Show that the coefficient of volume expansion for a solid substan… - Vidyadip

Question

Show that the coefficient of volume expansion for a solid substance is three times its coefficient of linear expansion.

✓

Answer

Consider a solid in the form of a rectangular parallelopiped of sides a , b , and c respectively $\therefore$ its volume $V=a b c$.
If the solid is heated so that its temperature rises by $\Delta T$, then increase in its sides will be
$\Delta a=a \cdot \alpha \cdot \Delta T, \Delta b=b \cdot \alpha \cdot \Delta T$ and $\Delta c=c \cdot \alpha \cdot \Delta T$ or $a^{\prime}=a+\Delta a=a(1+\alpha \cdot \Delta T), b^{\prime}=b+\Delta b, b(1+\alpha . \Delta T)$ and $c ^{\prime}= c +\Delta c= c (1+\alpha . \Delta T)$
$\therefore$ New volume $V^{\prime}=V+\Delta V=a^{\prime} b^{\prime} c^{\prime}=a b c(1+\alpha . \Delta T)^3$
$\therefore$ Increase in volume $\Delta V=V^{\prime}-V=\left[a b c(1+\alpha \cdot \Delta T)^3-a b c\right]$
$\therefore$ Coefficient of volume expansion $\gamma=\frac{\Delta V}{V \cdot \Delta T}=\frac{a b c(1+\alpha \cdot \Delta T)^3-a b c}{a b c \cdot \Delta T}$
$
\begin{aligned}
& \therefore \gamma=\frac{(1+\alpha \cdot \Delta T)^3-1}{\Delta T} \\
& =\frac{\left(1+3 \alpha \cdot \Delta T+3 \alpha^2 \cdot \Delta T^2+\alpha^3 \cdot \Delta T^3\right)-1}{\Delta T} \\
& =3 \alpha+3 \alpha^2 \Delta T+\alpha^3 \cdot \Delta T^2
\end{aligned}
$
As we know that $\alpha$ has an extremely small value for solids
$
\therefore \gamma=3 \alpha
$
$\Rightarrow$ the coefficient of volume expansion of a solid is three times of its coefficient of linear expansion.

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