The gas inside of a container exerts #12 Pa# of pressure and is at a temperature of #85 ^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 · 2017-11-27T07:06:04

The new pressure is #=8.47Pa#

Apply Gay Lussac's Law, which is

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

The initial pressure is #P_1=12Pa#

The initial temperature is #T_1=85K#

The final temperature is #T_2=60K#

The final pressure is

#P_2=T_2/T_1*P_1=60/85*12=8.47Pa#