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{aligned} & = V _0\left(1+49 \times 10^{-5} \times 30\right) \\ V _{ t } & = V _0(1+0.01470)=1.0147070 V _0 \\ \Rightarrow \frac{ V _0}{ V _{ t }} & =\frac{1}{1.01470} \end{aligned} $
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{aligned} \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{aligned} $
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} $