The gas inside of a container exerts #18 Pa# of pressure and is at a temperature of #60 ^o K#. If the pressure in the container changes to #32 Pa# with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2016-12-29T14:07:58

The temperature is #=106.7K#

We use Gay Lussac's Law

#P_1/T_1=P_2/T_2#

#P_1=18Pa#

#T_1=60K#

#P_2=32Pa#

#T_2=(P_2T_1)/P_1#

#=32*60/18=106.7K#