The gas inside of a container exerts #8 Pa# of pressure and is at a temperature of #35 ^o K#. If the temperature of the gas changes to #80 ^oK#, what is the new pressure of the gas?

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Answer 1 · 2018-01-09T10:33:38

The new pressure is #=18.3Pa#

Apply Gay Lussac's Law

#P_1/T_1=P_2/T_2#, #"at constant volume"#

The initial pressure is #=P_1=8Pa#

The initial temperature is #T_1=35K#

The final temperature is #T_2=80Pa#

Therefore,

The final pressure is

#P_2=T_2/T_1*P_1=80/35*8=18.3Pa#