The gas inside of a container exerts #84 Pa# of pressure and is at a temperature of #420 ^o K#. If the pressure in the container changes to #72 Pa# with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2017-01-02T07:55:40

The temperature is #=360K#

We use Gay Lussac's Law

#P_1/T_1=P_2/T_2#

#P_1=84Pa#

#T_1=420K#

#P_2=72Pa#

#T_2=(P_2T_1)/P_1#

#=72*420/84=360K#