Chapter 3.
These energy level diagrams show a single microstate of the system, which is meant to be representative of the macrostate (which is the ensemble of microstates, all of the same total energy).
3- Process 2 is best described as
A) free expansion B) heat added to system
C) work done on system
D) none of these descriptions is appropriate
Answer: Heat added to system. (Process 1 is work done on the system.)
3- In which process(es) does the entropy of the system increase?
A) 1 only B) 2 only C) both 1 and 2 D) neither 1 nor 2
Answer: 2 only Entropy is the log of the number of arrangements(multiplicity) of the dots among the lines, keeping the total energy fixed (multiplicity= number of microstates corresponding to the macrostate of fixed energy) In process 1, the configuration of dots and lines is unchanged (so no entropy change).
3-
Consider 3 processes:
I. slow, adiabatic expansion (Q = 0)
II. slow, isothermal expansion (T = constant)
III. free expansion (Q = W = 0)
The process shown could be
A) I only B) II only C) III only
D) II or III E) I, II, or III
Answer: II or III. The process shown has the total energy of the system unchanged (sum of the energies of all the dots is the same). In both processes II and III, the total energy U is constant. Process I can be eliminated because adiabatic expansion always results in a lower energy.
3- Consider the following graph of entropy S vs. energy U for a particular system. How does the temperature at point 1 compare to the temperature at point 2?
A) T1 = T2 B) T1 > T2 C) T1 < T2
Answer: T1 = T2 Temperature is a function of the slope of S vs. U: (1/T) = dS/dU|V,N The curve shown has constant slope, so the temperature is constant. This is an unusual situation: the temperature remains constant as the energy of the system increases. Such a situation occurs during a phase transition, for instance melting of ice.
3- According to the following graph of entropy vs. energy, the temperature of the system, in the limit U ® 0, is approaching
A) Zero B) a non-zero, finite constant C) infinity
D) a negative value(!)
Answer: T = infinity. (1/T) = dS/dU|V,N As U goes to zero, the slope dS/dU is approaching zero. The inverse of the slope (=T) is diverging. This is a case in which the temperature decreases as energy is added to the system. A collection of gravitationally bound particles acts like this.
3- A small amount of heat flows from a system A at higher temperature to a system B at lower temperature.
What happened to Stot = SA + SB as a result of the heat transfer?
A) Stot increased B) Stot decreased C) Stot remained constant
Answer: Stot increased. DStot = DSA + DSB =
The increase in SB is larger than the decrease in SA. The reason that heat spontaneously transfers from a hotter object to a colder object is that this increases the total entropy.
3. As a result of this heat flow, which is larger, |DSA| or |DSB| ?
A) |DSA| B)|DSB| C) Neither, |DSA|=|DSB|
Answer: |DSB| is larger. DS = Q/T T is smaller for system B, so |Q/T| is larger.
3. At the start of the heat flow, how do compare?
A) B) C)
Answer: The increase in SB must be larger than the decrease in SA in order for the total entropy to increase.
The entropies vs. energy for two systems in thermal contact are shown in the graph. UA+UB = Utot =fixed
3. Where on the graph is the equilibrium value of UA?
A) somewhere in region 1
B) at point 2, where the curves cross
C) somewhere in region 2
Answer: Somewhere in region 1, the slopes are equal and so the temperatures of A and B are equal and that is the point of equilibrium.
3. In region 1, which system is hotter?
A) System A is hotter.
B) System B is hotter.
C) Impossible to tell.
Answer: Impossible to tell, or rather, the relative temperatures depend on exactly where in region 1 you are. At an particular energy U, the curve with the steeper slope has the lower temperature.
3. An ideal gas (system A) is in thermal contact with a heat bath (system B, B for bath), which maintains constant temperature. The gas slowly expands, isothermally.
System B = heat bath (a system with such a large heat capacity that its temperature remains constant when heat Q is added or removed). System A = ideal gas
What happens to Stot = SA + SB during this process?
A) Stot increases B) Stot decreases
C) Stot remains constant
Answer: Stot remains constant. For any quasi-static process, DS = Q/T. Here, the T’s and |Q|’s are the same for system A and B. The entropy gained by A is equal to the entropy lost by B.
In a quasi-static process, entropy enters or leaves a system along with heat. In non-quasi-static processes (like free expansion), new entropy can be created inside the system “on
the spot”.
3. The entropy at temperature T is related to the heat capacity by .In order for the integral to be finite, we must have ..
A) CV goes to some constant, non-zero value as T ®0.
B) CV ® 0 as T ® 0.
C) There are no constraints on possible values of CV as T ®0.
Answer: CV ® 0 as T ® 0. Otherwise the integral will diverge, since ln(0) = -¥.
3. A gas container is divided into two sections (A and B) with a movable partition.
For this particular system, . The system will evolve so that side _____ increases its volume. (Fill in the blank.)
A) side A B) side B C) impossible to tell
Answer: side A. Since dS/dV is greater for side A, that side will have a large increase in entropy if its volume is increased, while side B will have a small decrease in entropy when its volume is decreased. Large increase in SA , plus small decrease in SB = overall increase in Stot. The 2nd Law says: Isolated systems always evolve in such a way as to increase Stot
3- The thermodynamic identity is dU = T dS – p dV, which implies that U = U(S,V) (N is assumed fixed.)
What is the relationship between pressure p and energy U?
A) B) C) Neither of these is correct.
Answer: If we fix S, dS = 0 and dU = -p dV (fixed S). Therefore .
3-
Consider two points in the S,V plane (N fixed) very close to each other. Is DU = Ufinal - Uinitial the same for all three paths?
A) Yes, DU is path-independent
B) No, DU depends on the path.
Answer: DU is path-independent. U is a “function of state”. It’s value is uniquely defined by the macrostate of the system: U = U(S,V). Consequently, DU is determined entirely by the initial and final states.
3- A ball rolls back and forth in a valley. Eventually, the ball slows and stops. We never observe the reverse: A ball at rest at the bottom of a valley is never seen to start rolling back and forth (unless some force from outside the system acts)
Consider
I: conservation of energy(1st Law)
II: conservation of momentum
III: 2nd Law: Entropy of an isolated system always increases
The reverse process is never observed because this would violate
A) the 2nd Law only
B) I, II, and III
C) I and III only
D) I and II only
E) II and III only
Answer: A) The reverse process would violate the 2nd Law only. In the forward process, energy and momentum are conserved. So by “playing the movie backwards”, energy and momentum are still conserved. However, in the forward process, the entropy increased (mechanical energy mgh was converted to heat Q, so DS = Q/T). So in the reverse process, the entropy would decrease; entropy decrease is forbidden by the second law.
3-
Two systems, A and B, are in thermal contact. The systems have S vs. U curves shown and initially have the same energy UA,init = UB,init. Which system will gain energy as (A+B) evolves toward thermal equilibrium.
A) Neither, they are already in thermal equilibrium
B) System A will gain
C) System B will gain
Answer: System B will gain. Energy will flow in the direction which causes an increase in total entropy. Since , a transfer of energy from A to B (DUA negative and DUB positive) will result in a large increase in SB and a not-so-large decrease in SB. This will give an overall increase in Stot = SA + SB .
3.13 The thermodynamic identity: dU = T dS – p dV + m dN
At constant energy and volume, what is the relation between chemical potential and entropy?
A) B) C) something else
Answer:
Constant energy and volume implies dU = 0 and dV = 0.
So 0 = T dS + m dN (constant U, V)
3- The two halves of a sealed container are separated by a fixed semi-permeable membrane. There are two species of molecules in the container, C (cubes) and D(disks). The membrane is permeable to the disks only. Each half of this system has two different chemical potentials, one for disks and one for cubes: mD and mC.
Given the constraints imposed, does this system appear to be in equilibrium? (Hint: How do the mC’s on the right and left compare? What about the mD’s?)
A) Yes, it looks close to equilibrium
B) No, it is obviously way out of equilibrium, so the system will look different a short time later.
Answer: Yes, it looks close to equilibrium. Approximately equal numbers of disks are on the two sides, so the mC’s on right and left are approximately equal.
3- An Einstein solid has multiplicity and energy U = q e
W(N=30,q=40) = 2.37 ´ 1019
W(31,40) = 5.54 ´ 1019
W(31,39) = 3.16 ´ 1019
W(31,38) = 1.79 ´ 1019
W(31,37) = 1.00 ´ 1019
What is at N = 30, q = 40?
The value is
A) greater than +e B) between 0 and -e
C) between -e and -2e D) less than -2e
Answer: This one is tricky! The chemical potential is somewhere between -e and -2e.
If a single particle is added (N increases from 30 to 31, DN = 1), but the energy is kept fixed at q = 40, then entropy increases. But we have to keep the entropy fixed when computing chemical potential. The only way to keep the entropy fixed when we add a particle is to decrease U (DU negative). If we decrease q by 1 (DU = -e) , the entropy is still higher than initially. If we decrease q by 2 (DU = -2e), then entropy is lower than initially. So to keep S constant, we must decrease U by some amount between one and two energy units.
3- Two paramagnets (labeled A and B) are brought into thermal contact. Before contact, paramagnet A is at a very high (positive) temperature and B has a negative temperature. After the two reach equilibrium, the entropy of B has ..
A) increased B) decreased
C) might have increased, decreased, or remained constant depending on the initial temperature of A
Answer: The entropy of B will increase regardless of the initial temperature of A (so long as A has a positive temperature). As in the diagram below, if energy transfers from B to A, so UA increases, but UB decreases, then both SA and SB increase (it’s a win-win situation). On the other hand, if the reverse occurs, if energy transfers from A to B, then both SA and SB will decrease --- impossible by the 2nd Law. So heat will always transfer from B to A, from the negative temperature to the positive temperature; hence we conclude that negative temperatures are “hotter” than positive temperatures.