Assume we have a gas in a container with a movable piston under atmospheric pressure. As we increase the temperature of the gas it expands, its volume increases. The piston moves out. As we cool the gas the piston moves in. The ideal gas law gives us the volume of the gas as a function of temperature.

PV = Nk_{B}T, V = (Nk_{B}/P)T.

The volume is proportional to the **absolute** temperature of
the gas. Assume the temperature change from 0 ^{o}C to
10^{o}C. We have

(ΔV/V) = (ΔT/T) = 10/273 = 0.037 = 3.7%.

By monitoring the volume of the gas we can monitor its temperature. We have a gas thermometer.

Solids and liquids also expand as the temperature
increases. For most solids we can define an **average coefficient of linear expansion**, α. The
change in length Δ*l* of a solid is proportional to its length
*l* and the change in temperature ΔT. The proportional
constant is α.

Δ*l = *α*l*ΔT.

The **average volume expansion coefficient**** **β is defined through
ΔV *= *βVΔT. We have β = 3α.

Linear expansion coefficients α: (per ^{o}C)

Aluminum | 23 * 10^{-6} |
---|---|

Copper | 17 * 10^{-6} |

Glass (ordinary) | 9 * 10^{-6} |

Glass (Pyrex) | 3.2 * 10^{-6} |

Steel | 11 * 10^{-6} |

Invar | 0.7 * 10^{-6} |

For some applications it is important to choose materials that have a small coefficient of linear expansion. Temperature variations can destroy a precision alignment. For other applications it is important that all materials have similar coefficients of linear expansion. When different parts of a structure expand by different amounts, the structure buckles.

How much taller does the Eiffel Tower become at the end of a day
when the temperature has increased by 15 ^{o}C? Its original
height is 321 m and you can assume it is made of steel.

Solution:

Δ*l =
*α*l*ΔT = (11*10^{-6}/^{o}C)*(321
m)*(15 ^{o}C) = 5.3*10^{-2} m = 5.3 cm.

The height of the tower increases by 5.3 cm.

A copper telephone wire has essentially no sag
between poles 35 m apart on a winter day when the temperature is -20 ^{o}C.
How much longer is the wire on a summer day, when the temperature is 35 ^{o}C?

Solution:

Δ*l = *α*l*ΔT. For copper α = 17*10^{-6}
per ^{o}C. Δ*l = *(17*10^{-6})(35 m)55 = 0.0327 m = 3.27
cm.

A square hole 8 cm along each side is cut in a sheet of copper. Calculate the change in the area of the hole if the temperature of the sheet is increased by 50 K.

Solution:

ΔA *= *2αAΔT. For copper α = 17*10^{-6} per ^{o}C. ΔA *
= *(34*10^{-6})(64 cm^{2})50 = 0.109 cm^{2}.

How large an expansion gap should be left between steel railroad rails if
they may reach a maximum temperature 35.0 ^{o}C greater than when
they were laid? Their original length is 10.0 m.

Solution:

Δ*l =
*α*l*ΔT = (11*10^{-6}/^{o}C)*(10
m)*(35 ^{o}C) = 3.85*10^{-3} m ~ 4 mm.

If we want to build a thermometer, we want to use a
material with a large coefficient of volume expansion. For
most liquids, the coefficient of volume expansion is on the order of
10^{-4}. For each degree Celsius temperature change,
the volume changes by (ΔV/V) = β = 10^{-4 }= 0.0001 = 0.01%.
If we tried to observe this change by monitoring the height of water
in a glass, we would probably be unsuccessful.

Assume you have 1 liter = 1000 cm^{3} of liquid with β =
10^{-4} in a container with bottom area A = 100 cm^{2}.

The height of the liquid is 1 cm. (Volume = area * height.)

You increase the temperature of the liquid by 20 ^{o}C.

ΔV = βVΔT = 10^{-4} * 1000 cm^{3 }* 20 = 2 cm^{3}.

The change in height is Δh = ΔV/A = 2 cm^{3}/(100 cm^{2})
= 0.02 cm = 0.2 mm.

If, however, you put a tight lid on the container at a height of
10 cm with a hole connected to a fine capillary, then the liquid
will rise in the capillary to a much greater height. Assume
you connected a tube with a cross sectional area of 1 cm^{2}.
Then Δh = ΔV/A = 2 cm^{3}/(1 cm^{2}) = 2 cm.
You have constructed a usable thermometer.

Thermometers can also be constructed from **bimetallic
strips**. Two strips of metal, one of steel and one of aluminum,
that have the same length at 0 ^{o}C, will have different length a 200 ^{
o}C. Assume the strips are 10 cm long at 0 ^{o}C. The
change in length of the steel strip at 200 ^{o}C is Δ*l = *α*l*ΔT
= 11*10^{-6}*10 cm*200 = 0.022 cm = 0.22 mm and the change in length of
the aluminum strip at 200 ^{o}C is Δ*l = *α*l*ΔT = 23*10^{-6}*10
cm*200 = 0.046 cm = 0.46 mm. If the two strips are bonded together, the
bimetallic strip will buckle or curl. Long bimetallic coils will uncoil
slightly as the temperature rises if the inner strip has the larger α. If
one end is fixed and a pointer is attached to the other end the pointer will
move. The position of the pointer indicates the temperature.

Link: Bimetallic strip demo (video)

Some liquid crystals change color when their temperature changes. They selectively only reflect one color of the white light falling onto them. The color that is reflected back depends on the temperature and can be used as a temperature indicator.

In a constant volume gas thermometer the pressure at 20 ^{o}C is 0.98
atm.

(a) What is the pressure at 45^{o }C?

(b) What is the temperature if the pressure is 0.5
atm?

Solution:

(a) PV/T = constant. Since V is constant, P/T = constant. P_{1}/T_{1
}= P_{2}/T_{2}. P_{1}T_{2}/T_{1 }=
P_{2}.

P_{2 }= (0.98 atm)(318 K/293 K) =1.064 atm.

(b) P_{2}T_{1}/P_{1 }= T_{2}. T_{2 }=
(0.5 atm)(293 K)/(0.98 atm) = 149.5 K.