Maths-
General
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Question


During mineral formation, the same chemical compound can be come different minerals depending on the temperature and pressure at the time of formation. A phase diagram is a graph that shows the conditions that are needed to form each mineral. The graph above is a portion of the phase diagram for aluminosilicates, with the temperature t , in degrees
Celsius open parentheses blank to the power of ring operator straight C close parentheses , on the horizontal axis, and the pressure p , in gigapascals (G Pa), on the
Which of the following systems of inequalities best describes the region where sillimanite can form?

  1. table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell P greater or equal than 0.0021 T minus 0.67 end cell row cell P greater or equal than 0.0013 T minus 0.25 end cell end table
  2. table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell P less or equal than 0.0021 T minus 0.67 end cell row cell P greater or equal than negative 0.0015 T plus 1.13 end cell end table
  3. table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell P less or equal than 0.0013 T minus 0.25 end cell row cell P greater or equal than negative 0.0015 T plus 1.13 end cell end table
  4. table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell P less or equal than 0.0013 T minus 0.25 end cell row cell P less or equal than negative 0.0015 T plus 1.13 end cell end table

hintHint:

Hint:
We need to find the system of inequalities which describe the region where sillimanite can form. First we find the equation of the boundary line of the region where sillimanite can be formed in the phase diagram. Then we put the inequalities sign in the appropriate equation by observing if the lower or upper part of the line is the region where the sillimanite can be formed.

The correct answer is: table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell P less or equal than 0.0021 T minus 0.67 end cell row cell P greater or equal than negative 0.0015 T plus 1.13 end cell end table


    The boundaries of the region where sillimanite can be formed are given by two lines: one is the boundary line between kyanite and sillimanite and the other is the boundary between andalusite and sillimanite.
    First we find the equation of the boundary line between kyanite and sillimanite:
    There are two points lying on this line left parenthesis 500 comma 0.38 right parenthesis:  and left parenthesis 795 comma 1.0 right parenthesis.
    Let these points be denoted by
    left parenthesis a comma b right parenthesis equals left parenthesis 500 comma 0.38 right parenthesis and
    In an xy plane, equation of a line passing through two points  (a , b ) and  (c , d ) is given by
    fraction numerator y minus d over denominator d minus b end fraction equals fraction numerator x minus c over denominator c minus a end fraction
    Using the above points, we have
    fraction numerator P minus 1 over denominator 1 minus 0.38 end fraction equals fraction numerator T minus 795 over denominator 795 minus 500 end fraction
    Simplifying the equation, we have
    fraction numerator P minus 1 over denominator 0.62 end fraction equals fraction numerator T minus 795 over denominator 295 end fraction
    Cross multiplying, we get
    295 left parenthesis P minus 1 right parenthesis equals 0.62 left parenthesis T minus 795 right parenthesis
    We use the calculator to expand the above equation
    295 P minus 295 equals 0.62 T minus 462.9
    Taking all the terms in the left hand side, we get
    295 P minus 0.62 T plus 462.9 minus 295 equals 0
    not stretchy rightwards double arrow 295 P minus 0.62 T plus 197.9 equals 0
    Dividing by 295 throught, we get
    P minus 0.0021 T plus 0.67 equals 0
    As the region which forms sillimanite is below this line, we change the equality to less thanor equal to inequality,
    P minus 0.0021 T plus 0.670 less or equal than 0
    Next,
    We find the equation of the boundary line between andalusite and sillimanite:
    There are two points lying on this linecolon left parenthesis 500 comma 0.38 right parenthesis andleft parenthesis 760 comma 0 right parenthesis
    Let these points be denoted by
    left parenthesis a comma b right parenthesis equals left parenthesis 500 comma 0.38 right parenthesis text  and  end text left parenthesis c comma d right parenthesis equals left parenthesis 760 comma 0 right parenthesis
    In an xy plane, equation of a line passing through two points (a , b )  and (c, d )  is given by
    fraction numerator y minus d over denominator d minus b end fraction equals fraction numerator x minus c over denominator c minus a end fraction
    Using the above points, we have
    fraction numerator P minus 0 over denominator 0 minus 0.38 end fraction equals fraction numerator T minus 760 over denominator 760 minus 500 end fraction
    Simplifying the equation, we have
    fraction numerator P over denominator negative 0.38 end fraction equals fraction numerator T minus 760 over denominator 260 end fraction
    Cross multiplying, we get
    260 P equals negative 0.38 left parenthesis T minus 760 right parenthesis
    We use the calculator to expand the above equation
    260 P equals negative 0.38 T plus 288.8
    Taking all the terms in the left hand side, we get
    260 P plus 0.38 T minus 288.8 equals 0
    Dividing by 260 throught, we get
    P plus 0.0014 T minus 1.11 equals 0
    As the region which forms sillimanite is above this line, we change the equality to greater than or equal to inequality,
    P plus 0.0014 T minus 1.11 greater or equal than 0
    Finally, we get
    The inequalities which describe the region where sillimanite is formed are
    P minus 0.0021 T plus 0.67 less or equal than 0
    P plus 0.0014 T minus 1.11 greater or equal than 0
    Rewriting these inequalities, we get
    P less or equal than 0.0021 T minus 0.67
    P greater or equal than negative 0.0014 T plus 1.11
    Option B)
    P less or equal than 0.0021 T minus 0.67
    P greater or equal than negative 0.0015 T plus 1.13
    best describes the region where silliminate can form.

    Note:
    We do not have the exact inequalities in the option, so we choose the one from the options which is the closest approximation of the inequalities that we have calculated. There are a couple of formulas and concepts used here, such as the equation of a line passing through two points and the concept that the region below a line is given by replacing the equality sign with less than or equal to in the standard form of the equation and vice versa.

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