SOL5-2011cpl

SOL5-2011cpl - ECH 152A, Fall 2011, Roland Faller Solutions...

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Unformatted text preview: ECH 152A, Fall 2011, Roland Faller Solutions for Homework 5, October 31, 2011 Problem 1 An ideal gas is initially at 600 K and 1 Mpa. It undergoes a reversible process in four steps. We first reduce pressure isothermally to 300 kpa to arrive at state 2. We then reduce pressure isochorically to 200 kpa to arrive at state 3. Then we reduce volume at constant pressure to state 4. Finally we return adiabatically to the initial state 1. (a) Sketch the complete cycle in a PV diagram. (b) Determine p and T at states 2- 4 (c) Calculate Q, W, U, H for every step. Solution (a) see last page (b) State 2: p 2 = 300 kpa, T 2 = 600 K State 3: p 3 = 200 kpa, pV = RT as V is const T 3 = p 3 p 2 T 2 = 400 K State 4: p 4 = p 3 = 200 kpa. At the same time it has to be adiabatically connected to state 1 so dQ=0 and consequently dU=dW or c v dT = pdV V = NRT p dV =- NRT p 2 dp c v dT =- NRT p dp c V dT NRT =- dp p c V NR ln T 1 T 4 = ln p 1 p 4 We assume c v = 2 . 5 NR (diatomic), monatomic would be fine as well or you...
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SOL5-2011cpl - ECH 152A, Fall 2011, Roland Faller Solutions...

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