Question

The coefficient of volume expansion of mercury is $5.4\times {10}^{-4}{}^{\circ }C$. What is the fractional change in its density for a ${80}^{\circ }C$ rise in temperature?

Answer 1

Let the initial volume be V and mass be $m.$

Density , $\rho =\frac{m}{V}$ ....(1)

$\gamma =5.4\times {10}^{-4}$

Change in volume when temperature is increased by ${80}^{o}C$

$\Delta V=V\gamma \Delta T=V\times 5.4\times {10}^{-4}\times 80$

New volume is $V_1=V+\Delta V=V\times 5.4\times {10}^{-4}\times 80+V$

New density is ${\rho }_1=\frac{m}{V_1}=\frac{m}{V+V\times 5.4\times {10}^{-4}\times 80}=\frac{m}{V\times 1.0432}=\frac{\rho }{1.0432}$

Fractional change $C=\frac{\rho -{\rho }_1}{\rho }$

$C=.0414$