# Question afbca

Dec 30, 2016

Relevant equation to be used is

ΔV=V_iβΔT
where $\Delta V$ is change of volume, ${V}_{i}$ is initial volume, $\beta$ is Coefficient of Volume Expansion and $\Delta T$ is change in temperature.

$\beta$ for [aluminum $6.9 \times {10}^{-} 5 \mathmr{and}$ gasoline 9.5xx10^-4K^-1 or "^@C^-1](https://en.wikipedia.org/wiki/Thermal_expansionVolume_expansion) at 20^@C. Assuming it to be valid over the given temperature range.

1. For aluminum cylinder
ΔV_c=1xx6.9xx10^-5xx(65-5)
=>ΔV_c=4.14ml
2. For gasoline
ΔV_c=1xx9.5xx10^-4xx(65-5)
ΔV_c=57ml

Boiling point of gasoline ranges between $104 \mathmr{and} {392}^{\circ} F$ which corresponds to $40 \mathmr{and} {200}^{\circ} C$. The wide range of boiling points is due to different blends of components. In the given question it is not given.
All the samples of gasoline which have boiling points ${65}^{\circ} C$ or lower would have evaporated. As such there is no spillage! in such cases.
For other samples, ignoring evaporation losses, the spillage is the difference between the volume expansion of gasoline and of aluminum cylinder.

Gasoline Spillage$= 57 - 4.14 = 53 m l$, rounded to nearest $m l$