The gas inside of a container exerts #18 Pa# of pressure and is at a temperature of #360 ^o K#. If the pressure in the container changes to #16 Pa# with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2017-03-02T17:13:53

The new temperature is #=320K#

We apply Gay Lussac 's Law

#P_1/T_1=P_2/T_2#

#P_1=18Pa#

#T_1=360K#

#P_2=16Pa#

#T_2=P_2/P_1*T_1#

#=16/18*360#

#=320K#