Question

A container is filled to the brim with 1.5 L of mercury at 20°C. As the temperature of the container and mercury is increased to 56°C, a total of 7.8 mL of mercury spill over the brim of the container. Determine the linear expansion coefficient of the material that makes up the container.

Answer 1

The linear expansion coefficient of the container is one- third its coefficient of volume expansion.

α_c = β_c/3    or      α_Hg = β_Hg/3    -------------(1)      

After heating change in volume of the container = ∆V_c = β_cV_c∆T   -----------------(2)

After heating change in volume of the mercury = ∆V_Hg = β_HgV_Hg∆T   -----------------(3)

∆V_Hg - ∆V_c = 7.8*10^-3L --------------------(4)

From eqn (2), (3) and (4)

β_HgV_Hg∆T   - β_cV_c∆T   = 7.8*10^-3L

β_c = (β_HgV_Hg∆T   - 7.8*10^-3L)/( V_c∆T)

but V_Hg = V_c = V

β_c = (β_HgV∆T   - 7.8*10^-3L)/( V∆T)

β_c = β_Hg    - (7.8*10^-3L)/( V∆T)

From eqn (1),

α_c = β_Hg/3   - (7.8*10^-3L)/( 3V∆T)

Now plugging values β_Hg = (0.18*10^-3K^-1)

α_c = (0.18*10^-3K^-1)/3   - (7.8*10^-3L)/( 3(1.5)(36)) = 1.2*10^-5 K^-1