The gas inside of a container exerts #7 Pa# of pressure and is at a temperature of #90 ^o K#. If the pressure in the container changes to #48 Pa# with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2017-03-07T06:41:18

The new temperature is #=617.1K#

We apply Gay Lussac's Law

#P_1/T_1=P_2/T_2#

#P_1=7Pa#

#T_1=90K#

#P_2=48Pa#

#T_2=P_2/P_1*T_1#

#=48/7*90=617.1K#