Question

The coefficient of volume expansion of glycerine is $49\times 10^{- 5}K^{- 1}$ . The fractional change in its density for a $30^{\circ} C$ rise in temperature will be

• A. $1.45\times 10^{- 2}$
• B. $2\times 10^{- 4}$
• C. $3.5\times 10^{- 3}$
• D. $2.5\times 10^{- 2}$

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 $V_{t}=V_{0}\left(1 + \gamma \Delta t\right)$
$= \, \, V_{0}\left(1 + 49 \times \left(10\right)^{- 5} \times 30\right) \, $
$V_{t}=V_{0}\left(1 + 0.01470\right)=1.01470 \, V_{0}$
$\Rightarrow \, \, \frac{V_{0}}{V_{t}}=\frac{1}{1.01470}$
Let $\rho _{o}$ 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
$\frac{\Delta \rho }{\rho _{0}}=$ fractional change in the density
$\frac{\left(\rho \right)_{t} - \left(\rho \right)_{0}}{\left(\rho \right)_{0}}=\frac{m \left(\frac{1}{V_{t}} - \frac{1}{V_{0}}\right)}{\frac{m}{V_{0}}}=\left(\frac{V_{0}}{V_{t}} - 1\right)$
$\frac{\Delta \rho }{\left(\rho \right)_{0}}=\left(\frac{1}{1.01470} - 1\right)=-0.0145$
Here, negative sign shows that density decreases with rise in temperature
$\left|\frac{\Delta \rho }{\rho _{0}}\right|=0.0145=1.45\times 10^{- 2}$