Problem Set 1 Solutions - Illinois State University 360/Homework...Chemistry 360 Dr. Jean M....

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  • Chemistry 360 Dr. Jean M. Standard

    Problem Set 1 Solutions 1. Determine the first derivatives of each of the following functions of one variable.

    a.)

    f (x) = 3x2ex ( is a constant)

    " f (x) = 6x 3 x2( )ex b.)

    Y (x) = A cos x( ) (A is a constant)

    " Y (x) = A sin x( )

    c.)

    g(y) = 1 2y2

    " g (y) = 2y 1 2y2( )1 / 2

    d.)

    H (T ) = a + bT + cT 2 + dT

    (a, b, c, and d are constants)

    " H (T ) = b + 2cT dT 2

    e.)

    u(r) = Ar12

    Br6

    (A and B are constants)

    " u (r) = 12Ar13

    + 6Br7

    f.)

    s(t) = e3t 1 n(t)[ ]

    " s (t) = 3e3t 1 n(t)[ ] e3t

    t

  • 2

    2. Determine the indefinite or definite integrals of each of the following functions of one variable.

    Note: For the solutions below involving indefinite integrals, the constant C is used to correspond to an overall arbitrary constant of integration. a.) kA2 dA (k is a constant)

    kA2 dA =k A2 dA

    =k A3

    3+C .

    b.) 3V dV

    V1

    V2

    3V dVV1

    V2 =3 V dVV1V2

    =3 V2

    2

    "

    #$

    %

    &'

    V1

    V2

    .

    Here, to complete the solution, we must evaluate the result at the limits of integration. The final result is therefore

    3V dVV1

    V2 =3 V22

    2V1

    2

    2

    #

    $%

    &

    '(.

    c.) 1V + b dV (b is a constant)

    Here, we can let u =V + b . Then, du = dV (since b is a constant). Substituting, the integral becomes 1u du=ln u( )+C.

    Expressing the result in terms of V gives

    1V + b dV =ln V + b( )+C.

  • 3

    2. continued d.) eaRT dT (a and R are constants)

    Here, we can let u = aRT . Then, du = aRdT , or alternately, dT = 1aRdu . Substituting, the integral

    becomes

    eaRT dT = 1aR eu du

    = 1aReu +C .

    Expressing the result in terms of T gives

    eaRT dT = 1aR eaRT +C .

    e.) aT 2+ b

    T 3!

    "#

    $

    %&dT300

    500 (a =250, b=5.0104)

    The easiest way to tackle this integral is to break it into the sum of two integrals,

    aT 2+ b

    T 3!

    "#

    $

    %&dT300

    500 =a 1T 2 dT300

    500 +b 1T 3 dT300

    500

    =a T 2 dT300

    500 +b T 3 dT300

    500

    aT 2+ b

    T 3!

    "#

    $

    %&dT300

    500 =a T 1( )

    300

    500+b 12 T

    2( )300

    500.

    The next step is to evaluate the result at the limits and substitute the numerical values of a and b,

    aT 2+ b

    T 3!

    "#

    $

    %&dT300

    500 =a T 1( )

    300

    500+b 12 T

    2( )300

    500

    =a 1500

    + 1300

    !

    "#

    $

    %& +

    b2 1

    500( )2+ 1300( )2

    !

    "##

    $

    %&&

    =250 1500

    + 1300

    !

    "#

    $

    %&+2.5104

    1500( )2

    + 1300( )2

    !

    "##

    $

    %&&

    =250 1.333103( )+2.5104 7.111106( )=0.333+0.178

    aT 2+ b

    T 3!

    "#

    $

    %&dT300

    500 =0.511.

  • 4

    3. For each of the following functions of x and y, determine the partial derivatives

    fx#

    $ %

    &

    ' ( y, f

    y#

    $ %

    &

    ' ( x,

    2 fx2#

    $ % %

    &

    ' ( ( y

    , 2 fy2#

    $ % %

    &

    ' ( ( x

    , 2 f

    xy

    #

    $ % %

    &

    ' ( ( , and

    2 fyx

    #

    $ % %

    &

    ' ( ( .

    a.)

    f (x, y) = x2y + 3y

    f x

    #

    $ %

    &

    ' ( y

    = 2xy f y

    #

    $ %

    &

    ' ( x

    = x2 + 3

    2 f x2#

    $ % %

    &

    ' ( ( y

    = 2y 2 f

    y2#

    $ % %

    &

    ' ( ( x

    = 0

    2 f x y

    #

    $ % %

    &

    ' ( ( = 2x

    2 f y x

    #

    $ % %

    &

    ' ( ( = 2x

    b.)

    f (x, y) = 5ex y + y

    f x

    #

    $ %

    &

    ' ( y

    = 5ex y f y

    #

    $ %

    &

    ' ( x

    = 5ex + 1

    2 f x2#

    $ % %

    &

    ' ( ( y

    = 5ex y 2 f

    y2#

    $ % %

    &

    ' ( ( x

    = 0

    2 f x y

    #

    $ % %

    &

    ' ( ( = 5e

    x 2 f

    y x

    #

    $ % %

    &

    ' ( ( = 5e

    x

    c.)

    f (x, y) = y n(x) + x n(x)

    f x

    #

    $ %

    &

    ' ( y

    = yx

    + 1 + n(x) f y

    #

    $ %

    &

    ' ( x

    = n(x)

    2 f x2#

    $ % %

    &

    ' ( ( y

    = yx2

    + 1x

    2 f

    y2#

    $ % %

    &

    ' ( ( x

    = 0

    2 f x y

    #

    $ % %

    &

    ' ( ( =

    1x

    2 f

    y x

    #

    $ % %

    &

    ' ( ( =

    1x

  • 5

    3. continued

    d.)

    f (x, y) = 6x3

    f x

    #

    $ %

    &

    ' ( y

    = 18x2 f y

    #

    $ %

    &

    ' ( x

    = 0

    2 f x2#

    $ % %

    &

    ' ( ( y

    = 36x 2 f

    y2#

    $ % %

    &

    ' ( ( x

    = 0

    2 f x y

    #

    $ % %

    &

    ' ( ( = 0

    2 f y x

    #

    $ % %

    &

    ' ( ( = 0

    e.)

    f (x, y) = xy( )1/ 2

    f x

    #

    $ %

    &

    ' ( y

    = 12 x1 / 2 y1 / 2 f

    y

    #

    $ %

    &

    ' ( x

    = 12 x1 / 2 y1 / 2

    2 f x2#

    $ % %

    &

    ' ( ( y

    = 14 x3 / 2 y1 / 2

    2 f y2#

    $ % %

    &

    ' ( ( x

    = 14 x1 / 2 y3 / 2

    2 f x y

    #

    $ % %

    &

    ' ( ( =

    14 x

    1 / 2y1 / 2 2 f

    y x

    #

    $ % %

    &

    ' ( ( =

    14 x

    1 / 2y1 / 2

    f.)

    f (x, y) = 3x2 cosy + xy3

    f x

    #

    $ %

    &

    ' ( y

    = 6x cos y + y 3 f y

    #

    $ %

    &

    ' ( x

    = 3x 2 sin y + 3xy2

    2 f x 2#

    $ %

    &

    ' ( y

    = 6 cos y 2 f

    y 2#

    $ %

    &

    ' ( x

    = 3x 2 cos y + 6xy

    2 f x y

    #

    $ %

    &

    ' ( = 6x sin y + 3y 2

    2 f y x

    #

    $ %

    &

    ' ( = 6x sin y + 3y 2

  • 6

    4. For each of the following functions of two variables, evaluate the two first partial derivatives. [Where it appears in the expressions below, R corresponds to the gas constant.]

    a.)

    H (T ,P) = 32 R nT PnP + 3T2P

    HT

    #

    $ %

    &

    ' ( P

    = 3R2T

    + 32P

    H P

    #

    $ %

    &

    ' ( T

    = 1 nP 3T2P2

    b.)

    s(v, t) = 12 vt2 + vev

    s v#

    $ %

    &

    ' ( t

    = 12 t2 + ev vev

    s t#

    $ %

    &

    ' ( v

    = vt

    c.)

    g(x, y) = e3x 1 x2( ) y3ny

    g x

    #

    $ %

    &

    ' ( y

    = e3x 3 1 x2( ) 2x* + , - . / y3ny

    g y

    #

    $ %

    &

    ' ( x

    = e3x 1 x2( ) 3y2ny + y2( )

    d.)

    P(V ,T ) = RTV

    1 + bV( )

    PV#

    $ %

    &

    ' ( T

    = RTV 2

    PT#

    $ %

    &

    ' ( V

    = RV

    1 + bV( )

    e.)

    u(r, ) = 32 r2cos rersin

    u r#

    $ %

    &

    ' (

    = 3rcos sin er + rer( )

    u

    $

    % &

    '

    ( ) r

    = 32 r2sin rercos

  • 7

    4. continued

    f.)

    H (T ,P) = 32 RT + RT2Pe3P

    HT

    #

    $ %

    &

    ' ( P

    = 32 R + 2RTPe3P

    H P

    #

    $ %

    &

    ' ( T

    = RT 2 e3P 3Pe3P( )

    g.)

    P(V ,T ) = RT + RTVnV

    PV#

    $ %

    &

    ' ( T

    = RT nV + 1( )

    PT#

    $ %

    &

    ' ( V

    = R + RVnV

    5. For each of the following functions of three variables, evaluate the requested partial derivatives.

    a.)

    r = x2 + y2 + z2 ; evaluate

    rx#

    $ %

    &

    ' ( y,z

    .

    r x

    #

    $ %

    &

    ' ( y,z

    = x x2 + y2 + z2( )1 / 2 = xr

    b.)

    y = r sin cos ; evaluate

    y

    $

    % &

    '

    ( ) r,

    .

    y

    $

    % &

    '

    ( ) r,

    = r sin sin

  • 8

    6. Evaluate the following expressions using the ideal gas equation of state.

    a.)

    PT#

    $ %

    &

    ' ( Vm

    The partial derivative required involves P and also requires

    Vm to be held constant. Therefore, the ideal gas equation of state should be solved for P and written in terms of

    Vm before the partial derivative is evaluated,

    P = nRTV

    = RTVm

    .

    Then, the partial derivative may be evaluated,

    PT#

    $ %

    &

    ' ( Vm

    = RVm

    .

    b.)

    PVm

    #

    $ %

    &

    ' (