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4 July, 23:39

A sample of gas has a volume of 0.600 L at a temperature of 303 k. If the volume is increased to 0.900 L, what is the new temperature?

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  1. 5 July, 02:12
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    Charles' Law states: V1/T1=V2/T2.

    So you have:

    0.600L/303°K=0.900L/x°K

    x°K = (0.900L) / (0.600L/303°K)

    x°K=454.5°K
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