Question

A glass flask of volume one litre at $0{}^\circ C$ is filled full with mercury at this temperature. The flask and mercury are now heated to $100{}^\circ C$. How much mercury will spill out, if coefficient of volume expansion of mercury is $1.82\times {{10}^{-4}}/{}^\circ C$ and linear expansion of glass is $0.1\times {{10}^{-4}}/{}^\circ C$ respectively?

Answer 1

When both the flask and mercury are heated then both of them will expand and since it is given in the sentence that mercury spills out of the flask, then the flask must have been less expanded as compared to the expansion of mercury and the amount of mercury spilled out can be found by finding the difference in volume expansion of the flask and mercury.
Formula used: $\Delta V=V\gamma \Delta \theta$

Complete answer:
To solve this question, we will have to find out the volume expansion of mercury as well as the flask. So, let’s start by finding out the expansion of mercury first. The expansion of mercury can be found out by the volume expansion formula which is as follows:
$\Delta {{V}_{m}}=V{{\gamma }_{m}}\Delta \theta$
Here, the volume of mercury expanded is given as $\Delta {{V}_{m}}$,
$V$ is the initial volume of mercury and as given $V=1\text{ L}$,
${{\gamma }_{m}}$ is the coefficient of volume expansion of mercury which is given as is $1.82\times {{10}^{-4}}/{}^\circ C$,
$\Delta \theta$ is the change in temperature. As it is given in the question that the initial temperature is $0{}^\circ C$ and the final temperature is $100{}^\circ C$ hence $\Delta \theta =100{}^\circ -0{}^\circ$. Substituting all the values, we get:

& \Delta {{V}_{m}}=1\times 1.82\times {{10}^{-4}}\times \left( 100{}^\circ -0{}^\circ \right) \\\ & \Rightarrow \Delta {{V}_{m}}=1\times 1.82\times {{10}^{-4}}\times 100{}^\circ \\\ & \Rightarrow \Delta {{V}_{m}}=1.82\times {{10}^{-2}}\text{ L} \\\ \end{aligned}$$ Similarly, the volume of flask expanded whose coefficient linear expansion is $0.1\times {{10}^{-4}}/{}^\circ C$ will be as follows: $\Delta {{V}_{f}}=V{{\gamma }_{g}}\Delta \theta $ But here we have to use the coefficient of volume expansion of glass instead of coefficient of linear expansion, hence we will use the relation: $\begin{aligned} & {{\gamma }_{g}}={{\alpha }_{g}} \\\ & \Rightarrow {{\gamma }_{g}}=3\times 0.1\times {{10}^{-4}}/{}^\circ C \\\ & \Rightarrow {{\gamma }_{g}}=0.3\times {{10}^{-4}}/{}^\circ C \\\ \end{aligned}$ Substituting all the values, we get: $$\begin{aligned} & \Delta {{V}_{f}}=0.3\times {{10}^{-4}}\times \left( 100{}^\circ -0{}^\circ \right) \\\ & \Rightarrow \Delta {{V}_{f}}=0.3\times {{10}^{-4}}\times 100{}^\circ \\\ & \Rightarrow \Delta {{V}_{f}}=0.3\times {{10}^{-2}}\text{ L} \\\ \end{aligned}$$ The amount of mercury that will spill out will be: $\begin{aligned} & \Delta V=\Delta {{V}_{m}}-\Delta {{V}_{f}} \\\ & \Rightarrow \Delta V=\left( 1.82-0.3 \right)\times {{10}^{4}} \\\ & \therefore \Delta V=1.52\times {{10}^{4}}\text{L} \\\ \end{aligned}$ Thus, the answer is $1.52\times {{10}^{4}}\text{L}$. **Note:** In the solution we have used the relation between coefficient of volume expansion and coefficient of linear expansion because the relation between coefficient of linear expansion $\left( \alpha \right)$, coefficient of areal expansion $\left( \beta \right)$ and coefficient of volume expansion $\left( \gamma \right)$ is as follows: $\dfrac{\alpha }{1}=\dfrac{\beta }{2}=\dfrac{\gamma }{3}$