Show that β ≈ 3α, by calculating the change in volume ΔV of a cube with sides of length L.

We calculate the change in length by the formula for linear expansion in solids and find new length and we further calculate change in volume and find new volume.

Formula for length expansion of solids $∆\mathrm{L}=\mathrm{\alpha L}∆\mathrm{T}$

Formula for volume expansion of solids $∆\mathrm{V}=\mathrm{\beta V}∆\mathrm{T}$

Here,$\alpha ,\beta$are the coefficient of linear and volume expansion respectively, L, V is original length and volume respectively, and $∆\mathrm{T}$is the change in temperature.

New length = $\mathrm{L}+∆\mathrm{L}$

New volume = $\mathrm{V}+∆\mathrm{V}$

But for cube,

Volume = (Length)³

$\begin{array}{rcl}{\left(L+∆L\right)}^{\mathbf{3}}& \mathbf{=}& \mathbf{V}\mathbf{+}\mathbf{∆}\mathbf{V}\\ {\left(L+\mathrm{\alpha L}∆T\right)}^{\mathbf{3}}& \mathbf{=}& \mathbf{V}\mathbf{+}\mathbf{\beta V}\mathbf{∆}\mathbf{T}\\ {\mathbf{L}}^{\mathbf{3}}{\left(1+\alpha ∆T\right)}^{\mathbf{3}}& \mathbf{=}& \mathbf{V}\left(1+\beta ∆T\right)\\ {\mathbf{L}}^{\mathbf{3}}\left(1+{\alpha }^3∆T^3+3\alpha ∆T+3{\alpha }^2∆T^2\right)& \mathbf{=}& \mathbf{V}\left(1+\beta ∆T\right)\end{array}$

Since α is very small, second and fourth terms in bracket on LHS are negligible. Also, wkt.V=L³. So the equation becomes

$\begin{array}{rcl}\left(1+3\alpha ∆T\right)& \mathbf{=}& \left(1+\beta ∆T\right)\\ \mathbf{3}\mathit{\alpha }\mathbf{∆}\mathbf{T}& \mathbf{=}& \mathbf{\beta }\mathbf{∆}\mathbf{T}\\ \mathbf{3}\mathbf{\alpha }& \mathbf{=}& \mathbf{\beta }\end{array}$

Therefore, it is proved that$\mathrm{\beta }=3\mathrm{\alpha }$