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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?

Answer 1

Sebastian Acosta

#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#

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Answer 2

Benjamin Asiedu

To find the new temperature (T2) when the pressure changes from P1 to P2 at constant volume, you can use Gay-Lussac's Law:

[ \frac{P1}{T1} = \frac{P2}{T2} ]

[ T2 = \frac{P2 \times T1}{P1} ]

Substitute the given values:

[ T2 = \frac{36 , Pa \times 60 , K}{15 , Pa} ]

[ T2 = 144 , K ]

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Answer from HIX Tutor

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