The gas inside of a container exerts #12 Pa# of pressure and is at a temperature of #120 ^o K#. If the temperature of the gas changes to #60 ^oK# with no change in the container's volume, what is the new pressure of the gas?

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Answer 1 · 2018-01-21T18:05:56

The new pressure is #=6Pa#

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=12Pa#

The initial temperature is #T_1=120K#

The final temperature is #T_2=60K#

The final pressure is

#P_2=T_2/T_1*T_2#

#=60/120*12#

#=6Pa#