Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN: 9781259696527
Author: J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher: McGraw-Hill Education
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Chapter 15, Problem 15.5P

(a)

Interpretation Introduction

Interpretation:

To graph the figures for A=975T18.4+3lnT

Concept Introduction:

Plot A v/s T by varying T and finding respective values of A

  A=975T18.4+3lnT

Plot T v/s x,by varying T, find respective values of x using the below equation,

  A(12x)=ln(1xx)

Where,

T = TemperaturE

x = Mole fraction

UCST = Upper Critical Solution TemperaturE

LCST = Lower Critical Solution TemperaturE

(a)

Expert Solution
Check Mark

Answer to Problem 15.5P

The corresponding figures of A v/s T and T v/s x are plotted.

Explanation of Solution

  A=975T18.4+3lnT

Using the above equation and varying T,

Now, Plot A v/s T, we get,

    T A T A T A T A
    250 2.064383 305 1.957657 355 1.962832 405 2.019069
    255 2.04732 310 1.954878 360 1.966645 410 2.02652
    260 2.032045 315 1.952956 365 1.970925 415 2.034233
    265 2.018435 320 1.951838 370 1.975644 420 2.042193
    270 2.006377 325 1.951476 375 1.980778 425 2.050385
    275 1.995768 330 1.951823 380 1.986303 430 2.058797
    280 1.986512 335 1.952839 385 1.992198 435 2.067417
    285 1.97852 340 1.954484 390 1.99844 440 2.076233
    290 1.971712 345 1.95672 395 2.005012 445 2.085234
    295 1.966011 350 1.959514 400 2.011894 450 2.094409
    300 1.961347 455 2.103749

Introduction to Chemical Engineering Thermodynamics, Chapter 15, Problem 15.5P , additional homework tip  1

From this graph of A v/s T, we get,

A =2 at T = 272.92 K and 391.209 K

Hence, we get that, UCST = 272.92 K

and

LCST = 391.2 K

Since, There are 2 points for A =2, We can plot T v/s x

Varying T from

225 K to UCST

And from

LCST to 482 K

For finding x,

  A(12x)=ln(1xx)

  A=1(12x)ln(1xx)

  A=975T18.4+3lnT

Varying, T, we find the values of x respectively upto 3 decimals.

We can see that the values of T are same for x and (1-x).

We find the corresponding values, for T v/s x, as below,

    x A x A A T
    0.273 2.157389 0.727 2.157389 2.157711656 229.2
    0.275 2.154223 0.725 2.154223 2.15444794 229.8
    0.277 2.151103 0.723 2.151103 2.15122162 230.4
    0.279 2.148026 0.721 2.148026 2.148032352 231
    0.281 2.144992 0.719 2.144992 2.144879797 231.6
    0.295 2.124933 0.705 2.124933 2.125260129 235.5
    0.297 2.122228 0.703 2.122228 2.122372431 236.1
    0.299 2.119563 0.701 2.119563 2.119518699 236.7
    0.311 2.104361 0.689 2.104361 2.10441677 240
    0.315 2.099584 0.685 2.099584 2.099167952 241.2
    0.321 2.092682 0.679 2.092682 2.092783653 242.7
    0.325 2.08825 0.675 2.08825 2.089045662 243.6
    0.339 2.073769 0.661 2.073769 2.073625685 247.5
    0.345 2.068035 0.655 2.068035 2.06802134 249
    0.357 2.057374 0.643 2.057374 2.057334882 252
    0.361 2.054052 0.639 2.054052 2.054261227 252.9
    0.367 2.04928 0.633 2.04928 2.049270664 254.4
    0.373 2.044756 0.627 2.044756 2.044442293 255.9
    0.383 2.037752 0.617 2.037752 2.037948522 258
    0.391 2.032618 0.609 2.032618 2.032623149 259.8
    0.401 2.026769 0.599 2.026769 2.026683214 261.9
    0.411 2.021534 0.589 2.021534 2.02181967 263.7
    0.427 2.014395 0.573 2.014395 2.014166319 266.7
    0.435 2.011382 0.565 2.011382 2.011259023 267.9
    0.447 2.007542 0.553 2.007542 2.007745467 269.4
    0.463 2.003663 0.537 2.003663 2.003702412 271.2
    0.487 2.000451 0.513 2.000451 2.000474405 272.7
    x A x A A T
    0.273 2.157389 0.727 2.157389 2.1576682 482.5
    0.275 2.154223 0.725 2.154223 2.1556408 481.5
    0.277 2.151103 0.723 2.151103 2.1515998 479.5
    0.279 2.148026 0.721 2.148026 2.1483803 477.9
    0.281 2.144992 0.719 2.144992 2.1465746 477
    0.295 2.124933 0.705 2.124933 2.1248316 466
    0.297 2.122228 0.703 2.122228 2.1221099 464.6
    0.299 2.119563 0.701 2.119563 2.1190127 463
    0.311 2.104361 0.689 2.104361 2.1046921 455.5
    0.315 2.099584 0.685 2.099584 2.0990594 452.5
    0.321 2.092682 0.679 2.092682 2.0925609 449
    0.325 2.08825 0.675 2.08825 2.0888838 447
    0.339 2.073769 0.661 2.073769 2.0735686 438.5
    0.345 2.068035 0.655 2.068035 2.0682904 435.5
    0.357 2.057374 0.643 2.057374 2.057098 429
    0.361 2.054052 0.639 2.054052 2.0545646 427.5
    0.367 2.04928 0.633 2.04928 2.0495558 424.5
    0.373 2.044756 0.627 2.044756 2.0446266 421.5
    0.383 2.037752 0.617 2.037752 2.0373882 417
    0.391 2.032618 0.609 2.032618 2.0326704 414
    0.401 2.026769 0.599 2.026769 2.0265203 410
    0.411 2.021534 0.589 2.021534 2.0212758 406.5
    0.427 2.014395 0.573 2.014395 2.0147294 402
    0.435 2.011382 0.565 2.011382 2.0118936 400
    0.447 2.007542 0.553 2.007542 2.0077282 397
    0.463 2.003663 0.537 2.003663 2.003672 394
    0.5 2 0.5 2 2 391.2092

Introduction to Chemical Engineering Thermodynamics, Chapter 15, Problem 15.5P , additional homework tip  2

(b)

Interpretation Introduction

Interpretation:

To graph the figures for A=540T17.1+3lnT

Concept Introduction:

Plot A v/s T by varying T and finding respective values of A

  A=540T17.1+3lnT

Plot T v/s x, by varying T, find respective values of x using the below equation,

  A(12x)=ln(1xx)

Where,

T = TemperaturE

x = Mole fraction

UCST = Upper Critical Solution TemperaturE

LCST = Lower Critical Solution TemperaturE

(b)

Expert Solution
Check Mark

Answer to Problem 15.5P

The corresponding figures of A v/s T and T v/s x are plotted.

Explanation of Solution

  A=540T17.1+3lnT

Using the above equation and varying T,

Now, Plot A v/s T, we get,

    T A T A T A T A
    250 1.624382754 305 1.831427133 355 2.037480129 405 2.244994535
    255 1.641437694 310 1.851652376 360 2.058312094 410 2.26554465
    260 1.65896797 315 1.872003631 365 2.079144116 415 2.28604038
    265 1.676925327 320 1.892462987 370 2.099968476 420 2.30647842
    270 1.695265877 325 1.913014009 375 2.120778078 425 2.326855742
    275 1.713949657 330 1.9336416 380 2.14156639 430 2.34716958
    280 1.732940238 335 1.954331894 385 2.162327405 435 2.367417404
    285 1.752204383 340 1.975072147 390 2.183055602 440 2.387596908
    290 1.771711734 345 1.995850642 395 2.203745902 445 2.407705993
    295 1.791434544 350 2.016656606 400 2.224393641 450 2.427742748
    300 1.811347424 455 2.447705444

Introduction to Chemical Engineering Thermodynamics, Chapter 15, Problem 15.5P , additional homework tip  3

From this graph of A v/s T, we get,

A =2 at T = 346 K

Hence, we get that,

LCST = 391.2 K

Since, there is only 1 point for A =2, we can plot T v/s x

And from

LCST to 482 K

For finding x,

  A(12x)=ln(1xx)

  A=1(12x)ln(1xx)

  A=540T17.1+3lnT

Varying, T, we find the values of x respectively upto 3 decimals.

We can see that the values of T are same for x and (1-x).

We find the corresponding values, for T v/s x, as below,

    x A x A A T
    0.15 2.478002 0.85 2.478002 2.4783 462.7
    0.17 2.402466 0.83 2.402466 2.402485 443.7
    0.19 2.338726 0.81 2.338726 2.338239 427.8
    0.21 2.284354 0.79 2.284354 2.284403 414.6
    0.23 2.237613 0.77 2.237613 2.237938 403.286
    0.25 2.197225 0.75 2.197225 2.197543 393.5
    0.27 2.162223 0.73 2.162223 2.162327 385
    0.29 2.131867 0.71 2.131867 2.131591 377.6
    0.31 2.105577 0.69 2.105577 2.105381 371.3
    0.33 2.082897 0.67 2.082897 2.082893 365.9
    0.35 2.063464 0.65 2.063464 2.063729 361.3
    0.37 2.046988 0.63 2.046988 2.046771 357.23
    0.39 2.033237 0.61 2.033237 2.033314 354
    0.41 2.02203 0.59 2.02203 2.022069 351.3
    0.43 2.013223 0.57 2.013223 2.013326 349.2
    0.45 2.006707 0.55 2.006707 2.006251 347.5
    0.47 2.002405 0.53 2.002405 2.002922 346.7
    0.49 2.000267 0.51 2.000267 2.000634 346.15
    0.5 2 0.5 2 2.00001 346

Introduction to Chemical Engineering Thermodynamics, Chapter 15, Problem 15.5P , additional homework tip  4

(c)

Interpretation Introduction

Interpretation:

To graph the figures for A=1500T19.9+3lnT

Concept Introduction:

Plot A v/s T by varying T and finding respective values of A

  A=1500T19.9+3lnT

Plot T v/s x, by varying T, find respective values of x using the below equation,

  A(12x)=ln(1xx)

Where,

T = TemperaturE

x = Mole fraction

UCST = Upper Critical Solution TemperaturE

LCST = Lower Critical Solution TemperaturE

(c)

Expert Solution
Check Mark

Answer to Problem 15.5P

The corresponding figures of A v/s T and T v/s x are plotted.

Explanation of Solution

  A=1500T19.9+3lnT

Using the above equation and varying T,

Now, Plot A v/s T, we get,

    T A T A T A T A
    250 2.664382754 305 2.178968117 355 1.941705481 405 1.815364905
    255 2.606143577 310 2.14842657 360 1.924978761 410 1.807008064
    260 2.551275662 315 2.119622678 365 1.909281102 415 1.799293392
    265 2.499566836 320 2.092462987 370 1.894563071 420 1.792192705
    270 2.450821433 325 2.066860162 375 1.880778078 425 1.785679271
    275 2.404858748 330 2.042732509 380 1.867882179 430 1.779727719
    280 2.361511667 335 2.020003536 385 1.855833899 435 1.774313955
    285 2.320625436 340 1.998601559 390 1.844594064 440 1.76941509
    290 2.282056562 345 1.978459338 395 1.834125649 445 1.765009363
    295 2.245671832 350 1.959513749 400 1.824393641 450 1.761076082
    300 2.211347424 455 1.757595554

Introduction to Chemical Engineering Thermodynamics, Chapter 15, Problem 15.5P , additional homework tip  5

From this graph of A v/s T, we get,

A =2 at T = 339.66 K

Hence, we get that,

UCST = 339.66 K

Since, there is only 1 point for A =2, we can plot T v/s x

And from

250 K to UCST

For finding x,

  A(12x)=ln(1xx)

  A=1(12x)ln(1xx)

  A=1500T19.9+3lnT

Varying, T, we find the values of x respectively up to 3 decimals.

We can see that the values of T are same for x and (1-x).

We find the corresponding values, for T v/s x, as below,

    x A x A A T
    0.11 2.680437 0.89 2.680437 2.68011 248.7
    0.13 2.568863 0.87 2.568863 2.56848 258.4
    0.15 2.478002 0.85 2.478002 2.47874 267.1
    0.17 2.402466 0.83 2.402466 2.40219 275.3
    0.19 2.338726 0.81 2.338726 2.33801 282.838
    0.21 2.284354 0.79 2.284354 2.28431 289.7
    0.23 2.237613 0.77 2.237613 2.23725 296.2
    0.25 2.197225 0.75 2.197225 2.19687 302.2
    0.27 2.162223 0.73 2.162223 2.16225 307.7
    0.29 2.131867 0.71 2.131867 2.13151 312.9
    0.31 2.105577 0.69 2.105577 2.10552 317.56
    0.33 2.082897 0.67 2.082897 2.08204 322
    0.35 2.063464 0.65 2.063464 2.0634 325.7
    0.37 2.046988 0.63 2.046988 2.0464 329.22
    0.39 2.033237 0.61 2.033237 2.03348 332
    0.41 2.02203 0.59 2.02203 2.02266 334.4
    0.43 2.013223 0.57 2.013223 2.01345 336.5
    0.45 2.006707 0.55 2.006707 2.00658 338.1
    0.47 2.002405 0.53 2.002405 2.00278 339
    0.49 2.000267 0.51 2.000267 2.00194 339.2
    0.5 2 0.5 2 339.663

Introduction to Chemical Engineering Thermodynamics, Chapter 15, Problem 15.5P , additional homework tip  6

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Introduction to Chemical Engineering Thermodynamics

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