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EXAM 3 CHEM 230 TEXTBOOK 7TH EDITION SOLUTIONS MANUAL CH 5...

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SVNAS 8th Edition Annotated Solutions

Chapter 5

Shown at the bottom is a PV diagram with two adiabatic lines 1 → 2 and 2 → 3, assumed to intersect at point 2. A cycle is formed by an isothermal line from 3 → 1. An engine traversing this cycle would produce work. For the cycle U = 0, and therefore by the first law, Q + W = 0. Since W is negative, Q must be positive, indicating that heat is absorbed by the system. The net result is therefore a complete conversion of heat taken in by a cyclic process into work, in violation of Statement 1a of the second law (Pg. 160). The assumption of intersecting adiabatic lines is therefore false.

W   QH

 1 

TC TH

 T W 1  C TH 

  323 K  -1 -1  QH   1   250 kJ s  148.8 kJ s 798 K   

QC  QH  W  250 kJ s-1 148.8 kJ s-1 101.2 kJ s -1

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 1 

Chapter 5

300 K TC 1  0.6 750 K TH

thermal  1 

QC QH

 1

TC 303  1  0.514 TH 623 



thermal  0.55 1  

 303  TC    0.55 1    0.35 TH  TH  

The energy balance for the over-all process is written: Q = Ut + EK + EP Assuming the egg is not scrambled in the process, its internal-energy change after it returns to its initial temperature is zero. So too is its change in kinetic energy. The potential-energy change, however, is negative, and by the preceding equation, so is Q. Thus heat is transferred to the surroundings. The total entropy change of the process is: Stotal = St + Stsurr Just as Ut for the egg is zero, so is t The

Since Q is negative, Stotal is positive, and the process is irreversible.

By Eq. (5.9) the thermal efficiency of a Carnot engine is: Differentiate: Since TC/TH is less unity, the efficiency changes more rapidly with TC than with TH. So in theory it is Updated 4/5/2017

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SVNAS 8th Edition Annotated Solutions

Chapter 5

more effective to decrease TC. In practice, however, TC is fixed by the environment, and is not subject to control. The practical way to increase η is to increase TH. Of course, there are limits to this too.



 1 

113.7 K TC 1   0.625 303.15 K TH



 









  



 

  Updated 4/5/2017

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SVNAS 8th Edition Annotated Solutions

Chapter 5

 

  

S  C p ln

 P T  R ln   T0  P0 

T2



Q  U  n Cv dT t

T1

Q  U t  nCv T

T 

15000 J Q   500 K nCv 1.443 mol  2.5  8.31451J mol -1 K -1

S  C p ln





 P 7 T 5 -1 -1  R ln    R ln 2  R ln 2  R ln 2  1.733R  14.4 J mol K T0 P 2  0 2 



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SVNAS 8th Edition Annotated Solutions









Chapter 5

 







 







 

  



 

  

For an ideal gas with constant heat capacities, and for the changes T1 → T2 and P1 → P2, Eq. (5.14) can be rewritten as:

V2 = V1, then Whence,

Since CP > CV ,

ra es that SP > SV . V2 = V1, then

Whence,

This demonstrates that the signs for ST and SV are opposite. Updated 4/5/2017

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SVNAS 8th Edition Annotated Solutions

Chapter 5

Start with the equation just preceding Eq. (5.10):

For an ideal gas PV = RT, and lnP

+ l

.

********************** As an additional part of the problem, one could ask for the following proof, valid for constant heat capacities. Return to the original equation and substitute dT/T = dP/P + dV/V:

As indicated in the problem statement the dif ntial a are: − ( A) − ( B) where QC and QH refer to the reservoirs.

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(a) With

Chapter 5

and

Whence, d ln TC = − d ln TH

, Eq. (B) becomes:

where

≡

Integration from TH0 and TC0 to TH and TC yields:

With

and

e

Eliminate TC by the boxed equation of

rt (a) and rearrange sl

For infinite time, TH =TC ≡T, and the answe of

Because /(

Updated 4/5/2017

+

− 1 = 1/(

:

t (a) beco es:

+ 1), then:

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SVNAS 8th Edition Annotated Solutions

Because TH =T , substitution of thes quan ties in the answer of

Chapter 5

rt (b)

elds:

As indicated in the problem statement the dif ntial a are: − ( A) − ( B) where QC and QH refer to the reservoirs. (a) With

, Eq. (B)

Substitute for dQH and dQC in Eq. (A):

Integrate from TC0 to TC:

For infinite time, TC = TH, and the boxed equation bove becomes:

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SVNAS 8th Edition Annotated Solutions

Chapter 5

Write Eqs. (5.9) in rate form and combine to eliminate |

|

H|=

H|:

this becomes:

Differentiate, noting that the quantity in square brackets is constant:

Equating this equation to zero, leads immediately to: 4r = 3 or r = 0.75 Hence

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SVNAS 8th Edition Annotated Solutions

thermal  1 

QC QH

 1

Chapter 5

TC 300  1  0.5 TH 600



 





 



QH T 300 W  1  H  1  1  0.2 250 QC QC TC



S  C p ln

P  T2  R ln  2  T1  P1 

S  C p ln

P  T2  R ln  2  T1  P1 

 7 T P S  R  ln 2  ln  2  2 T1  P1 

 -1 -1   8.314 3.5ln1.5  ln 5  1.58 J mol K 

 7 T P S  R  ln 2  ln  2  P1  2 T1

 -1 -1   8.314 3.5ln1.667  ln 5  1.48 J mol K 





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SVNAS 8th Edition Annotated Solutions

5 T P S  R  ln 2  ln  2  2 T1  P1 

Chapter 5

 -1 -1   8.314 2.5ln 0.667  ln 0.2  4.96 J mol K 



S  C p ln

P  T2  R ln  2  T1  P1 

 9 T  P  -1 -1 S  R  ln 2  ln  2    8.314  4.5 ln 0.75  ln 0.2  2.62 J mol K  P1    2 T1



S  C p ln

P  T2  R ln  2  T1  P1 

  P T S  R  3ln 2  ln  2 T  P1 1 

 -1 -1   8.314 4 ln 0.6  ln 0.2   3.61 J mol K 

This cycle is shown below. Assuming constant specific heats, the efficiency is given by:

Temperature T4 is not given and must be calculated. The following equations are used to derive and expression for T4.

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SVNAS 8th Edition Annotated Solutions

Chapter 5

For adiabatic steps 1 to 2 and 3 to 4:

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SVNAS 8th Edition Annotated Solutions

Chapter 5

Because W =0, Eq. (2.3) here becomes: Q=

Ut = m CV T

A necessary condition for T to be zero when Q is non-zero is that m =∞. This is the reason that natural bodies (air and water) that serve as heat reservoirs must be massive (oceans) or continually

renewed (rivers).



T  P  S ad , rev  C p ln  rev   R ln  2   0  T1   P1  T  P C p ln  rev   R ln  2   T1   P1  Trev  P2    T1  P1 

R

Cp

P  Trev  T1  2   P1 



R

Cp



2

7   298.15    2

7

 426.46 K

 

T  P S  C p ln  2  R ln  2  T1   P1  Updated 4/5/2017

  471.4  7    7 / 2 * R ln   R ln   0.3506R  298.15  2 

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SVNAS 8th Edition Annotated Solutions

T S ad , rev  C p ln  rev  T1

Chapter 5

  P2    R ln    0   P1  

 T S  Sad , rev  S  C p ln  2  T1  T S  C p ln  2  T1

 T rev  C p ln    T1

  P2    Trev   R ln    C p ln  P   1    T1

  P2    R ln     P1 

  T2    C p ln    Trev 

An appropriate energy balance here is: Q = Ht = 0 Applied to the process described, with T as the final temperature, this becomes:

If m1 = m2 = m,

.

Since

St is positive

Isentropic processes are not necessarily reversible and adiabatic. The term isentropic denotes a process for which the system does not change in entropy. There are two causes for entropy changes in a system: The process may be internally irreversible, causing the entropy to increase; heat may be transferred between system and surroundings, causing the entropy of the system to increase or decrease. For processes that are internally irreversible, it is possible for heat to be transferred out of the system in an amount such that the entropy changes from the two causes exactly compensate each other. One can imagine irreversible processes for which the state of the system is the same at the end as at the Updated 4/5/2017

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SVNAS 8th Edition Annotated Solutions

Chapter 5

beginning of the process. The process is then necessarily isentropic, but neither reversible nor adiabatic. More generally, the system conditions may change in such a way that entropy changes resulting from temperature and pressure changes compensate each other. Such a process is isentropic, but not necessarily reversible. Expansion of gas in a piston/cylinder arrangement is a case in point. It may be reversible and adiabatic, and hence isentropic. But the same change of state may be irreversible with heat transfer to the surroundings. The process is still isentropic, but neither reversible nor adiabatic. An isentropic process must be either reversible and adiabatic or irreversible and non-adiabatic.

By definition, By inspection, one sees that for both T > T0 and T0 > T the numerators and denominators of the above fractions have the same sign. Thus, for both cases H is positive. Similarly,

By inspection, one sees that for both T > T0 and T0 > T the numerators and denominators of the above fractions have the same sign. Thus, for both cases S is positive. When T = T0, both the numerators and denominators of the above fractions become zero, and the fractions are indeterminate. Application of l’Hˆopital’s rule leads to the result: H = S = CP.

Step 1-2: Volume decreases at constant P. Heat flows out of the system. Work is done on the system.

Step 2-3: Isothermal compression. Work is done on the system. Heat flows out of the system.

Step 3-1: Expansion process that produces work. Heat flows into the system. Since the PT product is constant,

Combine with (A) this becomes Updated 4/5/2017

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SVNAS 8th Edition Annotated Solutions

Chapter 5

Moreover

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T



S  C p T0

Chapter 5

 P dT  R ln   T  P0 

T

 P S D A    B  CT  3  dT  ln   R T T   P0 



T0

 T  P S C 2 D 2  Aln    B T  T0   T  T02  T  T0 2  ln   R 2 2  T0   P0 



T1 (K) 473.15

T2 (K) 1373.15 T2

ICPS 



T1

T1 (K) 523.15

ICPS 

T dT  Aln  2 RT  T1

Cp

T2 (K) 1473.15 T2

A 5.699







D (K2) C (1/K2) B (1/K) 8.01E-04 0.00E+00 -1.02E+00

 C 2 D 2 T2  T1 2  T2  T1 2   B  T2  T1   2 2 

A 1.213

 T2 





T1

Updated 4/5/2017

1



C (1/K2) D (K2) B (1/K) 2.88E-02 -8.82E-06 0.00E+00

 RT dT  Aln T   B T  T   2 T Cp

ICPS 6.793

C

2

1

2 2 2  T1

  D2  T



ICPS 20.234

2 2 2  T1

S (J/(mol K)) 56.48

S (J/(mol K)) 168.23



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SVNAS 8th Edition Annotated Solutions

Chapter 5

T



S  C p

dT T

T0

T2



t

Q  H  n C pdT T1

800000 J Q   9622 K= nR 10 mol  8.314 J mol -1 K -1

T2



Cp R

dT

473 K

T1 (K) 473.15

T2 (K) 1374.5 T2

ICPH 

2

2

D (K ) C (1/K ) B (1/K) 1.44E-02 -4.39E-06 0.00E+00

A 1.424

ICPH (K) 9.622E+03

 1

 R dT  A T  T   2 T  T  3 T  T  D T Cp

B

2

2 2

1

2 1

C

3 2

3 1

2

T1

S  R

1374.5 K





1  T1 

C p dT R T

473.15 K

T1 (K) 473.15

T2 (K) 1374.5 T2

ICPS 

A 1.424

 T2 

2

2

C (1/K ) D (K ) B (1/K) 1.44E-02 -4.39E-06 0.00E+00

 RT dT  Aln T   B T  T   2  T  T  2  T Cp

T1

Updated 4/5/2017

1

C

2

1

2 2

2 1

D

ICPS 10.835

2 2 2  T1



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SVNAS 8th Edition Annotated Solutions

S  R

1374.5 K



Chapter 5

Cp dT 10.835 R T

473.15 K



T2



2500000 J Q   20045 K= nR 15 mol  8.314 J mol -1 K -1

Cp R

dT

533.15 K

T1 (K) 533.15

T2 (K) 1413.7 T2

ICPH 

Cp

R

A 1.967

dT  A T2  T1 

T1

T1 (K) 533.15

T2 (K) 1413.7 T2

ICPS 

A 1.967

 T2 

2

2

D (K ) C (1/K ) B (1/K) 3.16E-02 -9.87E-06 0.00E+00

ICPH (K) 2.004E+04

 1 1 B 2 C 3 T2  T12  T2  T13  D   2 3  T2 T1 









2

2

C (1/K ) D (K ) B (1/K) 3.16E-02 -9.87E-06 0.00E+00

ICPS 21.307

 RT dT  Aln T   B T  T   2  T  T  2  T Cp

1

2 2

2 1

D

2 2 2  T1

1

T1

S  R

C

2

1413.7 K





C p dT  21.307 R T

533.15 K



T2

Q 1000000 Btu  nR 40 lbmol 1.986 Btu lbmol-1

 12588

993 K =



Cp R

dT

533.15 K

 Updated 4/5/2017

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T1 (K) 533.15

T2 (K) 1202...