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For the cyclic process shown, W is:D A] 0, because it’s a loop B] p 0 V 0 C] - p 0 V 0 D] 2 p 0 V 0 E] 6 p 0 V 0 For the cyclic process shown, U is: A For the cyclic process shown, Q is:D For ONE cycle:

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Is ALL heat that we add converted into work? In other words, is there any part of the cycle where heat is removed from the gas? A] No heat is removed B] Segment 3-4 only C] Segment 4-1 only D] Segments 3-4-1 E] Segments 3-4-1-2

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An ADIABATIC PROCESS is one in which no heat transfer occurs. Q=0. If we expand a gas from V A to V B adiabatically, what will be the final pressure? A] More than P B B] Less than P B C] Equal to P B D] cannot determine

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We saw that there is always heat “rejected” in a thermodynamic cycle = Q c

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Otto cycle -- in your “Ottomobile”. http://www.youtube.com/watch?v=E0PIdWdw15U

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THIS SIDE OF THE CYCLE IS ACTUALLY AN ENTIRE COMPRESSION & EXPANSION TO TAKE IN NEW AIR/FUEL.

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How much work is done by the gas in the cycle shown? D A] 0 B] p 0 V 0 C] 2p 0 V 0 D] -2p 0 V 0 E] 4 p 0 V 0 How much total heat is added to the gas in the cycle shown? D If “negative heat” is added to the gas, this means more heat is expelled from the gas than taken in. (The difference is the work done on the gas.)

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In one (ccw) Carnot cycle shown, the work done by the gas is: A] + B] - C] 0

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A] 1-2 B] 2-3 C] 3-4 D] 4-1 E] none Work < 0, so Q < 0. Along which paths is heat expelled from the gas?

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Heat is expelled from the gas during isothermal compression 3- 4. Heat is added to the gas during isothermal expansion 1-2. More heat is expelled than added. The net effect is to take heat from a cold reservoir, and add it to a hot reservoir (along with some extra heat from the work done on the gas.) This is a fridge!

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Monday 9/13 The laws of mechanics (and E&M, etc.) are time-reversal invariant. So how come this looks funny? http://www.youtube.com/watch?v=mGZjCUKowIs

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Heat flows from a hot object to a cold object in contact with it because: A] the hot object has more total internal energy - and heat flows until both objects have the same internal energy B] the hot object has more total energy per molecule- and heat flows until both objects have the same energy per molecule C] the hot object has more translational kinetic energy per molecule - and heat flows until both objects have the same translational kinetic energy per molecule.

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Ideal, monatomic gas goes around the cycle shown. Is this an engine or a fridge? A] engine B] fridge

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What is the temperature at c? A] T a B] 2T a C] 3T a D] T a /3 E] cannot determine

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T b =T a. A] p 0 V 0 B] - (2/3) p 0 V 0 C] - p 0 V 0 ln(3) D] - p 0 V 0 ln(1/3) E] cannot determine So T c = 3T a How much work does the gas do a-b? Use paper & pencil…

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A] p 0 B] 2p 0 C] -2p 0 D] 3p 0 E] cannot determine T c = 3T a What is the pressure at b? W ab = - p 0 V 0 ln(3)

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A] p 0 V 0 B] 3 p 0 (1/3) V 0 C] 3 p 0 (2/3) V 0 D] 0 E] cannot determine T c = 3T a What is the work done by the gas b-c? W ab = - p 0 V 0 ln(3)P b = 3P 0

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A] p 0 V 0 B] 3 p 0 (1/3) V 0 C] 3 p 0 (2/3) V 0 D] 0 E] cannot determine T c = 3T a What is the work done by the gas c-a? W ab = - p 0 V 0 ln(3)P b = 3P 0 W bc = 3 p 0 (2/3) V 0

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A] ab B] bc C] ca D] ab & bc E] bc & ca T c = 3T a Along which segments is heat added? W ab = - p 0 V 0 ln(3) P b = 3P 0 W bc = 2 p 0 V 0 W ca = 0

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A] 2p 0 V 0 B] nC v T a C] nC p T a D] nC p 2T a E] nC p 3T a T c = 3T a Heat is added only along bc. How much heat is added? W ab = - p 0 V 0 ln(3) P b = 3P 0 W bc = 2 p 0 V 0 W ca = 0

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T c = 3T a Now p 0 V 0 = nRT a. Let’s find the efficiency! W ab = - p 0 V 0 ln(3) P b = 3P 0 W bc = 2 p 0 V 0 W ca = 0 Q bc = nC p 2T a e= W/Q added

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A cylinder containing an ideal gas is heated at constant pressure from 300K to 350K by immersion in a bath of hot water. Is this process reversible or irreversible? A] reversible B] irreversible

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A hot piece of metal is placed in an insulating box filled with a polyatomic gas. When thermal equilibrium has been reached: A] the metal and the gas have equal total energy B] the average energy per atom in the metal is equal to the average energy per molecule in the gas C] the average kinetic energy per atom in the metal is equal to the average translational kinetic energy per molecule in the gas D] the average kinetic energy per atom in the metal is equal to the average kinetic energy per atom in the gas

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W 9/15 What is the work done by the gas in the reversible isothermal expansion shown? A] p 0 V 0 ln(2) B] p 0 V 0 C] 2 p 0 V 0 D] 0 E] none of these What is the heat added, Q? A

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No change in internal energy, so W=Q= p 0 V 0 ln(2). What is the entropy change of the gas? A] p 0 V 0 ln(2) B] nRln(2) C] nRln(1/2) D] 0 E] cannot determine What is the entropy change in the hot reservoir which is adding heat to the gas? S = Q/T for an isothermal process. Use p 0 V 0 =nRT along with Q= p 0 V 0 ln(2) to find S = nRln(2).

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A] p 0 V 0 ln(2) B] nRln(2) C] nRln(1/2) D] 0 E] cannot determine What is the entropy change in the hot reservoir which is adding heat to the gas? In a reversible process, S = 0. So the entropy change in the hot reservoir (which is at the same temperature T as the gas) is -nRln(2). Answer C.

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We showed, for a Carnot cycle, that Q H /T H = |Q c |/T C = -Q c /T c What is the change in entropy of the gas around the entire Carnot cycle? A] p 0 V 0 ln(2) B] nRln(2) C] nRln(1/2) D] 0 E] cannot determine

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Any reversible process consists of “adjoining” Carnot cycles. S for adjoining segments cancels. So: Entropy, like Internal Energy, is a “state” variable, and depends only on the state of a system (p, V for a gas). -> You can calculate entropy changes for irreversible processes by taking a reversible path to the same endpoint. Example: free expansion to double the volume. T f = T i.

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Entropy changes in non-isothermal processes Example 1: heating water Example 2a/b: heating a gas at constant V/p

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