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24 June, 21:53

A sealed, expandable container is initially at 2.000 L and 25°C. To what temperature must it be changed to have a volume of 6.00 L?

... I used V1/T1 = V2/T2 to get 75 degrees Celsius, but for some reason that's wrong. I'd love to know what I did wrong.

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  1. 25 June, 00:29
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    The new temperature is 894 K or 621 °C

    Explanation:

    Step 1: Data given

    Initial volume of the container = 2.000L

    Initial temperature = 25.0 °C = 298 K

    Volume increased to 6.00 L

    Step 2: Calculate the new temperature

    V1/T1 = V2/T2

    ⇒with V1 = the initial volume = 2.00L

    ⇒with T1 = the initial temperature = 25 °C = 298 K

    ⇒with V2 = the increased volume 6.00 L

    ⇒with T2 = the new temperature

    2.00 L / 298 K = 6.00 L / T2

    T2 = 894 K = 621 °C

    The new temperature is 894 K or 621 °C
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