Note 3 Effects of Temperature!
Increasing temperatures make most material expand. This is called thermal expansion.
Expansion for solids and liquids
Increasing the temperature of a object increases its length.
Li (Ti)
Lf (Tf)
∆L
The fractional change in the length happens to be proportional to the temperature change and the
proportionality constant is called the coefficient of linear expansion.
ΔL
= αΔT
Li
This is usually written like this.
ΔL = αLi ΔT
One way to understand this behavior is to model the atoms a balls connected by springs. When
the temperature increases, the spacing between the balls increases by some amount. The more
springs to begin with, the more the final increase in length.
initial length, Li
∆L
initial length, Li
∆L
For a two-dimensional object, the change in the area looks like this.
Li,x (Ti)
∆Lx
Lf,x (Tf)
(0, 0)
Li,y (Ti)
∆Ly
Lf,y (Tf)
The change in the area is
ΔA = ΔLx Li,y + ΔLy Li,x + ΔLx ΔLy
The initial area is
Ai = Li,x Li,y
page 1
The fractional change in the area is
ΔLx ΔLy
ΔLx ΔLy ΔLx ΔLy
ΔA ΔLx Li,y ΔLy Li,x
=
+
+
=
+
+
Ai
Li,x Li,y
Li,x Li,y
Li,x Li,y
Li,x
Li,y
Li,x Li,y
This reduces to
ΔA
= αΔT + αΔT + α2ΔT 2
Ai
Now, the value of the coefficient is small, about 10-5 K-1. Furthermore, the change in the
temperature is 1000 K at most since most things melt before this temperature. Therefore, we
have this.
αΔT ≤ 10−2 and α2ΔT 2 ≤ 10−4
We may neglect the squared term as being at much less than 1%.
ΔA
= 2αΔT
Ai
This is usually written like this.
ΔA = 2αAi ΔT
Using the same argument, the fractional volume change is this.
ΔV
= 3αΔT
Vi
This is usually written like this.
ΔV = 3αVi ΔT
The coefficient of volume expansion has its own symbol. Surprisingly, it is beta.
ΔV = βVi ΔT
This equation works for solids, but liquids do not have a value for the linear coefficient. Therefore,
their volume coefficients are just measured directly. Most coefficients are around 10-4 K-1.
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Example 1 (Problem 19.41)
Mercury just fills a pyrex glass bulb. The diameter of the bulb is 0.25 cm and the diameter of the
capillary is 0.004 cm. The initial height of the mercury is zero.
What is the height of the mercury when the temperature is increased by 30 °C?
Analysis
When the temperature increases, the volume of the bulb increases thus lowering the mercury
level. However, the volume of the mercury itself increases as well, thus raising the mercury level.
You probably know that the mercury level goes up as the temperature goes up. We can
reasonably assume that the mercury’s volume increase is larger than that of the bulb. However
much extra volume that the bulb can not contain goes to the capillary level.
The way to approach these problem is to just find the initial and final values then compare them.
Solution
The volumes of the bulb are
Vi =
4
πR 3 and Vf = Vi (1 + 3αΔT )
3
The volumes of mercury in the capillary, for whatever the final height h, are
vi = 0 and v f = Ai (1 + 2αΔT ) ⋅ h
The volumes of the mercury are
Vi,Hg =
4
πR 3 and Vf ,Hg = Vi,Hg (1 + βΔT )
3
Assuming the final volume of the mercury is larger than that of the bulb, the extra volume of
mercury is
Vf ,Hg −Vf = Vi,Hg (1 + βΔT ) −Vi (1 + 3αΔT )
The initial volumes are the same.
Vf ,Hg −Vf = Vi ΔT(β − 3α)
This extra volume of mercury goes into the capillary to some height, h
Vi ΔT(β − 3α) = Aih(1 + 2αΔT )
h=
h=
h=
Vi ΔT(β − 3α)
Ai (1 + 2αΔT )
3
4 ⎛⎜ 0.25 cm ⎞⎟
⎟⎟ (30 K)(1.82 ×10−4 K −1 − 3 ⋅ 3.2 ×10−6 K −1 )
π⎜
⎟⎠
3 ⎜⎝
2
⎛ 0.004 cm ⎞⎟2
⎟⎟ (1 + 2 ⋅ 3.2 ×10−6 K −1(30 K))
π ⎜⎜
⎜⎝
⎟⎠
2
(8.181×10−3 cm 3 )(30 K)(1.724 ×10−4 K −1 )
(1.257 ×10−5 cm 2 )(1.000)
= 3.366 cm or 3.37 cm
page 3
Notice that the change in the area of the capillary is too small to show up. The factor that tells you
how much this change in the area contributes to the volume is this.
1 + 2αΔT ≅ 1.000
Seeing that the coefficient is about 10-6 and ∆T is only 30, we could have just ignored it.
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Expansion for gases
A solid is constrained to its dimensions. A liquid can be contained by an open container. Most
gases must be constrained by closed containers.
What do we know about the characteristics of gases? Around 1800, several behaviors of gases
were experimental found.
(1)!Boyle’s law says that the pressure of a gas is inversely proportional to the volume given that
the amount of gas and the temperature are constant.
P∝
1
V
(2)!Charles’ law says that the volume of a gas is proportional to the temperature given that the
amount of gas and the pressure are constant.
V ∝T
(3)!Gay-Lussac’s law says that the pressure is proportional to the temperature given that the
amount of gas and the volume are constant.
P ∝T
(4)!Avogadro’s law says that the volume proportional to the amount of gas (the number of
molecules in moles) given that the pressure and temperature are constant.
V ∝n
you can combine them to form a single statement.
P
1
∝
or PV ∝ nT
nT V
The proportionality constant is called “R”, the universal gas constant (or the molar gas constant).
PV = nRT
This is called the ideal gas law. This is how a gas responds to change to its states.
universal gas constant has been measured to be (according to NIST)
R = 8.314 462 1
The
J
mol⋅K
In certain other applications, you might end up with different units.
R = 0.082 057
L⋅atm
mol⋅K
If you convert the moles to the number of molecules, you get this.
⎛ N ⎞⎟
⎛ R ⎞⎟
⎟⎟ = N ⎜⎜
⎟
PV = nRT ⎜⎜⎜
⎜⎝ N ⎟⎟⎠T = NkBT
⎝ n ⋅ N A ⎟⎠
A
⇒ PV = NkBT
The constant NA is avogadro’s number, and its combination with the universal gas constant is the
Boltzmann constant.
kB =
J
8.314 462 1 mol⋅K
R
=
= 1.380 648 8 ×10−23
1
NA
6.022 141 29 ×1023 mol
J
K
Note that the temperature must be in the unit of kelvin. The pressure is definitely not zero when
the temperature is at 0 °C or at 0 °F.
page 5
What is an ideal gas?
There are assumptions that have to be met to called a gas ideal. The first is that the molecules of
an ideal gas “do not” interact or if they do, the interaction is weak. The interaction here refers to
condensation and bonding, not just the fact that they colliding.
The second assumption is that the volume taken up by the molecules is low.
page 6
Example 2 (Problem 19.29)
Air at 1 atm and 10 °C is used to fill a tire. In the tire, the volume of the air is reduced to 28% of
original volume. The temperature is increased to 40 °C.
(a)!What is the tire pressure?
(b)!After some driving, what is the tire pressure if the temperature rises to 85 °C and the volume of
the tire increases by 2%?
Analysis
The ideal gas law tells you what the relationship is between the various properties of a gas,
pressure, volume, amount, and temperature. This means that before you do anything to a gas,
there is one set of unique values for the variables. After you do something to the gas, there is
another set of unique values. The information given are just these sets and how the sets are
related to each other.
The order of business here is to look at the sets and then relate across them.
Solution
(a)!Some volume of air was used to fill the tire. Let me call it V. This air have a pressure of 1 atm.
Its temperature is 10 °C = 283.15 K. The number of air molecules is also fixed, call it N.
PiVi = N ikBTi
⇒ (1 atm)V = NkB (283.15 K)
When inside the tire, the volume becomes 0.28V. The temperature becomes 40 °C = 313.15 K.
The number of molecules is the same as before.
PfVf = N f kBTf
⇒ Pf (0.28)V = NkB (313.15 K)
Let’s take the ratio of the two equations.
Pf (0.28)V
(1 atm)V
=
NkB (313.15 K)
NkB (283.15 K)
⇒ Pf =
1 atm 313.15
= 3.9498 atm = 3.95 atm
0.28 283.15
(b)!After driving, the final volume is increased by 2% to (0.28)V(1.02). The temperature is 85 °C =
358.15 K.
Pf (0.28)V(1.02)
(1 atm)V
=
NkB (358.15 K)
NkB (283.15 K)
⇒ Pf =
1 atm 358.15
= 4.4288 atm = 4.43 atm
0.28 ⋅ 1.02 283.15
page 7