Question

The coefficient of volume expansion of glycerine is $49\times 10^{-5}/{}^{\circ }C$. What is the fractional change in its density (approx.) for $30^{\circ }C$ rise in temperature?

• A. $1.5\times 10^{-2}$
• B. $2.5\times 10^{-2}$
• C. $2.0\times 10^{-2}$
• D. $2.8\times 10^{-2}$

Answer 1

Answer: (A)

Let $V_0$ be the initial volume of glycerine, i.e., at $0^{\circ }C$ (dry). If $V_t$ be its volume at $30^{\circ }C$.
Then $V_t=V_0(1+v\Delta t)$
$\begin{array}{rl}& =V_0\left(1+49\times 10^{-5}\times 30\right)\\ V_t& =V_0(1+0.01470)=1.0147070V_0\\ \Rightarrow \frac{V_0}{V_t}& =\frac{1}{1.01470}\end{array}$
Let ${\rho }_0$ and ${\rho }_t$ be the initial and final densities of glycerine then initial density, ${\rho }_0=\frac{m}{v_0}$ and final density, ${\rho }_t=\frac{m}{V_t}$ where, $m=$ mass of glycerine
$\begin{array}{rl}\frac{\Delta \rho }{{\rho }_0}& =\\ \frac{{\rho }_t-{\rho }_0}{{\rho }_0}& =\frac{m\left(\frac{1}{V_t}-\frac{1}{V_0}\right)}{\frac{m}{V_0}}=\left(\frac{V_0}{V_t}-1\right)\\ \Rightarrow \frac{\Delta \rho }{\Delta {\rho }_0}& =\left(\frac{1}{1.01470}-1\right)=-0.0145\end{array}$
Here, negative sign shows that density decreases with rise in temperature.
$\begin{array}{l}\frac{\Delta \rho }{{\rho }_0}=0.0145=1.45\times 10^{-2}\\ \Rightarrow \frac{\Delta \rho }{\Delta {\rho }_0}=1.5\times 10^{-2}\end{array}$