A driver fills an 18.9L steel gasoline can with gasoline at 15.0°C right up to the top. He forgets to replace the cap and leaves the can in

Question

A driver fills an 18.9L steel gasoline can with gasoline at 15.0°C right up to the top. He forgets to replace the cap and leaves the can in the back of his truck. The temperature climbs to 30.0°C by 1pm. How much gasoline spills out of the can​

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Thiên Hương 6 months 2021-08-09T07:18:26+00:00 1 Answers 17 views 0

Answers ( )

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    2021-08-09T07:20:06+00:00

    Answer:

    ΔV = 0.98 L

    Explanation:

    First, we will calculate the increased volume using Charles’ Law:

    \frac{V_1}{T_1} = \frac{V_2}{T_2}

    where,

    V₁ =initial volume = 18.9 L

    V₂ = final volume = ?

    T₁ = initial temperature = 15°C + 273 = 288 k

    T₂ = final temperature = 30°C + 273 = 303 k

    Therefore,

    \frac{18.9\ L}{288\ k} = \frac{V_2}{303\ k}

    V₂ = 19.88 L

    Now, we calculate the change in volume:

    ΔV = V₂ – V₁ =  19.88L – 18.9 L

    ΔV = 0.98 L

    This is the volume of gasoline that will spill out.

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