mid_term10-31-06-solution

mid_term10-31-06-solution - Mid Term 105A 10/31/06 Mid Term...

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Mid Term 105A 10/31/06 1/5 Mid Term #1 105A Fall 2006 Question 1 One mole of gas in a chamber is compressed and decompressed using a piston which slides without friction. The initial state is at p 1 = 2x10 5 Pa and V 1 =14x10 -3 m 3 . This gas goes successively through: 1. An isobaric decompression, which double its volume; 2. An isothermal compression, which bring the volume back to V 1 ; 3. An isochoric (constant volume) cooling, which brings it back to p 1 . We suppose this gas ideal. a- What is the temperature at which the compression happens? b- Draw the different processes on a (p,V) diagram. How is this type of transformation called? c- What is the work and heat exchanged by this system during one of these transformations? a- (5 pts) We start with one mole of gas at 2x10 5 Pa and 14x10 -3 m 3 . Using the ideal gas law the temperature is found to == = 22 1 1 21 2 2 p Vp V T RR or 2 = 673.56 K. b- (10 pts) Using the ideal gas law at constant temperature for closed system we also find that = 33 . Since the isothermal process brings us back to V 1 w e h a v e t h a t V 3 = V 1 . So = 31 2 1 2 and because p 2 = p 1 we finally find = 2 Since we come back to the original state this transformation is called a cycle. c- (15 pts) Using the first law of thermodynamics we have for the whole cycle 1 2 3 p V V 1 2V 1 p 1 2p 1
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Mid Term 105A 10/31/06 2/5 ∆= cycle UQ W .Since 0 U , Q = . The elementary work for a simple compressional system such as this one is pdV and the total work is →→→ =++ 12 23 31 WWW . So =+ + ∫∫∫ 231 123 pdV . Since the process 3 1 is isochoric = 0 , since the process 1 2 is isobaric =− = 1 2 1 11 () pV V pV and since 2 3 is isothermal == ∫∫ 33 22 1 1 2 dV p V VV or 3 2 2l n 2 l n 2 V . The total work is =−< l n 2 0 or = -1081.6 J = - Question 2 A closed chamber has adiabatic wall (no exchange of heat with the surroundings).
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mid_term10-31-06-solution - Mid Term 105A 10/31/06 Mid Term...

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