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

The amount of gasoline that spills out due to thermal expansion can be calculated using the coefficient of volume expansion for gasoline and the change in temperatures. The change in volume is found by multiplying the coefficient of expansion by the initial volume and the temperature change.

Step-by-step explanation:

The question is related to the concept of thermal expansion, where a gasoline's volume increases more than the steel tank's volume due to temperature rise. To calculate the amount of gasoline that spills out of the tank when the temperature increases from 4.5 ℃ to 27 ℃, we need to calculate the change in volume of gasoline using the coefficient of volume expansion for gasoline (β = 0.000950 K⁻¹) and the initial volume of the gasoline.

The equation for volume expansion, ΔV = βVΔT, can be used to determine the increase in volume of the gasoline, where ΔV is the change in volume, V is the initial volume, β is the coefficient of volume expansion, and ΔT is the change in temperature.

With an initial volume of 53 L, we calculate the increase in volume as follows:

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Answer 2

`\Delta V=1.0911\ L`

Step-by-step 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* 10^(-6)\ K^(-1)`

• volumetric coefficient of thermal expansion for steel,
  `\beta_s=35* 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* 35* 10^(-6)* (27-4.5)`

`\Delta V_s=41737.5* 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* 950* 10^(-6)* (27-4.5)`

`\Delta V_g=1132875* 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)* 10^(-6)`

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