View
219
Download
1
Embed Size (px)
Citation preview
For the cyclic process shown, W is:D
A] 0, because it’s a loop
B] p0V0
C] - p0V0
D] 2 p0V0
E] 6 p0V0
For the cyclic process shown, U is: A
For the cyclic process shown, Q is:DFor 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 VA to VB adiabatically, what will be the final pressure?
A] More than PB
B] Less than PB
C] Equal to PB
D] cannot determine
Otto cycle -- in your “Ottomobile”.
http://www.youtube.com/watch?v=E0PIdWdw15U
How much work is done by the gas in the cycle shown? D
A] 0
B] p0V0
C] 2p0V0
D] -2p0V0
E] 4 p0V0
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 isexpelled from the gas than taken in. (The difference is the work done on the gas.)
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 9/13The 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.
Tb=Ta.
A] p0V0
B] - (2/3) p0V0
C] - p0V0 ln(3)
D] - p0V0 ln(1/3)
E] cannot determine
€
TcTb=VcVb
So Tc = 3Ta
How much work does the gas do a-b? Use paper & pencil…
A] p0
B] 2p0
C] -2p0
D] 3p0
E] cannot determine
Tc = 3Ta
What is the pressure at b?
Wab = - p0V0 ln(3)
A] p0V0
B] 3 p0 • (1/3) V0
C] 3 p0 • (2/3) V0
D] 0
E] cannot determine
Tc = 3Ta
What is the work done by the gas b-c?
Wab = - p0V0 ln(3) Pb = 3P0
A] p0V0
B] 3 p0 • (1/3) V0
C] 3 p0 • (2/3) V0
D] 0
E] cannot determine
Tc = 3Ta
What is the work done by the gas c-a?
Wab = - p0V0 ln(3) Pb = 3P0
Wbc = 3 p0 • (2/3) V0
A] ab
B] bc
C] ca
D] ab & bc
E] bc & ca
Tc = 3Ta
Along which segments is heat added?
Wab = - p0V0 ln(3)
Pb = 3P0Wbc = 2 p0V0
Wca = 0
A] 2p0V0
B] nCvTa
C] nCpTa
D] nCp •2Ta
E] nCp •3Ta
Tc = 3Ta
Heat is added only along bc. How much heat is added?
Wab = - p0V0 ln(3)
Pb = 3P0Wbc = 2 p0V0
Wca = 0
Tc = 3Ta
Now p0V0 = nRTa. Let’s find the efficiency!
Wab = - p0V0 ln(3)
Pb = 3P0Wbc = 2 p0V0
Wca = 0
Qbc= nCp •2Ta
e= W/Qadded
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 9/15What is the work done by the gas in the reversible
isothermal expansion shown?
A] p0V0ln(2)
B] p0V0
C] 2 p0V0
D] 0
E] none of these
What is the heat added, Q? A
No change in internal energy, so W=Q= p0V0ln(2).What is the entropy change of the gas?
A] p0V0ln(2)
B] nRln(2)
C] nRln(1/2)
D] 0
E] cannot determine
What is the entropy change in the hot reservoir which isadding heat to the gas?
S = Q/T for an isothermal process. Use p0V0=nRT along with Q= p0V0ln(2) to find S = nRln(2).
A] p0V0ln(2)
B] nRln(2)
C] nRln(1/2)
D] 0
E] cannot determine
What is the entropy change in the hot reservoir which isadding 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 QH/TH = |Qc|/TC= -Qc/Tc
What is the change in entropy of the gas around the entire Carnot cycle?
A] p0V0ln(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. Tf = Ti.