Question

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?

Answer 1

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