The gas inside of a container exerts #9 Pa# of pressure and is at a temperature of #240 ^o K#. If the pressure in the container changes to #3 Pa# with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2016-03-12T20:08:49

#T_2=80 ^o K#

#"use The Gay-Lussac's law"#
#P_1/T_1=P_2/T_2#
#9/240=3/T_2#
#T_2=(240*cancel(3))/cancel(9)#
#T_2=cancel(240)/cancel(3)#
#T_2=80 ^o K#