The gas inside of a container exerts #9 Pa# of pressure and is at a temperature of #190 ^o K#. If the pressure in the container changes to #34 Pa# with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2017-01-10T06:38:55

The temperature is #=717.8K#

We apply Gay Lussac's Law

#P_1/T_1=P_2/T_2#

#P_1=9Pa#

#T_1=190K#

#P_2=34Pa#

#T_2=(P_2*T_1)/P_1#

#=34*190/9=717.8K#