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23 June, 04:13

If 600.0 ml of air is heated from 293 k to 333 k what volume will it occupy

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  1. 23 June, 05:08
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    V1 = 600.0 mL

    T1 = 293 K

    V2 unknown

    T2 = 333K

    Charles law V1*T2=V2*T1

    where V1 and T1 are the initial volume and temperature. V2 and T2 are the final temperature and volume.

    V2 = (V1*T2) / T1 = (600.0*333) / 293 = 682 mL (with significant figures) (681.911 mL before signifcant figures.)
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