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Guide for the second edition sixth chapter in Chemical thermodynamics....

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Chapter 6 Solutions Engineering and Chemical Thermodynamics 2e

Milo Koretsky Wyatt Tenhaeff School of Chemical, Biological, and Environmental Engineering Oregon State University [email protected]

6.1 At atmospheric pressure water boils at 100 oC – this is the state of the system when we put the lid on. As the water boils, the pressure will increase as more vapor forms. If the evaporation rate is constant, the number of moles entering the vapor is constant and the pressure almost increases linearly with time (some effect from increase in T on P as well). We are essentially moving along the liquid/vapor equilibrium line as water continues to boil, but at a higher and higher pressure. The relation between T and P can be modeled by the Clausius – Clapeyron equation:

ln

Δ hvap ⎡ 1 1 ⎤ P2sat = − ⎢ − ⎥ P1sat R ⎣ T2 T1 ⎦

T increases as the ln of P so less than linear. A sketch is shown below:

2

6.2 at less than 300K. Equilibrium is given when the molar Gibbs energy of the liquid and vapor are equal:

g vA = g lA Substituting in the definition of Gibbs energy

hvA − TsvA = hlA − TslA With attractive interactions in the vapor phase hvi would decrease relative to an ideal gas. That would cause the Gibbs energy in the vapor to be less than that of the liquid and all the liquid would evaporate. To return to phase equilibrium, the temperature needs to be lowered (to lessen the effect of entropy). An alternative approach is to look at the Clapeyron equation and make an argument about vv.

3

6.3 greater than 10 bar. Equilibrium is given when the molar Gibbs energy of the liquid and vapor are equal:

gAv = g lA Substituting in the definition of Gibbs energy

hvA − TsvA = hlA − TslA With attractive interactions in the vapor phase hvi would decrease relative to an ideal gas. That would cause the Gibbs energy in the vapor to be less than that of the liquid and all the liquid would evaporate. To return to phase equilibrium, the pressure needs to be raised (to lessen the effect of entropy). An alternative approach is to look at the Clapeyron equation and make an argument about vv.

4

6.4 (a) The quality is 0.1 so the system has 10% by mass vapor and 90% liquid. From the steam tables: vˆv = 0.039441

m3 m3 and vˆl = 0.001286 kg kg

So the ratio is V l 0.9 × vˆl = = 0.003 V v 0.1× vˆ v

Conclusion: eventhough the system has much more mass in the liquid – the volume is mostly vapor. (b) Before the valve is opened, the temperature is at the boiling point for steam at 5 MPa, which is 264 oC. If the valve is opened to the atmosphere, the steam will leave the system decreasing the pressure. This will lower the boiling point. Ultimately the system will reach equilibrium at 1 atm and 100 oC. If the heat transfer is really fast, it will follow the PT coexistence curve. If the heat transfer is really slow, the temperature change may be negligible. In reality, it is probably between these two extremes but closer to the slow heat transfer case. 6

5

Pressure [Mpa]

4

Equilibrium From Steam Tables Fast heat transfer

3

Slow heat transfer

2

1

0 100

150

200 Temperature [C]

5

250

300

6.5 Starting with the Clapeyron equation Δh dP = dT Δ vT Using the mathematical relation d ln P d ln P dP 1 dP = = dT dP dT P dT

We can get dP d ln P Δh =P = dT dT Δ vT Using two of the three assumptions for the Clausisus-Clapeyron equation (that the molar volume of solid or liquid is insignificant compared to gas and that the gas can be assumed to be ideal) we get the following: Δh d ln P = ⎛ RT ⎞ dT ⎜ P ⎟T ⎝ ⎠ which simplifies and rearranges to d ln P Δh = dT RT 2 P

Finally Δh =

d ln P RT 2 dT

and 15800 − 0.76 ln T + 19.25) T RT 2 = (15800 − 0.76 T ) R Δ h1 = dT 15300 d (− − 1.26 ln T + 21.79) T Δ h2 = RT 2 = (15300 − 1.26T ) R dT d(−

Since vapors of a species have higher enthalpy than their liquids, and their liquids have higher enthalpy than their solids, then Δhs ub should be larger than Δhvap . The first one starts out higher and decreases slower with temperature and will always be higher than the second one. Therefore the first is sublimation and the second is evaporation.

6

6.6 (a)

For a single component system:

µ = Gi = g i = g From the fundamental property relation given by Equation 5.9: dg = − sdT + vdP

We can identify a phase transition from the vertical line of the g vs. T plot, as indicated below. Since this transition is vertical, i.e., the temperature is constant, the pressure must also be constant. Thus, we can differentiate the Gibbs energy with respect to temperature at constant pressure to get:

⎛ ∂g ⎞ ⎜ ⎟ = −s ⎝∂ T ⎠P Hence the slope of a plot of g (or ฀) vs. T at any temperature must be the negative of the value of entropy on the plot for s vs. T. The resulting curve is sketched below.

β

s

phase transition; vertical line indicates P = const

α

T β µ

α Thick line denotes lowest value of µ

α Slope = the negative value of s at the same T on the curve above

β T*

T

7

(b) For a single component system the fundamental property relation, Equation 5.9, gives: dg = − sdT + vdP

We can identify a phase transition from the vertical line of the g vs. P plot, as indicated below. Since this transition is vertical, i.e., the pressure is constant, the temperature must also be constant. Thus, we can differentiate the Gibbs energy with respect to pressure at constant temperature to get:

⎛ ∂g ⎞ ⎜ ⎟ =v ⎝∂ P ⎠T Hence the slope of a plot of g vs. P must have a slope that matches the plot for v vs. T. Since the molar volume of phase α is about twice the value of phase β, its slope should be twice as big. The resulting curve is sketched below.

v

phase transition; vertical line indicates T = const

α

vα vβ

β P

g

Straight line since v is constant

α Slope of top line is β

about twice as big as the slope of bottom line

β Thick line denotes lowest value of g

α P*

P

8

6.7

The ferrite phase has stronger bonds. At room temperature, iron is in the ferrite phase. The heating to 912 ºC has the effect of increasing the entropy contribution to the Gibbs energy. At a high enough temperature, the austenite phase becomes stable, so that its entropy must be greater than the ferrite phase. If the entropy of the austenite phase is greater, the enthalpy of the ferrite phase must be greater or else the austenite phase would be stable over the entire temperature range. Hence, the ferrite phase has stronger bonds.

9

6.8 (a) The freezing point occurs where there is a discontinuity in the g vs. T plot, as indicated below. The liquid is at a temperature higher than the freezing point and the solid at lower temperature. These are demarked below. The melting temperature is 250 K, which occurs at a value g = 3,000 [J/mol]

Pure species "a" 6000

Gibb's Energy, g (J/mol)

1 unit up New Freezing point

4,000 solid Freezing point

2,000

liquid

100

200

1.2 units up

300

Temperature, T (K) (b) At constant pressure, the entropy can be found from Equation 5.14. For the solid we have:

Δ g 1,000 ⎛ ∂g ⎞ ⎡ J ⎤ s = −⎜ = = 10 ⎢ ⎟ =− ∂ Δ T 10 ⎝ T ⎠P ⎣ mol K ⎦⎥ And for the liquid, we get:

Δ g 2,000 ⎛ ∂g ⎞ ⎡ J ⎤ s = −⎜ = = 40 ⎢ ⎟ =− ΔT 50 ⎝ ∂T ⎠ P ⎣ mol K ⎥⎦

10

(c) As we change pressure, we can see how the Gibbs energy changes at any given temperature by Equation 5.14:

⎛ ∂g ⎞ ⎜ ⎟ =v ⎝∂ P ⎠T Assuming the molar volumes of the liquid and vapor stay constant over the temperature range around the melting point, we see that the Gibbs energy of the liquid increases by 1.2 times the Gibbs energy of the solid, since the molar volume of the liquid is 20% larger. The Gibbs energy of the new freezing point at higher pressure is schematically drawn on the plot above. For convenience, we choose the solid to increase by 1 unit on the plot. Thus, the liquid increases by 1.2 units. As the sketch shows, the freezing point, where the two lines intersect, will shift to higher temperature.

11

6.9

Along the coexistence line, the Gibbs energy of solid must equal that of liquid – which leads to the Clapeyron equation:

dP hl − hs = l i si dT vi − vi T

(

)

Since solids are stabilized by bonds, hil > his and the numerator is always positive. In the denominator T is also always positive. Therefore the slope of the coexistence line corresponds to the relative magnitude of the molar volumes of the liquid. If the liquid volume is larger than the solid volume, all terms are positive and the slope is positive. Conversely, species that expand upon freezing the term is negative and the slope of the coexistence line is negative.

12

6.10 (a) The like interactions are stronger – when you replace A-A and B-B with A-B interactions the enthalpy goes up, suggesting the A-B is less strong (b) Increase. If they mix adiabatically, all the enthalpy stays in the system. Energy is released as the stronger pure species interactions are replaced by weaker unlike interactions. This energy goes into increasing the kinetic energy (speed) of the molecules, and the temperature goes up. (c) Negative. For the final temperature to return to the initial temperature, energy must leave the system via heat.

13

6.11

1 atm

Tm

1 atm

T=?

Liquid

Liquid l

l 1

n , n , n3l , n4l

n1

l 2

Solid

Solid s 1

s 1

n

n

Answer: T < Tm. For the case on the left:

g1s = g1l or

h1s − Ts1s = h1l − Ts1l Species 1 in the liquid mixture (right) has a higher entropy than pure 1 in the liquid (left). Thus, at Tm the equivalent term on the right will be more negative, and the liquid will have a lower molar Gibbs energy than the solid. This will cause all the solid to melt. To return to phase equilibrium, we must lower the temperature.

14

6.12 T < Tb. For the case of pure species: l

v

g1 = g1 or l

l

v

v

h1 − Ts1 = h1 − Ts1 Species 1 in the vapor mixture has a higher entropy than pure 1 in the vapor. Thus, at Tb the equivalent term on the right will be more negative, and the vapor will have a lower molar Gibbs energy than the liquid. This will cause all the liquid to evaporate. To return to phase equilibrium, we must lower the temperature.

15

6.13 T < Tm. For the case of pure species:

g1s = g1l or s

s

l

l

h1 − Ts1 = h1 − Ts1 Species 1 in a liquid of species 2 has a higher entropy than pure 1 in the liquid. Thus, at Tm the equivalent term on the right will be more negative, and the liquid will have a lower molar Gibbs energy than the solid. This will cause all the solid to melt. To return to phase equilibrium, we must lower the temperature.

16

6.14 The entropy of the liquid increases – so the melting temperature must decrease.

17

6.15 1 atm

1 atm

300 K

300 K Liquid

Liquid l 1

l 1

n , n , n3l , n4l

n

System I

l 2

System II

(a) H 1I = H1II If 1-1 and 1-2, 1-3, and 1-4 interactions are the same, there is no difference in species 1’s contribution to the energy of the mixture vs. its pure species behavior. I II (b) S1 < S1 More configurations in System II

(c) G1I > G1II From the definition of Gibbs energy and answers to part (a) and (b) above I II (d) g 1 = g 1 gi just depends on T and P

(e) H 1I = h1 < H1II Stronger interactions leads to lower partial molar enthalpy.

18

6.16

The following can be shown with the Gibbs-Duhem equation 0 = x1V1 + x2V2 Differentiation with respect to x1:

0 = x1

dV1 dV + x2 2 dx1 dx1

If the partial molar volume of species 1 is constant, the Gibbs-Duhem equation simplifies to

0=

dV2 dx1

Therefore, the partial molar volume of species 2 is also constant. Note that in this case, since the partial molar volume of species 1 is constant:

V1 = v1 and similarly for species 2: `

V2 = v2

Hence, the molar volume can be written:

v = x1V1 + x2V2 = x1v1 + x2v2 This is known as Amagat’s law.

19

6.17 (a)

The Clausius-Clapeyron equation: dPisat Pisat

vap

=

Δhi

dT

RT 2

vap ⎛ P sat ⎞ ⎡1 Δh 1 ⎤ i ln⎜ ⎟=− i ⎢ − R ⎣ T 373 [K]⎥⎦ ⎝ 101 [kPa]⎠

or

so Pi

sat

⎧ Δh vap ⎡ 1 1 ⎤⎫ = (101 [kPa])exp ⎨− i ⎢ − ⎬ R ⎣ T 373 [K] ⎥⎦⎭ ⎩

(b) and (c)

Using

⎡ kJ ⎤ Δh vap = 40.626 ⎢ ⎣mol ⎥⎦ we obtain the following table

T [K]

Eqn 6.24 [kPa]

Steam Tables [kPa]

% Difference

273.156 278.15 283.15 288.15 293.15 298.15 303.15 308.15 313.15 318.15 323.15 328.15 333.15 338.15 343.15 348.15 353.15 358.15 363.15 368.15 373.15

0.84 1.16 1.58 2.13 2.84 3.76 4.93 6.40 8.24 10.54 13.36 16.82 21.04 26.13 32.26 39.58 48.28 58.56 70.67 84.84 101.35

0.6113 0.87 1.23 1.71 2.34 3.17 4.25 5.63 7.38 9.59 12.35 15.76 19.94 25.03 31.19 38.58 47.39 57.84 70.14 84.55 101.35

37.30% 33.02% 28.30% 24.51% 21.51% 18.62% 15.94% 13.68% 11.71% 9.86% 8.19% 6.75% 5.50% 4.40% 3.42% 2.58% 1.87% 1.25% 0.75% 0.34% 0.00%

20

The logarithmic trend is well-represented. However, at lower temperatures the Clausius⎡ kJ ⎤ Clapeyron equation is up to 37% off. The actual heat of vaporization changes from 2501.3 ⎢ ⎥ ⎣ kg ⎦

⎡ kJ ⎤ at 0.01 oC to 2257.0 ⎢ ⎥ at 100 oC, a difference of around 10%. ⎣ kg ⎦

(d)

For 100 ºC to 200 ºC, we obtain the following table:

T [K]

Eqn 6.24 [kPa]

Steam Tables [kPa]

% Difference

373.15 378.15 383.15 388.15 393.15 398.15 403.15 408.15 413.15 418.15 423.15 428.15 433.15 438.15 443.15

101.35 120.51 142.64 168.11 197.30 230.63 268.55 311.54 360.11 414.82 476.24 545.00 621.74 707.16 801.99

101.35 120.82 143.28 169.06 198.53 232.1 270.1 313 361.3 415.5 475.9 543.1 617.8 700.5 791.7

0.00% 0.26% 0.44% 0.56% 0.62% 0.63% 0.57% 0.47% 0.33% 0.16% 0.07% 0.35% 0.64% 0.95% 1.30%

21

448.15 453.15 458.15 463.15 468.15 473.15

906.98 1022.93 1150.68 1291.10 1445.10 1613.62

892 1002.2 1122.7 1254.4 1397.8 1553.8

1.68% 2.07% 2.49% 2.93% 3.38% 3.85%

Over this range the Clausius-Clapeyron equation represents the data well and is no more than 4 % off. The actual heat of vaporization changes from 2257.0 [kJ/kg] at 100 oC to 1940.7 [kJ/kg] at 200 oC, a difference of around 15%. (e)

The heat of vaporization can be corrected for temperature as follows Tb

Δhvap (T ) =

∫ T

T

clP dT

+ Δh vap(T b ) + ∫ c vPdT Tb

We can acquire heat capacity data from Appendix A.2, but to simplify the analysis, we will use an average heat capacity for the vapor.

Δhvap (T ) = 75.4(373.15 − T ) + 40626 + 34.13(T − 373.15) Δhvap (T ) = 56026 − 41.27T

22

Substitute this expression into the Clausius-Clapeyron equation

dPi sat Pisat

=

(56026 − 41.27 T) dT RT 2

Integrate: ⎡1 ⎡ ⎛1 ⎛ T ⎞⎤⎤ 1 ⎞ Pi sat = (101.35 kPa ) exp ⎢ ⎢− 56026⎜ − ⎟ ⎥⎥ ⎟ − 41.27 ln ⎜ ⎝ T 373.15 ⎠ ⎝ 373.15 ⎠ ⎦ ⎦ ⎣R ⎣

Now plot the data as before from 0.01 ºC to 200 ºC.

T [K]

Eqn 6.24 [kPa]

Steam Tables [kPa]

% Difference

273.16 278.15 283.15 288.15 293.15 298.15 303.15 308.15 313.15 318.15 323.15 328.15 333.15 338.15 343.15 348.15 353.15 358.15 363.15 368.15 373.15 378.15 383.15 388.15 393.15 398.15 403.15 408.15 413.15 418.15 423.15 428.15

0.64 0.91 1.28 1.78 2.43 3.29 4.39 5.81 7.60 9.86 12.66 16.12 20.35 25.49 31.68 39.10 47.91 58.32 70.54 84.80 101.35 120.46 142.40 167.48 196.00 228.29 264.70 305.56 351.26 402.16 458.64 521.10

0.6113 0.87 1.23 1.71 2.34 3.17 4.25 5.63 7.38 9.59 12.35 15.76 19.94 25.03 31.19 38.58 47.39 57.84 70.14 84.55 101.35 120.82 143.28 169.06 198.53 232.1 270.1 313 361.3 415.5 475.9 543.1

4.95 4.96 4.24 3.88 3.86 3.65 3.35 3.17 3.03 2.78 2.51 2.28 2.06 1.84 1.58 1.34 1.09 0.82 0.57 0.29 0.00 0.30 0.61 0.94 1.28 1.64 2.00 2.38 2.78 3.21 3.63 4.05

23

433.15 438.15 443.15 448.15 453.15 458.15 463.15 468.15 473.15

589.92 665.51 748.27 838.59 936.89 1043.55 1158.98 1283.56 1417.67

617.8 700.5 791.7 892 1002.2 1122.7 1254.4 1397.8 1553.8

4.51 4.99 5.49 5.99 6.52 7.05 7.61 8.17 8.76

The agreement between the two values at lower temperatures improves significantly at lower temperatures, but actually worsens at higher temperatures. The agreement could potentially be improved by not averaging the heat capacity.

24

6.18

We can find the required pressure by applying the Clapeyron equation: T

Δh fus dP = dT v s −v l T

(

)

To use this equation we need an initial condition on the solid-liquid equilibrium line. We can find the values using the triple point of water and then integrate to –5 oC. At the triple point, 0.01 ºC and 0.6113 kPa, Tables B.3 and B.1 give:

⎡m3⎤ vˆs = 1.0908 × 10−3 ⎢ ⎥ ⎣ kg⎦ ⎡m3 ⎤ vˆl = 1.00 ×10−3 ⎢ ⎥ ⎣kg ⎦

and

⎡ kJ ⎤ hˆ s = −333.4 ⎢ ⎥ ⎣kg ⎦

and

hl = 0

To calculate how the enthalpy of fusion changes with T we can use the following path:

liquid

T

solid

T Δh fus

273.16 l cP dT T

T

∫ cPsdT

∫

273.16

T = 273.16 [K] ⎡kJ ⎤ ⎢⎣kg ⎥⎦ ⎡ J ⎤ Δh fus = −6001.2 ⎣mol ⎦

liquid

273.16

Δh

T fus

=

∫ T

Δ hˆ fus = −333.4

T

c dT +Δ h l P

273.16 fus

+

∫ 273.16

solid

T

c dT = s P

∫

(csP− clP)dT +Δ h273.16 fus

273.16

25

Using data in Appendix A: T 273.16 Δh fus = −4.873R (T − 273.16) + Δh fus

which gives 273.16

dP −4.873R 1331R + Δh fus = s l + dT v s − vl T v −v

(

(

)

)

If we assume that (v s − v l )is independent of temperature and pressure, we can now separate variables in the Clapeyron equation and integrate.

P − Ptp =

−4.873 R

(T − Ttp )+

(vs − v l )

273.16 1331R + Δh fus