The gas inside of a container exerts #84 Pa# of pressure and is at a temperature of #420 ^o K#. If the pressure in the container changes to #64 Pa# with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2017-03-25T13:14:07

The new temperature is #=320K#

We apply Gay Lussac's Law

#P_1/T_1=P_2/T_2#

Pressure, #P_1=84Pa#

Temperature, #T_1=420K#

Pressure, #P_2=64Pa#

Temperature,

#T_2=P_2/P_1*T_1#

#=64/84*420#

#=320K#