Question

A one litre glass flask contains some mercury. It is found that at different temperatures the volume of air inside the flask remains the same. The volume of mercury in this flask, if coefficient of linear expansion of glass is $9\times {10}^{-6}{\left(oC\right)}^{-1}$ , while volume of expansion of mercury is $1.8\times {10}^{-4}{\left(oC\right)}^{-1}$ will be (in $c{m}^3$) (volume of air remains same).

Answer 1

It is given that the volume of air in the flask remains the same. This means that the expansion in volume of the vessel is exactly equal to the volume expansion of mercury. Since glass is isotropic, ${\gamma }_g=3{\alpha }_g$
$\Delta V_g=\Delta V_l\Rightarrow V_g{\gamma }_g\Delta T=V_l{\gamma }_l\Delta T\Rightarrow V_g\left(3\alpha \right)\Delta T=V_l{\gamma }_l\Delta T\Rightarrow V_l=\frac{\left(3\alpha \right)V_g}{{\gamma }_l}=\frac{1000\times \left(3\times 9\times {10}^{-6}\right)}{1.8\times {10}^{-4}}=150cc$