The gas inside of a container exerts #24 Pa# of pressure and is at a temperature of #90 ^o K#. If the pressure in the container changes to #42 Pa# with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2017-09-19T06:35:07

The temperature is #=157.5K#

We apply GayLussac's Law

#P_1/T_1=P_2/T_2# at constant Volume

The initial Pressure is #P_1=24Pa#

The initial Temperature is #T_1=90K#

The final Pressure is #P_2=42Pa#

The final temperature is

#T_2=P_2/P_1*T_1=42/24*90=157.5K#