The gas inside of a container exerts #16 Pa# of pressure and is at a temperature of #430 ^o K#. If the temperature of the gas changes to #120 ^oC# with no change in the container's volume, what is the new pressure of the gas?

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Answer 1 · 2018-03-01T10:43:16

#14.63"Pa"#

According to Gay-Lussac's Law, when volume is constant,

#PpropT#, and so:

#P/T=k#, and it so follows that:

#P_1/T_1=P_2/T_2#.

Here, we have:

#P_1=16"Pa"#

#T_1=430"K"#

#T_2=120+273.15=393.15"K"#

We rearrange to solve for #P_2#:

#P_2=(P_1T_2)/T_1#

Inputting:

#P_2=(16*393.15)/430#

#P_2=14.63"Pa"#

Answer 2 · 2018-03-01T10:47:04

The new pressure is #=14.62Pa#

Apply Gay Lussac's Law

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

The initial pressure is #P_1=16Pa#

The initial temperature is #T_1=430K#

The final temperature is #T_2=120+273=393K#

The final pressure is

#P_2=T_2/T_1*P_1=393/430*16=14.62Pa#