Question

A specific gravity bottle is completely filled by 271.92 g of mercury at ${0^ \circ }C$ the mass of mercury which would fill in specific gravity bottle at ${100^ \circ }C$ is (Coefficient of linear expansion of the material of the specific gravity bottle $8.0 \times {10^{ - 6}}{/^ \circ }C$ and coefficient of areal expansion of mercury = $0.000182{/^ \circ }C$)
A. 267 g
B. 277 g
C. 500 g
D. 300 g

Answer 1

Hint: The density is defined as the amount of mass per unit volume of the substance –
$\rho = \dfrac{M}{V}$
The specific gravity bottle is used to measure the specific gravity of a liquid, which is the measure of the number of times its density is higher than that of water.

Complete step by step solution:
Step 1: List the given data –
Mass of mercury at ${0^ \circ }C$, ${m_1}$= 271.92 g
Coefficient of linear expansion of the material of specific gravity bottle, ${\alpha _g} = 8.0 \times {10^{ - 6}}{/^ \circ }C$
Coefficient of areal expansion of mercury, \[\beta = 0.000182{/^ \circ }C\]
To find –
Mass of mercury after heating it to ${100^ \circ }C,{m_2} = ?$
The mass of mercury at ${100^ \circ }C,{m_2} = {\rho _2}{V_2}$
where ${\rho _2}\& {V_2}$ are density and volume of the mercury at the temperature respectively.
Thus, finding the value of the density and volume at ${100^ \circ }C$will give us the mass.
Step 2: Find the value of ${\rho _2}$
The mercury expands on heating from ${0^ \circ }C$to ${100^ \circ }C$. The new density after expansion is given by,
${\rho _2} = {\rho _1}\left( {1 - \beta \left( {\Delta \theta } \right)} \right)$
where $\Delta \theta \& \beta $ are the increase in temperature and coefficient of areal expansion, respectively.
Also,
$
  {\rho _1} = \dfrac{{{m_1}}}{{{V_1}}} \\
  Substituting, \\
  {\rho _1} = \dfrac{{271.92}}{{{V_1}}} \\
$
Substituting the values of ${\rho _1}$, $\Delta \theta \& \beta $ in the equation above;
\[
  {\rho _2} = {\rho _1}\left( {1 - \beta \left( {\Delta \theta } \right)} \right) \\
  {\rho _2} = \dfrac{{271.92}}{{{V_1}}}\left( {1 - \left( {0.000182} \right)\left( {100 - 0} \right)} \right) \\
  Solving \\
  {\rho _2} = \dfrac{{271.92}}{{{V_1}}}\left( {1 - 0.000182 \times 100} \right) \\
  {\rho _2} = \dfrac{{271.92}}{{{V_1}}}\left( {1 - 0.0182} \right) \\
  {\rho _2} = \dfrac{{271.92 \times 0.9818}}{{{V_1}}} \\
  {\rho _2} = \dfrac{{266.97}}{{{V_1}}} - \left( 1 \right) \\
 \]
Step 3: Find the value of ${V_2}$
There is a volumetric expansion on heating the specific gravity bottle from ${0^ \circ }C$to ${100^ \circ }C$. The new density after expansion is given by,
${V_2} = {V_1}\left( {1 + {\gamma _g}\left( {\Delta \theta } \right)} \right)$
where $\Delta \theta \& {\gamma _g}$ are the increase in temperature and coefficient of volumetric expansion, respectively.
Coefficient of volumetric expansion, ${\gamma _g} = 3 \times {\alpha _g}$
Hence, the equation becomes –
${V_2} = {V_1}\left( {1 + 3{\alpha _g}\left( {\Delta \theta } \right)} \right)$
Substituting the values, we get –
$
  {V_2} = {V_1}\left( {1 + 3{\alpha _g}\left( {\Delta \theta } \right)} \right) \\
  {V_2} = {V_1}\left( {1 + 3 \times 8 \times {{10}^{ - 6}}\left( {100 - 0} \right)} \right) \\
  Solving, \\
  {V_2} = {V_1}\left( {1 + 3 \times 8 \times {{10}^{ - 6}} \times 100} \right) \\
  {V_2} = {V_1}\left( {1 + 3 \times 8 \times {{10}^{ - 6}} \times {{10}^2}} \right) \\
  {V_2} = {V_1}\left( {1 + 3 \times 8 \times {{10}^{ - 4}}} \right) \\
  {V_2} = {V_1}\left( {1 + 24 \times {{10}^{ - 4}}} \right) \\
  {V_2} = {V_1}\left( {1 + 0.0024} \right) \\
  \therefore {V_2} = {V_1}\left( {1.0024} \right) - \left( 2 \right) \\
 $
Step 4: Combining $\left( 1 \right)$ and $\left( 2 \right)$, we get the mass of the mercury –
$
  {m_2} = {\rho _2}{V_2} \\
  Substituting, \\
  {m_2} = \left( {\dfrac{{266.97}}{{{{{V}}_1}}}} \right)\left( {{{{V_1}}}\left( {1.0024} \right)} \right) \\
  {m_2} = 266.97 \times 1.0024 \\
  {m_2} = 267.61g \\
 $
Therefore, the correct option is OPTION C.

Note:
There are three types of coefficients of expansion, namely, Linear $\left( \alpha \right)$, Superficial or Areal $\left( \beta \right)$ and Volumetric $\left( \gamma \right)$. The relationship between them is as follows:
$\dfrac{\alpha }{1} = \dfrac{\beta }{2} = \dfrac{\gamma }{3}$