Question

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

• A. $1.5\times {10}^{-2}$
• B. $2\times {10}^{-4}$
• C. $3.5\times {10}^{-3}$
• D. $2.5\times {10}^{-2}$

Answer 1

The correct option is A $1.5\times {10}^{-2}$

Let $V_0$ be the initial volume of glycerine, i.e., at $0^{o}C$ (dry). If $V_t$ be its volume at ${30}^{o}C$.
Then $V_t=V_0(1+v\Delta t)$
$=V_0(1+49\times {10}^{-5}\times 30)$
$V_t=V_0(1+0.01470)=1.0147070V_0$
$\Rightarrow \frac{V_0}{V_t}=\frac{1}{1.01470}$
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
$\frac{\Delta \rho }{{\rho }_0}=$
$\frac{{\rho }_t-{\rho }_0}{{\rho }_0}=\frac{m(\frac{1}{V_t}-\frac{1}{V_0})}{\frac{m}{V_0}}=(\frac{V_0}{V_t}-1)$
$\Rightarrow \frac{\Delta \rho }{\Delta {\rho }_0}=(\frac{1}{1.01470}-1)=-0.0145$
Here, negative sign shows that density decreases with rise in temperature.
$\frac{\Delta \rho }{{\rho }_0}=0.0145=1.45\times {10}^{-2}$
$\Rightarrow \frac{\Delta \rho }{\Delta {\rho }_0}=1.5\times {10}^{-2}$