The gas inside of a container exerts #5 Pa# of pressure and is at a temperature of #180 ^o K#. If the pressure in the container changes to #32 Pa# with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2018-07-13T08:35:27

The new temperature is #=879^@C#

Apply Gay Lussac's Law

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

#P_1/T_1=P_2/T_2#

The initial pressure is #P_1=5Pa#

The initial temperature is #T_1=180K#

The final pressure is #P_2=32Pa#

The final temperature is

#T_2=P_2/P_1*T_1=32/5*180=1152K#

The new temperature is #=1152-273=879^@C#