Question

Coefficient of volume expansion of mercury is $0.18 \times {10^{ - 3}}{\,^{\text{o}}}{{\text{C}}^{ - 1}}$. If the density of mercury at $0{\,^{\text{o}}}{\text{C}}$ is $13.6\,{\text{g}}\,{\text{c}}{{\text{c}}^{{\text{ - 1}}}}$, then its density at $200{\,^{\text{o}}}{\text{C}}$.
A. $13.11\,{\text{g}}\,{\text{c}}{{\text{c}}^{{\text{ - 1}}}}$
B. $52.11\,{\text{g}}\,{\text{c}}{{\text{c}}^{{\text{ - 1}}}}$
C. $16.11\,{\text{g}}\,{\text{c}}{{\text{c}}^{{\text{ - 1}}}}$
D. $26.11\,{\text{g}}\,{\text{c}}{{\text{c}}^{{\text{ - 1}}}}$

Answer 1

We are asked to find the density of mercury at a temperature of $200{\,^{\text{o}}}{\text{C}}$. Recall the concepts of thermal expansion and write the formula for density at a particular temperature in terms of its coefficient of volume expansion, put the given values and calculate the required density. Then check which option matches with the answer obtained.

Complete step by step answer:
Given, the coefficient of volume expansion of mercury is $\gamma = 0.18 \times {10^{ - 3}}{\,^{\text{o}}}{{\text{C}}^{ - 1}}$.
Density of mercury at $0{\,^{\text{o}}}{\text{C}}$ is ${\rho _o} = 13.6\,{\text{g}}\,{\text{c}}{{\text{c}}^{{\text{ - 1}}}}$.
We are asked to find the density of mercury at $T = 200{\,^{\text{o}}}{\text{C}}$.
Let us know what volume expansion is. Volume expansion is the increase in volume of a substance when there is a rise in temperature. As the volume expands or increases the density of the substance will reduce.We can write density of a substance at particular temperature in terms of its coefficient of volume expansion as,
$\rho = \dfrac{{{\rho _0}}}{{1 + \gamma \Delta T}}$ (i)
where $\rho$ is the density at a particular temperature,
${\rho _0}$ is the density at $0{\,^{\text{o}}}{\text{C}}$
$\gamma$ is coefficient of volume expansion
$\Delta T$ is the difference in temperature.
We will use equation (i) here to find the value of density of mercury at temperature $200{\,^{\text{o}}}{\text{C}}$.
Here, density of mercury at $0{\,^{\text{o}}}{\text{C}}$ is given and we are asked to find its density at $200{\,^{\text{o}}}{\text{C}}$, so temperature difference will be
$\Delta T = 200{\,^{\text{o}}}{\text{C}} - {0^{\text{o}}}{\text{C}}$
$\Rightarrow \Delta T = 200{\,^{\text{o}}}{\text{C}}$
Now, substituting the values of $\Delta T$, ${\rho _0}$ and $\gamma$ in equation (i) we get,
$\rho = \dfrac{{13.6}}{{1 + \left( {0.18 \times {{10}^{ - 3}}} \right)\left( {200} \right)}}$
$\Rightarrow \rho = \dfrac{{13.6}}{{1 + 0.18 \times {{10}^{ - 3}} \times 200}}$
$\Rightarrow \rho = \dfrac{{13.6}}{{1 + 36 \times {{10}^{ - 3}}}}$
$\Rightarrow \rho = \dfrac{{13.6}}{{1 + 0.036}}$
$\Rightarrow \rho = \dfrac{{13.6}}{{1.036}}$
$\therefore\rho = 13.11\,{\text{g}}\,{\text{c}}{{\text{c}}^{{\text{ - 1}}}}$
Therefore, the density of mercury at $200{\,^{\text{o}}}{\text{C}}$ is $13.11\,{\text{g}}\,{\text{c}}{{\text{c}}^{{\text{ - 1}}}}$.

Hence, the correct answer is option A.

Note: Thermal expansion is the change in shape, size, volume and density of matter due to change in temperature. There are three important types of thermal expansion; these are linear expansion, area expansion and volume expansion. Here we have discussed volume expansion. Linear expansion is the increase in length of a solid on heating and area expansion is the increase in surface area of a solid on heating.