The gas inside of a container exerts #18 Pa# of pressure and is at a temperature of #60 ^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 · 2016-06-10T17:10:49

#140K#.

From the general gas Laws, Boyle's Law, etc we get that :

#(P_1V_1)/T_1=(P_2V_2)/T_2#

#therefore 18/60=42/T_2#

#therefore T_2=(60xx42)/18=140K#.