The gas inside of a container exerts #12 Pa# of pressure and is at a temperature of #90K#. 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 · 2018-03-19T13:22:08

The new temperature is #=360K#

Apply Gay Lussac's Law

#"Pressure"/"Temperature"="Constant"#, #"Volume is constant"#

#P_1/T_1=P_2/T_2#

The initial pressure is #P_1=12Pa#

The initial temperature is #T_1=90K#

The final pressure is #P_2=48Pa#

The final temperature is

#T_2=P_2/P_1*T_1=48/12*90=360K#