The gas inside of a container exerts #7 Pa# of pressure and is at a temperature of #420 ^o K#. If the pressure in the container changes to #19 Pa# with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2018-01-09T07:44:34

The new temperature is #=1140K#

Apply Gay Lussac's Law

#P_1/T_1=P_2/T_2#, #"at constant volume"#

The initial pressure is #=P_1=7Pa#

The initial temperature is #T_1=420K#

The final pressure is #P_2=19Pa#

Therefore,

The final temperature is

#T_2=P_2/P_1*T_1=19/7*420=1140K#