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101

2.If the number of molecules in an ideal gas is doubled and the volume is doubled, by how much does the pressure change if the temperature is held constant ?

SOLUTION

The ideal gas law is

P V = NkT

If T is constant then

P / NV

If N is doubled and V is doubled then P does not change.

The ideal gas law is
P V = NkT
which implies
V / NPT
If N ! 2N, T ! 2T and P ! 12 P then V ! 21£=22 V = 8V . Thus the volume increases by a factor of 8.

102

CHAPTER 16. KINETIC THEORY OF GASES

3.If the number of molecules in an ideal gas is doubled, and the absolute temperature is doubled and the pressure is halved, by how much does the volume change ?

(Absolute temperature is simply the temperature measured in Kelvin.)

SOLUTION

Chapter 17

Review of Calculus

103

= lim 5
¢x!0
= 5 as expected because the slope
of the straight line y = 5x + 2 is 5.
¢x!0
¢x!0

104

CHAPTER 17. REVIEW OF CALCULUS

1. Calculate the derivative of y(x) = 5x + 2.

SOLUTION

y(x) = 5x + 2

y(x + ¢x) = 5(x + ¢x) + 2 = 5x + 5¢x + 2

dxdy =

=

lim y(x + ¢x) ° y(x) ¢x

lim 5x + 5¢x + 2 ° (5x + 2) ¢x

105

2.Calculate the slope of the curve y(x) = 3x2 + 1 at the points x = °1, x = 0 and x = 2.

SOLUTION

y(x) = 3x2 + 1

y(x + ¢x) = 3(x + ¢x)2 + 1

=3(x2 + 2x¢x + ¢x2) + 1

=3x2 + 6x¢x + 3(¢x)2 + 1

dy dx

=

 

lim

y(x + ¢x) ° y(x)

 

 

¢x!0

 

 

 

¢x

=

 

lim

3x2 + 6x¢x + 3(¢x)2 + 1 ° (3x2 + 1)

 

¢x!0

 

 

 

¢x

=

 

lim (6x + 3¢x)

 

¢x!0

 

 

 

 

 

=

6x

Øx=

 

 

 

 

 

 

dy

 

 

= °6

 

 

 

 

 

1

 

 

 

dx

°

 

 

 

 

Ø

 

 

 

 

 

 

dy

Ø

 

 

 

 

 

 

 

 

Ø

 

 

= 0

 

 

 

dx

 

 

 

 

 

Øx=0

 

 

 

 

 

 

Ø

 

 

 

 

 

 

 

dy

Ø

 

 

 

 

 

 

 

 

Ø

 

 

= 12

 

 

 

dx

 

 

 

 

 

Øx=2

 

 

 

 

 

 

Ø

 

 

 

 

 

 

 

 

Ø

 

 

 

 

 

 

 

 

Ø

 

 

 

 

 

106

 

CHAPTER 17. REVIEW OF CALCULUS

3.

Calculate the derivative of x4 using the formula

dxn

= nx1. Verify

dx

 

your answer by calculating the derivative from dy

 

 

y(xx)°y(x)

 

 

=

lim

.

 

 

dx

 

¢x!0

¢x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

SOLUTION

 

dxn

= nx1

 

 

dx

 

 

 

:::

dx4

= 4x4°1 = 4x3

 

dx

 

 

 

Now let's verify this.

y(x) = x4

y(x + ¢x) = (x + ¢x)4

= x4 + 4x3¢x + 6x2x)2 + 4xx)3 + (¢x)4

dy

=

lim

 

y(x + ¢x) ° y(x)

 

dx

¢x

 

¢x!0

 

 

=

lim

 

x4 + 4x3¢x + 6x2x)2 + 4xx)3 + (¢x)4 ° x4

 

 

¢x!0

 

 

¢x

 

=

lim [4x3 + 6x2¢x + 4xx)2 + (¢x)3]

 

 

¢x!0

 

 

 

 

=

4x3

which agrees with above

107

4. Prove that dxd (3x2) = 3dxdx2 .

SOLUTION

y(x) = 3x2

y(x + ¢x) = 3(x + ¢x)2 = 3x2 + 6x¢x + 3(¢x)2

dy

 

=

d

 

(3x2) =

lim

y(x + ¢x) ° y(x)

dx

dx

¢x

 

 

¢x!0

 

 

 

 

 

=

lim

3x2 + 6x¢x + 3(¢x)2 ° 3x2

 

 

 

 

 

¢x

 

 

 

 

 

 

¢x!0

 

 

 

 

 

=

lim 6x + 3¢x

 

 

 

 

 

 

¢x!0

 

 

 

 

 

 

=

6x

 

Now take

y(x) = x2 ) dxdy = 2x

Thus

dxd (3x2) = 6x

= 3dxd x2

108

CHAPTER 17. REVIEW OF CALCULUS

5.Prove that dxd (x + x2) = dxdx + dxdx2 .

SOLUTION

Take y(x) = x + x2

y(x + ¢x) = x + ¢x + (x + ¢x)2

=x + ¢x + x2 + 2x¢x + (¢x)2

dy

 

=

 

d

(x + x2) = lim

 

y(x + ¢x) ° y(x)

dx

 

¢x

 

 

dx

¢x!0

 

 

 

=

 

lim

x + ¢x + x2 + 2x¢x + (¢x)2 ° (x + x2)

 

 

 

¢x!0

 

¢x

 

 

=

 

lim (1 + 2x + ¢x)

 

 

 

 

 

¢x!0

 

 

 

 

=

2x

 

 

 

 

 

 

 

 

 

 

 

::: d

(x + x2) = dx

+ dx2

 

 

 

 

dx

dx

 

dx

dx

= 1 + 2x

 

 

=

1

 

 

 

 

dx dx2

dx

109

6.Verify the chain rule and product rule using some examples of your own.

SOLUTION

your own examples

110

CHAPTER 17. REVIEW OF CALCULUS

7.Where do the extremum values of y(x) = x2 ° 4 occur? Verify your answer by plotting a graph.

SOLUTION

y(x) = x2 ° 4

0 = dxdy = 2x

::: x = 0

y(0) = 0 ° 4 = °4

::: extreme occurs at (x; y) = (0; °4)

The graph below shows this is a minimum.