Gujarat BoardEnglish MediumSTD 11 SciencePhysicsThermal Properties of Matter3 Marks
Question
Show that the coefficient of volume expansion for a solid substance is three times its coefficient of linear expansion.
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Answer
Consider a solid in the form of a rectangular parallelopiped of sides a, b and c respectively so that its volume V = abc.
If the solid is heated so that its temperature rises by $\Delta\text{T,}$ T, then increase in its sides will be,
$\Delta\text{a}=\text{a}.\alpha.\Delta\text{T},\Delta\text{b}=\text{b}.\alpha.\Delta\text{T}$ and $\Delta\text{c}=\text{c}.\alpha.\Delta\text{T}$
or $\alpha'=\text{a}+\Delta\text{a}=\text{a}(1+\alpha.\Delta\text{T})$
$\text{b}'=\text{b}+\Delta\text{b}=\text{b}(1+\alpha.\Delta\text{T})$
and $\text{c}'=\text{c}+\Delta\text{c}=\text{c}(1+\text{a}.\Delta\text{T})$
$\because$ New volume, $\text{V}'=\text{V}+\Delta\text{V}=\text{a}'\text{b}'\text{c}'$
$=\text{abc}(1+\alpha.\Delta\text{T})^3$
$\therefore$ Increase in volume,
$\Delta\text{V}=\text{V}'-\text{V}=\big[\text{abc}(1+\alpha.\Delta\text{T})^3-\text{abc}\big]$
$\therefore$ Coefficient of volume expansion,
$\gamma=\frac{\Delta\text{V}}{\text{V}.\Delta\text{T}}=\frac{\text{abc}(1+\alpha.\Delta\text{T})^3-\text{abc}}{\text{abc}.\Delta\text{T}}$
$\therefore\gamma=\frac{(1+\alpha.\Delta\text{T})^3-1}{\Delta\text{T}}$
$=\frac{(1+3\alpha.\Delta\text{T}.\Delta\text{T}^2+\alpha^3.\Delta\text{T}^3)-1}{\Delta\text{T}}$
$=3\alpha+3\alpha^2\Delta\text{T}+\alpha^3.\Delta\text{T}^2.$
However, as a has an extremely small value for solids, hence terms containing higher powers of a may be neglected. Therefore, we obtain the relation $\gamma=3\alpha.$
i.e. coefficient of volume expansion of a solid is three times of its coefficient of linear expansion.
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