Question

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

Answer 1

Here, $\gamma =5.4\times {10}^{-4}{}^{o}C,\Delta T={80}^{o}C$
Let there be m grams of mercury and its initial volume be V. Suppose that the volume of the mercury becomes $V^{|}$ after a rise of temperature of ${80}^{o}C$. Then,
$V^1=V(1+\gamma \Delta T)$
$V=(1+5.4\times {10}^{-4}\times 80)$
$V^1=1.0432V$
Initial density of the mercury, $\rho =\frac{m}{V}$
Final density of the mercury,
${\rho }^{|}=\frac{m}{V^{|}}=\frac{m}{1.0432V}=\frac{\rho }{1.0432}$
$=0.9586\rho$
Therefore, fractional change in the value of density of mercury,
$\frac{\rho -rh{o}^{|}}{\rho }=\frac{\rho -0.9586\rho }{\rho }=0.0414$