The gas inside of a container exerts #15 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 · 2017-04-17T11:32:02

The new temperature is #=384K#

We apply Gay Lussac's Law

#P_1/T_1=P_2/T_2#

Initial pressure is #P_1=15Pa#

Initial temperature is #T_1=180K#

Final pressure is #P_2=32Pa#

Final temperature is

#T_2=P_2/P_1*T_1#

#=32/15*180=384K#