The gas inside of a container exerts #15 Pa# of pressure and is at a temperature of #60 ^o K#. If the pressure in the container changes to #36 Pa# with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2017-10-06T15:44:59

#144K#

#PV=nr∆T#
#V=(nr∆T)/P#

#P_1=15Pa#
#T_1=60K#
#P_2=36Pa#

#V_1=V_2#
#T_2=?#

And, #n" & " r# will remain the same. So,

#(cancel(nr)60K)/(15Pa)=(cancel(nr)T_2)/(36Pa)#

#=>T_2=(36cancel(Pa)*60K)/(15cancel(Pa))#

#=>T_2=144K#