The gas inside of a container exerts #8 Pa# of pressure and is at a temperature of #120 ^o K#. If the pressure in the container changes to #4 Pa# with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2018-03-16T19:03:03

The new temperature is #=60K#

Apply Gay Lussas's Law

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

#P_1/T_1=P_2/T_2#

The initial pressure is #P_1=8Pa#

The initial temperature is #V_1=120K#

The final pressure is #P_2=4Pa#

The final temperature is

#T_2=P_2/P_1*T_1=4/8*120=60K#