quiz samples for chapter 20 general physics i name:...

General Physics IQuiz Samples for Chapter 20

Entropy and the Second Law of ThermodynamicsJune 15, 2020

Name: Department: Student ID #:

Notice

� +2 (−1) points per correct (incorrect) answer.

� No penalty for an unanswered question.

� Fill the blank ( ) with � (8) if the statement iscorrect (incorrect).

� Textbook: Walker, Halliday, Resnick, Principlesof Physics, Tenth Edition, John Wiley & Sons(2014).

20-1 Entropy

1. (�) An irreversible process is one that cannotbe reversed by means of small changes in theenvironment.

2. (�) In a reversible process, the system is alwaysclose to equilibrium states.

3. (�) A slow (quasi-static) process is NOT reversibleif friction is present.

4. (�) Entropy S is a state property (or statefunction) of the system. It is independent of theway in which the system reached that state.

5. (�) Entropy PostulateIf an irreversible process occurs in a closed system,the entropy of the system always increases.

6. (�) For all reversible processes involving a systemand its environment, the total entropy of thesystem and its environment does not change.

7. (�) The change in entropy is zero for reversibleadiabatic processes.

8. (�) In any closed cycle, the change in entropy iszero.

9. (�) The entropy change ∆S,

∆S = Sf − Si,

for an irreversible process is the same as ∆S forany reversible process that has the common initial(i) and final (f) states.

∆S = Sf − Si =

∫ f

i

dQ

T.

10. (�) In an isothermal process,

∆S = Sf − Si =

∫ f

i

dQ

T=

Q

T.

11. (�) The unit of entropy is the same as that for theBoltzmann’s constant kB, joules per kelvin (J/K).

12. The first law of thermodynamics in differentialform is

dEinternal = dQ− dW.

(a) (�) dW and dEinternal can be expressed as

dW = pdV,

dEinternal = nCV dT.

(b) (�) The heat transfer to the system dQ,volume change dV , and temperature changedT are related as

dQ = pdV + nCV dT.

(c) (�) Ideal gas equation can be used to reducethe equation into

dQ

T= nR

dV

V+ nCV

dT

T.

(d) (�) The change in the entropy can becomputed as

∆S = nR lnVf

Vi+ nCV ln

Tf

Ti.

13. (�) The Second Law of ThermodynamicsIf a process occurs in a closed system, the entropyof the system increases for irreversible processesand remains constant for reversible processes. Itnever decreases,

∆S ≥ 0.

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14. (�) In a constant-volume process, the change inthe entropy is

∆S = nCV lnTf

Ti.

15. (�) In a constant-temperature process, the changein the entropy is

∆S = nR lnVf

Vi.

16. Consider a rubber band undergoing a smallincrease in length dx, as we stretch it between ourhands. We assume that the temperature isconstant. The change dE in the internal energy ofrubber is 0, if the total stretch of the rubber bandis not very much.

(a) (�) The force from the rubber band hasmagnitude F , is directed inward, and doeswork during length increase dx.

dW = −Fdx.

(b) (�) The first law of thermodynamics statesthat

dE = dQ− dW.

Because dQ = TdS and dW = −Fdx,

dE = TdS + Fdx.

(c) (�) Because the internal energy of the rubberband is invariant, dE = 0 and

F = −T dS

dx.

This tells that F is proportional to the rubberband’s entropy changes during a small changein the rubber band’s length.

17. (�) An ideal gas expands into a vacuum in a rigidvessel. As a result the entropy of the gas systemincreases.

18. (�) A hot object and a cold object are placed inthermal contact and the combination is isolated.They transfer energy until they reach a common

temperature. The change ∆SH in the entropy ofthe hot object, the change ∆SC in the entropy ofthe cold object, and the change ∆Stotal in theentropy of the combination are

∆SH < 0, ∆SC > 0, and ∆Stotal > 0.

20-2 Entropy in the Real World: Engines

1. For an engine that is a device operating in a cycle,

(a) (�) Input energy is the heat |QH| from ahigh-temperature reservoir.

(b) (�) Output energy is the work |W |.(c) (�) The efficiency of any engine is defined as

ε =output energy

input energy=|W ||QH|

.

2. In an ideal engine, all processes are reversible andthere is no wasteful energy transfer. A Carnotengine is one of the ideal engines.

(a) (�) At TH the engine absorbs heatQH = |QH| > 0 to make an isothermalexpansion: Va → Vb.

(b) (�) The gas makes an adiabatic expansion(without heat exchange):

Vb(TH) → Vc(TL).

(c) (�) At TL the engine loses heat(QL = −|QL| < 0) to make an isothermalcompression: Vc → Vd.




(d) (�) The gas makes an adiabatic compression(without heat exchange):

Vd(TL) → Va(TH).

(e) (�) Conservation of energy states that

W = |QH| − |QL|.

The work per cyclic is the area of the closedpath a→ b→ c→ d→ a on the p-V plot.

(f) (�) Reversible energy transfers as heat occursonly in the two isothermal processes a→ band c→ d. The other two processes b→ c andd→ a are adiabatic and, therefore, Q = 0 andthe entropy is invariant.

The net entropy change is

∆S = ∆SH + ∆SL

=|QH|TH− |QL|

TL.

(g) (�) S is a state function and, therefore,

∆S =|QH|TH− |QL|

TL= 0

after a cycle a→ b→ c→ d→ a. Thus

|QH||QL|

=TH

TL.

(h) (�) The efficiency of the Carnot engine is

ε =|W ||QH|

=|QH| − |QL||QH|

= 1− |QL||QH|

= 1− TL

TH.

(i) (�) The efficiency of the Carnot engineapproaches 100% as TH →∞.

3. (�) There are no perfect engines.No series of processes is possible whose sole resultis the transfer of energy as heat from a thermalreservoir and the complete conversion of thisenergy to work.

4. Figure shows the operating cycle of an idealStirling engine.

(a) (�) At TH the engine absorbs heatQH = |QH| > 0 to make an isothermalexpansion: Va → Vb. The engine does work,

Wa→b =

∫ Vb

Va

pdV

= nRTH

∫ Vb

Va

dV

V

= nRTH lnVb

Va.




Because T = TH, fixed, the internal energy ofthe gas is invariant. The heat absorbed by thegas is the same as the work,

Qa→b = Wa→b = nRTH lnVb

Va.

(b) (�) The gas is cooled down to emit the heatQ during a constant volume process Vb = Vc,

TH(Vb) → TL(Vc = Vb).

The engine does not work Wc→d = 0 becausedV = 0.

Qb→c = −nCV (TH − TL) < 0.

(c) (�) At TL the engine loses heat(QL = −|QL| < 0) to make an isothermalcompression: Vc(= Vb)→ Vd(= Va). Theengine does work

Wc→d =

∫ Va

Vb

pdV

= nRTL

∫ Va

Vb

dV

V

= −nRTL lnVb

Va.

Because T = TL, fixed, the internal energy ofthe gas is invariant. The heat absorbed by thegas is the same as the work,

Qc→d = Wc→d = −nRTL lnVb

Va.

Note that the negative sign represents the gasactually lost the heat.

(d) (�) The gas is heated up to absorb the heat Qduring a constant volume process Vd = Va,

TL(Vd) → TH(Va = Vd).

The engine does not work Wd→a = 0 becausedV = 0.

Qd→a = nCV (TH − TL) > 0.

(e) (�) The net work done by the engine during acycle is

W = Wa→b→c→d→a = nR(TH − TL) lnVb

Va.

(f) (�) The input heat absorbed by the engineduring a cycle is

Qin = Qa→b + Qd→a

= Wa→b + Q

= nRTH lnVb

Va+ nCV (TH − TL).

(g) (�) The efficiency of the Stirling engine is

εStirling =W

Qin

=nR(TH − TL) ln Vb

Va

nRTH ln VbVa

+ nCV (TH − TL)

=1− TL

TH

1 + CV

R lnVbVa

(1− TL

TH

)< 1− TL

TH= εCarnot.

20-3 Refrigerators and Real Engines

1. A refrigerator uses work W to absorbs heatQL = |QL| > 0 at lower temperature TL and emittsheat QH = −|QH| at higher temperature TH.

(a) (�) The coefficient of performance for anyrefrigerator is

K =absorbed heat at TL

input work

=|QL||W |

.

(b) (�) A Carnot refrigerator is a Carnot engineoperating in reverse.

(c) (�) For a Carnot refrigerator, the first law ofthermodynamics gives

|W | = |QH| − |QL|.

Here, |W | is not the work done by therefrigerator but the work applied to therefrigerator.




(d) (�) The coefficient of performance for aCarnot refrigerator is

KCarnot =|QL|

|QH| − |QL|=

1|QH||QL| − 1

.

(e) (�) For a Carnot refrigerator, heat transferoccurs at TH and TL and the entropy changeafter a cycle is

∆S = −|QH|TH

+|QL|TL

.

Note that the engine accepts heat at TL andemitts heat at TH which is in reverse incomparison with the Carnot engine.

(f) (�) After a cycle, the system returns to theinitial state. Thus the sate function S must beinvariant,

−|QH|TH

+|QL|TL

= 0.

Like the Carnot engine,

|QL||QH|

=TL

TH.

(g) (�) The coefficient of performance for aCarnot refrigerator is

KCarnot =TL

TH − TL.

20-4 A Statistical View of Entropy

1. (�) In 1877, Austrian physicist Ludwig Boltzmann(the Boltzmann of Boltzmann’s constant kB)derived a relationship between the entropy S of aconfiguration of a gas and the multiplicity W ofthat configuration. That relationship is

S = kB lnW,

where kB is the Boltzmann’s constant.

kB = 1.380 649× 10−23J/K.

2. (�) Stirling’s approximation for ln N !:

lnN ! ≈ N(lnN)−N.

3. Consider a system of N molecules that may bedistributed between the two halves of a box. Letn1 be the number of molecules in one half of thebox and n2 = N − n1 be that of the other box.The ordered pair (n1, n2) represents thatconfiguration of the state.

Initial state

Final state

(a) (�) The multiplicity for the state (n1, n2) isgiven by

W =N !

n1!n2!.

(b) (�) For any N , the multiplicity for the state(0, N) and (N, 0) is given by

W =N !

N !0!= 1.

The corresponding entropy is

S(0,N) = S(N,0) = 0.

(c) (�) The multiplicity for the state (12N, 12N) isgiven by

W =N !

(N2 !)(N2 !).

The corresponding entropy is

S(12N,

12N)

= kB[lnN !− 2 ln(12N !)

].




(d) (�) Applying Stirling’s approximation, wefind that

lnN ! ≈ N(lnN)−N,

2 ln(N2 !) ≈ N(lnN

2)−N.

(e) (�) If there are n moles of gas molecules, thenN = nNA and

S(12N,

12N)≈ kBN ln 2 = nR ln 2.

(f) (�) The free expansion from (N, 0) to(12N, 12N) leads to the entropy change,

∆S = S(12N,

12N)− S(N,0) = nR ln 2.

(g) (�) We recall that the entropy change is

∆S = nR lnVf

Vi+ nCV ln

Tf

Ti.

For a free expansion Vi = V to Vf = 2V ,Tf = Ti. Therefore,

∆S = nR ln 2.

Thus the microscopic approach is consistentwith the macroscopic computation of theentropy!


quiz samples for chapter 20 general physics i name:...

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