Question

Early in the morning, when the temperature is 4.5 ∘C, gasoline is pumped into a car's 53-L steel gas tank until it is filled to the top. Later in the day the temperature rises to 27 ∘C . Since the volume of gasoline increases more for a given temperature increase than the volume of the steel tank, gasoline will spill out of the tank, gasoline will spill out of the tank. How much gasoline spills out in this case?

Use β=0.000950K−1 for gasoline.

Answer 1

Answer:

$\Delta V=1.0911\ L$

Explanation:

Given:

• Initial temperature, $T_i=4.5^{\circ}C$

• initial volume, $V_i=53\ L$

• final temperature,  $T_f=27^{\circ}C$

• volumetric coefficient of thermal expansion for gasoline, $\beta_g=950\times 10^{-6}\ K^{-1}$

• volumetric coefficient of thermal expansion for steel, $\beta_s=35\times 10^{-6}\ K^{-1}$

Now the increment in the volume of the container:

$\Delta V_s=V_i.\beta_s.\Delta T$

$\Delta V_s=53\times 35\times 10^{-6}\times (27-4.5)$

$\Delta V_s=41737.5\times 10^{-6}\ L$

and the increment in the volume of the gasoline:

$\Delta V_g=V_i.\beta_g.\Delta T$

$\Delta V_g=53\times 950\times 10^{-6}\times (27-4.5)$

$\Delta V_g=1132875\times 10^{-6}\ L$

Hence the difference in the increment of the two volumes:

$\Delta V=\Delta V_g-\Delta V_s$

$\Delta V=(1132875-41737.5)\times 10^{-6}$

$\Delta V=1.0911\ L$ of the gasoline spills out of the container in this case.