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Section 20.3 Thermal Volume Expansion

Subsection 20.3.1 Volume Expansion of Solids and Liquids

The most general type of expansion will be the expansion of the volume. Suppose the expansion in volume \(\Delta V\) is small compared to the original volume \(V\text{.}\)

\begin{equation*} \Delta V \lt\lt V. \end{equation*}

We denote the coefficient of volume expansion by the Greek letter \(\beta\text{.}\)

\begin{equation} \Delta V = \beta V\Delta T.\label{eq-Volume-Expansion-of-Solids-and-Liquids}\tag{20.3.1} \end{equation}

By substracting the volume at \(T\) from the volume at \(T+\Delta T\text{,}\) just as we did for the change in area, we find that

\begin{equation} \Delta V = 3\alpha V\Delta T.\tag{20.3.2} \end{equation}

Therefore, the coefficient of volume expansion is 3 times as much as the coefficient of the linear expansion.

\begin{equation} \beta = 3 \alpha.\tag{20.3.3} \end{equation}

Bear in mind that this relation only works out when the volume change in small compared to the original volume. In the case of gases, the change is usually too great and this relation does not hold true any more. But, for solids and liquids, it is approximately correct.

The coefficients of volume expansion of some common fluids are given in Table 20.3.1.

Table 20.3.1. Coefficients of volume expansion of some common liquids at \(20^{\circ}\text{C}\) (Handbook of Chemistry and Physics, 82nd Volume, 2001-2002.
Material Coefficient of vol. exp.
\(\beta (\times 10^{-3} \textrm{per} ^{\circ}\text{C})\)
Acetone \(1.46 \)
Ethyl alcohol \(1.40 \)
Gasoline \(0.95\)
Glycerol \(0.520 \)
Mercury \(1.811 \)
Water \(0.206 \)

Subsection 20.3.2 Volume Expansion Gases

Gases respond much more strongly to temperature change. At low densities all gases tend to behave the same way, independent of their chamical compostion; we model dilute gases by the ideal gas equation. To study expansion at constant pressure, we will place the gas in a cylinder with movable piston. When we increase the temperature of the gas, the piston will move so that the pressure is same as before but the volume will change. You will find that the relative change in volume is given by the following formula.

\begin{equation} \dfrac{\Delta V}{V} = \dfrac{1}{T}\Delta T,\tag{20.3.4} \end{equation}

where temperature is in kelvin scale. Comparing this to Eq. (20.3.1), we notice that the coefficient of volume expansion of an ideal gas at constant pressure is highly dependent on the temperature.

\begin{equation} \beta_{\text{gas}} = \dfrac{1}{T}.\tag{20.3.5} \end{equation}

\(100\text{ ml}\) of water is placed in a closed container at \(27^\circ\text{C}\text{.}\) Water is heated to temperature of \(67^\circ\text{C}\text{.}\) What is the volume of water now?

Data: \(\beta_\text{water} = 2.06\times 10^{-4}/^\circ\text{C}\text{.}\)


Use \(\Delta V = \beta\,V\,\Delta T\text{.}\)


\(101\text{ ml}\text{.}\)


Using \(\Delta V = \beta\,V\,\Delta T\) we get the change in volume.

\begin{align*} \Delta V \amp = \beta\,V\,\Delta T,\\ \amp = 2.06\times 10^{-4}/^\circ\text{C}\times 100\text{ ml}\times (67-27)^\circ\text{C}= 1.03\text{ ml}. \end{align*}

Therefore, the volume will be \(101\text{ ml}\text{.}\)

(a) A steel ball of spherical shape has a radius of 20 cm at \(20^{\circ}\text{C}\text{.}\) It is heated in an oven till its temperature is \(400^{\circ}\text{C}\text{.}\) What is the radius at \(400^{\circ}\text{C}\text{?}\)

(b) A spherical shell of negligible thickness made of steel has a radius of 20 cm at \(20^{\circ}\)C. It is heated in an oven till its temperature is \(400^{\circ}\text{C}\text{.}\) What is the radius at \(400^{\circ}\text{C}\text{?}\)

Data: \(\alpha = 12\times 10^{-6} /^{\circ}\text{C} \text{.}\)


(a) You can follow volume change or radius change. (b) You could follow surface area change or radius change.


(a) and (b) \(20.09\text{ cm}\text{.}\)

Solution 1 (a)

(a) Let the new radius be \(r'\) at \(400^{\circ}\text{C}\text{.}\) The change in radius will just like other lengths.

\begin{equation*} r' = ( 1 + \alpha \Delta T)\, r, \end{equation*}

where \(r\) is the original radius at \(20^{\circ}\text{C}\text{.}\) Putting in the numbers we get

\begin{equation*} r' = \left( 1 + \frac{12\times 10^{-6}}{\ ^{\circ}\text{C}}\times 380^\circ\text{C} \right)\, 20\text{ cm} = 20.09\text{ cm}. \end{equation*}
Solution 2 (b)

(b) We now have a surface expansion. Therefore, we could apply the same idea to the surface area in place of volume.

\begin{equation*} \Delta A = 4\pi r'^2 - 4\pi r^2 = 2\alpha\times 4\pi r^2\times \Delta T. \end{equation*}

Solving for \(r'\) and putting in the numbers we obtain \(r' = 20.09\text{ cm}\text{.}\) This part could have been done directly at expansion of the radius, as we did for the volume expansion.