Question

A one litre glass flask containing some mercury. It is found that at different temperatures the volume of air inside the flask remains the same. What is the volume of mercury in the flask if the coefficient of linear expansion of gas is $9\times 10^{-6} / ^\circ C$ while of volume expansion of mercury is $1.8 \times 10^{-4} / ^\circ C$.

Answer 1

Hint: Whenever heat is given to a body, the atoms and molecules present inside the volume of the material starts vibrating with high amplitudes. This motion causes the molecules to get separated by a certain distance internally. Due to this reason, externally the body is appeared to be expanded. This expansion can be linear, areal or volumetric. Practically, all expansions are volumetric and all expansions are interrelated.

Formula used:
$V = V_\circ(1 + \gamma \Delta T) or \ \Delta V = V - V_\circ = V_\circ \gamma \Delta T$

Complete step-by-step answer:
Here, in the statement itself, it is given that at different temperatures, the volume of air inside the flask remains the same. This means that the amount of volume of mercury increased is the same as the volume increased of the material flask. Hence the concept is that an increase in volume of both mercury and flask is the same.
Hence: $\Delta V_m = \Delta V_c$
Or $V_m \gamma_m \Delta T = V_c \gamma_c \Delta T$
Or $V_m = \dfrac{\gamma_c}{\gamma_m}V_c$
Given, $\alpha_m = 1.8 \times 10^{-4}/\ \circ C \ and \ \alpha_c = 9\times 10^{-6}/\circ C$
Hence $\dfrac{\alpha_c}{\alpha_m} = \dfrac{\gamma_c}{\gamma_m} = \dfrac{9\times 10^{-6}}{1.8 \times 10^{-4}}=0.05$
Also, given $V_c = 1\ litre$
Therefore, $V_m = \dfrac{\gamma_c}{\gamma_m}V_c = 0.05 \times 1 = 0.05\ litre$
Hence the volume of mercury is 0.05 litre.

Additional Information: The coefficient of expansion gives us the measure of how much a material will expand in a particular direction if the temperature is raised by $1^\circ C$. In case of a 1-D rod, we assume that only its length is increasing. But practically when any material is heated, all its length, area and volume increases.

Note: In the above solution, we wrote that the ratio of volumetric expansion coefficient and linear expansion coefficient is the same. This is because we have a well known relation between different types of coefficients. $\dfrac{\alpha}{1} = \dfrac{\beta}{2} = \dfrac{\gamma}{3}$. Here $\alpha$ is the coefficient of linear expansion, $\beta$ is the coefficient of areal expansion and $\gamma$ is the coefficient of volumetric expansion. Thus when taken the ratio, ratios are equal.