The gas inside of a container exerts #9 Pa# of pressure and is at a temperature of #690 ^o K#. If the pressure in the container changes to #34 Pa# with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2017-01-27T16:34:44

The temperature is #=2606.7K#

We use Gay Lussac's Law

#P_1/T_1=P_2/T_2#

#P_1=9Pa#

#T_1=690K#

#P_2=34Pa#

Therefore,

#T_2=(P_2T_1)/P_1#

#=(34*690)/9=2606.7K#