Question

The fraction of the volume of a glass flask that must be filled with mercury so that the volume of the empty space may be the same at all temperatures is given by (${\alpha _{glass}} = 9 \times {10^{ - 6}}/^\circ C$, ${\gamma _{mercury}} = 18.9 \times {10^{ - 5}}/^\circ C$)
A. $\dfrac{1}{2}$
B. $\dfrac{1}{7}$
C. $\dfrac{1}{4}$
D. $\dfrac{1}{5}$

Answer 1

For the empty space to be the same at all temperatures, the increase in the volume of the flask must be equal to the increase in the volume of the mercury. Use the relation that the volume of liquid in a cylindrical flask is proportional to the height of the mercury in the flask.

Formula used: In this solution we will be using the following formulae;
$\gamma = \dfrac{{\Delta V}}{{V\Delta T}}$ where $\gamma$ is the coefficient of cubic (or volume) expansion of a substance, $\Delta V$ is the change in volume of the substance, $V$ is the initial volume, and $\Delta T$ is the change in temperature of the substance.
$\gamma = 3\alpha$ where $\gamma$ is the coefficient of cubic expansion of a solid, and $\alpha$ is the coefficient of linear expansion of the solid.

Complete step by step answer:
To solve the above, we note that for the empty space to have the same volume at all temperatures, the change in volume of the mercury be equal to the change in volume of the flask.
Hence, $\Delta {V_m} = \Delta {V_g}$
Now, coefficient of volume expansion of a substance can be given as
$\gamma = \dfrac{{\Delta V}}{{V\Delta T}}$ where $\Delta V$ is the change in volume of the substance, $V$ is the initial volume, and $\Delta T$ is the change in temperature of the substance.
Hence,
$\Delta V = \gamma V\Delta T$
Hence,
${\gamma _m}{V_m}\Delta T = {\gamma _g}{V_g}\Delta T$
$\dfrac{{{\gamma _m}}}{{{\gamma _g}}} = \dfrac{{{V_g}}}{{{V_m}}}$
But the flask is a cylinder, meaning that the volume is proportional to the height. Hence,
$\dfrac{{{\gamma _m}}}{{{\gamma _g}}} = \dfrac{{{h_g}}}{{{h_m}}}$
Now, we know that the cubic expansion of a solid can be given as
$\gamma = 3\alpha$ where $\alpha$ is the coefficient of linear expansion of the solid.
Hence,
$\dfrac{{{\gamma _m}}}{{3{\alpha _g}}} = \dfrac{{{h_g}}}{{{h_m}}}$
By insertion of values, we have
$\dfrac{{18.9 \times {{10}^{ - 5}}}}{{3\left( {9 \times {{10}^{ - 6}}} \right)}} = \dfrac{{{h_g}}}{{{h_m}}}$
$\Rightarrow \dfrac{{{h_g}}}{{{h_m}}} = 7$
Hence, by inverting,
$\dfrac{{{h_m}}}{{{h_g}}} = \dfrac{1}{7}$
Hence, the correct option is B.

Note: Observe that the $\Delta T$ in ${\gamma _m}{V_m}\Delta T = {\gamma _g}{V_g}\Delta T$ cancels out. This is because the temperature of the mercury will be the same as the temperature of the glass at the equilibrium condition. However, during heating, the flask absorbs heat before the mercury and hence the glass expands first due to having a higher temperature, which makes the condition not hold (during heating).