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04-05-2018
Chemistry

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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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taliahblom taliahblom

10-05-2018

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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