The gas inside of a container exerts #24 Pa# of pressure and is at a temperature of #420 ^o K#. If the pressure in the container changes to #52 Pa# with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2016-12-06T09:38:53

The new temperature is #=910K#

We use Gay-Lussac's Law

#P_1/T_1=P_2/T_2#

#P_1=24 Pa#

#T_1=420K#

#P_2=52Pa#

#T_2=(P_2T_1)/P_1#

#=(52*420)/24=910K#