A container with a volume of #48 L# contains a gas with a temperature of #210^o K#. If the temperature of the gas changes to #690 ^o K# without any change in pressure, what must the container's new volume be?

Question

1023 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-04T16:16:09

The volume is #=157.7#

We use Charles' Law.

#V_1/T_1=V_2/T_2#

#V_1=48 L#

#T_1=210K#

#T_2=690K#

#V_2=(V_1T_2)/T_2#

#=48*690/210=157.7L#