The gas inside of a container exerts #6 Pa# of pressure and is at a temperature of #120 ^o K#. If the pressure in the container changes to #42 Pa# with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2018-01-24T07:47:40

The new temperature is #=840K#

Apply Gay Lussac 's Law

#"Pressure"/"Temperature"="Constant"# at #"Constant Volume"#

#P_1/T_1=P_2/T_2#

The initial pressure is #P_1=6Pa#

The initial temperature is #T_1=120K#

The final pressure is #P_2=42Pa#

The final temperature is

#T_2=P_2/P_1*T_1#

#=42/6*120#

#=840K#