Presentation is loading. Please wait. # 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:

## Presentation on theme: "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:"— Presentation transcript:

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:

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

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

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

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

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

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.)

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

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?

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!

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

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.

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

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

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…

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)

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

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

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

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

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

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

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

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

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).

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.

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

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.

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

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