The gas inside of a container exerts #42 Pa# of pressure and is at a temperature of #270 ^o K#. If the pressure in the container changes to #36 Pa# with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2016-11-26T15:28:05

The new temperature is #=231.4K#

We use Gay Lussac's Law

#P_1/T_1=P_2/T_2#

#T_1=270K#

#P_1=42Pa#

#P_2=36Pa#

#T_2=P_2T_1/P_1=36*270/42#

#=231.4K=-41.6#ºC