91Ó°ÊÓ

The initial temperature is\(T = 0^\circ {\rm{C}}\).

The density of gasoline is\(\rho = 0.68 \times {10^3}\;{\rm{kg/}}{{\rm{m}}^3}\).

The final temperature is \(T' = 33^\circ {\rm{C}}\).

In this problem, the density of gasoline will be equivalent to the fraction between the mass of gasoline and its volume.

The relation to find the density can be written as:

\(\begin{aligned}{l}{\rho _{\rm{H}}} &= \frac{M}{{{V_0}}}\\{\rho _{\rm{H}}} &= \frac{M}{{V\left( {1 + \beta \left( {T' - T} \right)} \right)}}\\{\rho _{\rm{H}}} &= \frac{\rho }{{1 + \beta \left( {T' - T} \right)}}\end{aligned}\)

Here, Mis the mass,\({V_0}\)is the volume of gasoline, and\(\beta \)is the coefficient of volume expansion of gasoline.

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}{\rho _{\rm{H}}} &= \left( {\frac{{0.68 \times {{10}^3}\;{\rm{kg/}}{{\rm{m}}^3}}}{{1 + \left( {950 \times {{10}^{ - 6}}\;{\rm{/}}^\circ {\rm{C}}} \right)\left( {33^\circ {\rm{C}} - 0^\circ {\rm{C}}} \right)}}} \right)\\{\rho _{\rm{H}}} &= 0.659 \times {10^3}\;{\rm{kg/}}{{\rm{m}}^3}\end{aligned}\)

Thus, \({\rho _{\rm{H}}} = 0.659 \times {10^3}\;{\rm{kg/}}{{\rm{m}}^3}\) is the density on a hot day.

The relation to find the percentage change in the density can be written as:

\(\% N = \left( {\frac{{{\rho _{\rm{H}}} - \rho }}{\rho }} \right) \times 100\% \)

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}\% N &= \left( {\frac{{0.659 \times {{10}^3}\;{\rm{kg/}}{{\rm{m}}^3} - 0.68 \times {{10}^3}\;{\rm{kg/}}{{\rm{m}}^3}}}{{0.68 \times {{10}^3}\;{\rm{kg/}}{{\rm{m}}^3}}}} \right) \times 100\% \\\% N &= - 3\% \end{aligned}\)

Thus, \(3\% \) is the change in density.