The gas inside of a container exerts #12 Pa# of pressure and is at a temperature of #70 ^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 · 2018-02-02T14:54:02

The new temperature is #=280K#

Apply Gay Lussac's Law

#"Pressure"/"Temperature"="Constant"#

#P_1/T_1=P_2/T_2# at #" Constant Volume "#

The initial pressure is #P_1=12Pa#

The initial temperature is #T_1=70K#

The final pressure is #P_2=48Pa#

The final temperature is

#T_2=P_2/P_1*T_1#

#=48/12*70#

#=280K#