The gas inside of a container exerts #28 Pa# of pressure and is at a temperature of #270 ^o K#. If the pressure in the container changes to #60 Pa# with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2017-04-30T12:41:00

The new temperature is #=578.6K#

We apply Gay Lussac's Law

#P_1/T_1=P_2/T_2#

The initial pressure is #P_1=28Pa#

The initial temperature is #T_1=270K#

The final pressure is #P_2=60Pa#

The final temperature is

#T_2=P_2/P_1*T_1#

#=60/28*270=578.6K#