ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL)
ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL)
4th Edition
ISBN: 9781119540632
Author: FELDER
Publisher: WILEY
Question
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Chapter 2, Problem 2.38P
Interpretation Introduction

(a)

Interpretation:

The plot of ln y versus x on rectangular coordinates passes through (1.0, 693) and (2,0) should be drawn.

Concept introduction:

The straight-line plot has following equation:

ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  1

Here, ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  2 values are plotted at y axis, x values are plotted at x axis, b is slope of the reaction and ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  3 is intercept.

And, the following equation shows the exponential plot:

ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  4

Interpretation Introduction

(b)

Interpretation:

The semilog plot of y versus x passes through (1,2) and (2,1) should be drawn.

Concept introduction:

The straight-line plot has following equation:

ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  5

Here, ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  6 values are plotted at y axis, x values are plotted at x axis, b is slope of the reaction and ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  7 is intercept.

And, the following equation shows the exponential plot:

ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  8

Interpretation Introduction

(c)

Interpretation:

A log pot y versus x passes through (1,2) and (2,1), determine the equation.

Concept introduction:

The straight-line plot has following equation:

ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  9

Here, ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  10 values are plotted at y axis, x values are plotted at x axis, b is slope of the reaction and ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  11 is intercept.

And, the following equation shows the exponential plot:

ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  12

Interpretation Introduction

(d)

Interpretation:

A semilog plot of xy (logarithmic axis) versus y/x passes through (1.0. 40.2). Determine the equation.

Concept introduction:

The straight-line plot has following equation:

ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  13

Here, ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  14 values are plotted at y axis, x values are plotted at x axis, b is slope of the reaction and ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  15 is intercept.

And, the following equation shows the exponential plot:

ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  16

Interpretation Introduction

(e)

Interpretation:

The equation of log plot of y2/x versus (x-2) passes through (1.0, 40.2) and (2.0, 807.0)

Concept introduction:

The straight-line plot has following equation:

ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  17

Here, ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  18 values are plotted at y axis, x values are plotted at x axis, b is slope of the reaction and ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  19 is intercept.

And, the following equation shows the exponential plot:

ELEM.PRIN.OF CHEMICAL...ABRIDGED (LL), Chapter 2, Problem 2.38P , additional homework tip  20

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Problem 9.11 An 80 mm long line MN has its end M 15 mm in front of the V.P. The distance between the ends projector is 50 mm. The front view is parallel to and 20 mm above reference line. Draw the projections of the line and determine its inclination with the V.P. Also, locate the traces. Interpretation Front view of a line is parallel to xy, therefore, 1. The line is parallel to the H.P. 2. The top view of the line has true length. 3. The front view has projected length equal to the distance be- tween the projectors. Construction Refer to Fig. 9.11. 1. Draw a reference line xy. Mark point m' 20 mm above xy and point m 15 mm below xy. 2. Draw a 50 mm long line m'n' parallel to xy. 3. Draw an arc with centre m and radius 80 mm to meet projec- tor from point n' at point n. Join mn to represent the top view. Determine its inclination with xy as the inclination of line MN with the V.P. Here = 51°. 4. Traces Extend line mn to meet xy at point v. Project point v to meet m'n' produced at…
oh 30 20 D и D P 60 60 80
⑤ b Δε m ab C 40
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