V V V 2 ln RT w q 3. (a) This is a isobaric process, so you can use C V . T dT R T dT C T dq dS V rev 2 3 2 ln 2 3 2 ln 2 3 ln 2 3 ln 2 3 2 3 R R V P R V P R R V P R V P R T T R T dT R S A A A A A A B B A B T T B A = 8.644 J/K (b) Adiabatic process: S = 0
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(c) Isothermal process. C A V V V V V V A C rev A C rev V V R V RdV T PdV T PdV T dw T dq S A C A C A C ln  2 / 3 2 / 3 2 2
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Unformatted text preview: R V P R V P T T T T V V V V A A A A c B A c B C C B C A 2 ln 2 3 2 ln 2 / 3 R R S = - 8.644 J/K (d) 2 ln 2 3 2 ln 2 3 3 2 1 R R S S S S Or S since the initial and final states are identical. 4. (a) You need to calculate the works done on the left and right sides of the cylinder separately. 1 1 1 1 1 ) ( ) ( V P V P V P left w 2 2 2 2 2 ) ( ) ( V P V P V P right w So 2 2 1 1 ) ( ) ( V P V P right w left w w (b) w q w U since adiabatic 2 2 1 1 1 2 V P V P w U U U 1 1 1 2 2 2 V P U V P U 1 2 H H H...
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This note was uploaded on 10/24/2010 for the course CH 153K taught by Professor Wilson during the Fall '10 term at University of Texas at Austin.

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PracticeEx1_Soln - R V P R V P T T T T V V V V A A A A c B...

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