A glass flask whose volume is 1000.00 cm3 at 0.0_Cis completely filled with mercury at this temperature. When flask and mercury are warmed to 55.0ºC, 8.95 cm³ of mercury overflow. If the coefficient of volume expansion of mercury is $\mathbf{18}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}{\mathbf{K}}^{\mathbf{-}\mathbf{1}}$, compute the coefficient of volume expansion of the glass.

Volume expansion is the expansion of the dimensions of an object with respect to temperature changes and can be expressed as,

$\mathbf{\Delta V}\mathbf{=}{\mathbf{V}}_{\mathbf{0}}\mathbf{}\mathbf{\beta \Delta T}$ ...(i)

Here, V is the expansion in volume, V₀ is the unexpanded volume, β is the coefficient of volume expansion and T is the absolute temperature.

For mercury, the coefficient of volume expansion,

${\beta }_{Hg}=18\times {10}^{-5}\hspace{0.33em}{(C^0)}^{-1}$

The volume of mercury over flow is 8.95 cm³ because both the volume of mercury and the volume of glass expands under increase in temperature.

So, ${\mathrm{\Delta V}}_{\mathrm{Hg}}-{\mathrm{\Delta V}}_{\mathrm{glass}}=8.95{\mathrm{cm}}^3$ ...(ii)

Now, ${\mathrm{\Delta V}}_{\mathrm{Hg}}={\mathrm{V}}_0{\mathrm{\beta }}_{\mathrm{Hg}}\mathrm{\Delta T}$

Substitute all the values in the above equation,

$$\begin{gathered} {\mathrm{\Delta V}}_{\mathrm{Hg}}=\left(18\times {10}^{-5}{\left({\mathrm{C}}^0\right)}^{-1}\right)\left(1000.00{\mathrm{cm}}^3\right)\left(55.0{\mathrm{C}}^{°}\right) \\ =9.9{\mathrm{cm}}^3 \end{gathered}$$

Substitute the values in equation (ii),

$$\begin{gathered} ∆{\mathrm{V}}_{\mathrm{glass}}=∆{\mathrm{V}}_{\mathrm{Hg}}-8.95{\mathrm{cm}}^3 \\ =0.95{\mathrm{cm}}^3 \end{gathered}$$

For glass,

localid="1664358046376" $$\begin{gathered} ∆V_{glass}=V_0{\beta }_{glass}∆T \\ {\beta }_{glass}=\frac{∆V_{glass}}{V_0∆T} \\ =\frac{0.95c{m}^3}{\left(1000.00c{m}^3\right)\left(55.0C^{°}\right)} \\ =\left(1.7\times {10}^{-5}{\left(C^{°}\right)}^{-1}\right) \end{gathered}$$

Thus, coefficient of volume expansion of the glass is $\left(1.7\times {10}^{-5}{\left(C^{°}\right)}^{-1}\right)$.