Question

 Show that for small thermal expansions of a solid object the linear and 

volume coefficients of expansion are related by β ≈ 3α.

Answer 1

From the linear thermal expansion law we know that

$ΔL = αL_iΔT \\ dL = αL_idT$

Note that, every dimension of the cube will expand linearly and the final volume of the cube will equal to

$V_f = L_fw_fh_f$

The initial volume of the cube before raising its temperature is

$V_i = L_iw_in_i$

From the law of volume thermal expansion we know, that:

$ΔV = βV_iΔT \\ dV = βV_idT$

Note that, the cube’s dimensions are equal; therefore,

$L_i = w_i = h_i = L$

Because the cubic is made of one material so that, the three dimensions will expand by the same rate; therefore,

$L_f=w_f=h_f=L_f$

From all the above,

$V_i=L^3_i \\ V_f = L^3_f$

Note that,

$dL = L_f -L_i \\ dL +L_i = L_f \\ V_f = (L_i+dL)^3 \\ dL = αL_idT \\ V_f = (L_i+αL_idT)^3 = (L_i(1+α))^3 \\ V_i = L^3_i \\ V_f = V_i(1+α)^3 \\ V_f = V_i (1+αdT + 2αdT+2α^2dt^2)(1+αdT) \\ V_f = V_i(1+αdT + 2αdT+2α^2dT^2 + α^2dT + α^3dT^3) \\ V_f = V_i(1+3αdT + 3α^2dT^2+α^3dT^3)$

We are supposing that the change in the length dL is very infinitesimal. This means that, $dL^2$ is very close to zero and $dL^3$ is also very close to zero. So that, we will neglect themselves

$V_f = V_i(1+3αdT + 0+0) \\ V_f = V_i(1+3αdT) \\ V_f = V_i+3αV_idT \\ V_f- V_i = 3αV_idT \\ dV = 3αV_idT \\ dV = βV_idT \\ βV_idT = 3αV_idT \\ β = 3α$