Question

The volume of a glass vessel is 1000 cc at ${20}^{\circ }C$. What volume of mercury should be poured into it at this temperature so that the volume of the remaining space does not change with temperature ? Coefficients of cubical expansion of mercury and glass are $1.8\times {10}^{-4}{}^{\circ }{C}^{-1}$ and $9.0\times {10}^{-6}{}^{\circ }{C}^{-1}$ respectively.

Answer 1

$V_g=1000cc{V}_{Hg}=?$

$T_1={20}^{\circ }C$

${\gamma }_{Hg}=1.8\times {10}^{-4}{/}^{\circ }C$

${\gamma }_g=9\times {10}^{-6}{/}^{\circ }C$

Here $\Delta T$ remains constant

Volume of remaining space $=V_g^{\prime }-V_{Hg}^{\prime }$

Now $V_g^{\prime }=V_g(1+{\gamma }_g\Delta T)$ . . . (1)

$V_{Hg}^{\prime }=V_{Hg}(1+{\gamma }_{Hg}\Delta T)$ . .. (2)

Subtracting (2) from (1)

$V_g^{\prime }-V_{Hg}^{\prime }=V_g-V_{Hg}+V_g{\gamma }_{Hg}+V_g{\gamma }_g\Delta T-V_{Hg}{\gamma }_{Hg}\Delta T$

$\Rightarrow \frac{V_g}{V_{Hg}}=\frac{{\gamma }_{Hg}}{{\gamma }_g}$

$\Rightarrow \frac{1000}{V_{Hg}}=\frac{1.8\times {10}^{-4}}{9\times {10}^{-6}}$

$\Rightarrow V_{Hg}=\frac{9\times {10}^{-3}}{1.8\times {10}^{-4}}=50cc$