The gas inside of a container exerts #18 Pa# of pressure and is at a temperature of #450 ^o K#. If the pressure in the container changes to #27 Pa# with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2017-01-17T08:08:42

The temperature is #=675K#

We use Gay Lussac's Law

#P_1/T_1=P_2/T_2#

#P_1=18Pa#

#T_1=450K#

#P_2=27Pa#

#T_2=P_2/P_1*T_1#

#=27/18*450=675K#