Question

Coefficient of volume expansion of mercury is $0.18 \times {10^{ - 3}}{/^\circ }C.$ If the density of mercury at ${0^\circ }C$ is $13.6g/cc$ then its density at ${200^\circ }C$ is
(A) $13.11g/cc$
(B) $52.11g/cc$
(C) $16.11g/cc$
(D) $26.11g/cc$

Answer 1

Hint: Use the formula of density at a given temperature for thermal expansion. Thermal expansion is the process where the volume of the given material changes due to rise in temperature.

Formula used:
${\rho _t} = \dfrac{{{\rho _0}}}{{(1 + {\alpha _v}T)}}$
Where,
${\rho _0}$ is density at $0{}^0C$
${\rho _t}$ is density at temperature other than zero.
${\alpha _v}$ is the coefficient of volume expansion
$T$ is the temperature at which the density is to be calculated.

Complete step by step answer:
It is given in the question that, the coefficient of volume expansion of the mercury is ${\alpha _v} = 0.18 \times {10^{ - 3}}/{}^0C$
The density of mercury at $0{}^0C$ is ${\rho _0} = 13.6/cc$
Let the density of mercury at $200{}^0C$ is ${\rho _{200}}$
Then, we can use the formula
${\rho _{200}} = \dfrac{{{\rho _0}}}{{(1 + {\alpha _v}T)}}$
Where,
$T = 200{}^0C$ is the temperature at which the density of mercury is to be calculated.
Substitute the given values in the above equation. We get
${\rho _{200}} = \dfrac{{13.6}}{{(1 + 0.18 \times {{10}^{ - 3}} \times 200)}}$
On simplifying it, we get
$ = \dfrac{{13.6}}{{(1 + 36 \times {{10}^{ - 3}})}}$
Further simplifying it, we get
$ \Rightarrow {\rho _{200}} = \dfrac{{13.6}}{{(1 + 0.036)}}$
$ \Rightarrow {\rho _{200}} = \dfrac{{13.6}}{{1.036}}$
$ \Rightarrow {\rho _{200}} = 13.11g/cc$

Hence, the correct option is (A).

Note:It is a simple question of substituting the values in the formula. For such questions you need to know that formula. Be careful while doing the calculations. Do not make mistakes. You can use a log table for complex calculations.
In thermal expansion, the volume of the material changes due to rise in thermal expansion. But the number of molecules in the material does not change. That means, the number of molecules per unit area reduces due to increase in volume. Therefore, the density decreases. That is why we can observe that the value of density of mercury at ${200^0}C$ is less than the density at ${0^0}C.$