📂Thermal Physics

Linear Expansion Coefficient and Volumetric Expansion Coefficient

Coefficient of Linear Expansion

The coefficient of linear expansion refers to the change in length per unit length of a solid when it expands due to heat, as follows:

$$\alpha = \dfrac{\Delta L}{L} \dfrac{1}{\Delta T} \left[ ^\circ \mathrm{C} ^{-1} \right]$$

Here, $L$ is the original length of the solid, $\Delta T$ is the change in temperature, and $\Delta L$ is the change in length.

Derivation

Let’s assume that a solid with an original length of $L$ expands to a length of $L+\Delta L$ after heating. Then, the increase in length would be proportional to both the original length and the change in temperature, as follows:

$$\Delta L \propto L \Delta T$$

If we let the proportionality constant be $\alpha$, we obtain the following:

$$\Delta L = \alpha L \Delta T \implies \alpha = \dfrac{\Delta L}{L} \dfrac{1}{\Delta T} \left[ ^\circ \mathrm{C} ^{-1} \right]$$

Coefficient of Volume Expansion

The coefficient of volume expansion refers to the change in volume per unit volume of a solid when it expands due to heat, as follows:

$$\beta = 3\alpha$$

Here, $\alpha$ is the coefficient of linear expansion.

Derivation

Let’s say the initial volume is $V=L^3$ and the expanded volume is $V^{\prime}=(L+\Delta L)^3$. When $x$ is sufficiently small, the following approximation holds:

$$(1+x)^n \approx 1 + nx \quad (|x| \ll 1)$$

This is because, according to the binomial theorem,

$$(1+x)^n = \dfrac{n!}{0!n!}1+\dfrac{n!}{1!(n-1)!}x+\dfrac{n!}{2!(n-2)!}x^2+\dfrac{n!}{3!(n-3)!}x^3 + \cdots$$

If the magnitude of $x$ is sufficiently small, terms of second order and higher are too small to be considered. Therefore, we obtain the following:

$$\begin{align*} V^{\prime} &= (L+\Delta L)^3 \\ &= L^3 \left( 1+\dfrac{\Delta L}{L} \right)^3 \\ &\approx L^3 \left( 1+3\dfrac{\Delta L}{L} \right) \\ &= V\left( 1+ 3\dfrac{\Delta L}{L} \right) \end{align*}$$

Continuing the calculation yields:

$$\begin{align*} && V^{\prime} &= V+3V\dfrac{\Delta L}{L} \\ \implies && \Delta V = V^{\prime}-V &= 3V\dfrac{\Delta L}{L} \\ \end{align*}$$

Since the volume change would be proportional to both the original volume and the temperature change, we can set up the following proportionality:

$$\Delta V \propto V \Delta T$$

Here, if we let the proportionality constant be $\beta$, we obtain the following results:

$$\Delta V = \beta V \Delta T$$

$$\begin{align*} \implies \beta &= \dfrac{\Delta V}{V} \dfrac{1}{\Delta T} \\[1em] &= \dfrac{3V \frac{\Delta L}{L}}{V} \dfrac{1}{\Delta T} \\[1em] &= 3 \dfrac{\Delta L }{L} \dfrac{1}{\Delta T} \\[1em] &= 3\alpha \mathrm{} \end{align*}$$

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