The gas inside of a container exerts #9 Pa# of pressure and is at a temperature of #320 ^o K#. If the pressure in the container changes to #32 Pa# with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2017-03-02T09:32:38

The new temperature is #=1137.8Pa#

The ideal gas law is

#PV=nRT#

#P/T=nR/V=k#

Where #k=# a constant

This is Gay Lussac's Law

#P_1/T_1=P_2/T_2#

#P_1=9Pa#

#T_1=320K#

#P_2=32Pa#

#T_2=P_2/P_1*T_1#

#=32/9*320=1137.8Pa#