The gas inside of a container exerts #27 Pa# of pressure and is at a temperature of #150 ^o K#. If the pressure in the container changes to #40 Pa# with no change in the container's volume, what is the new temperature of the gas?

2 Answers
Nov 20, 2017

Answer:

The new temperature is #=222.2K#

Explanation:

Apply Gay Lussac's law

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

The initial pressure is #=P_1=27Pa#

The initial temperature is #=T_1=150K#

The final pressure is #P_2=40Pa#

The final temperature is

#T_2=P_2/P_1*T_1=40/27*150=222.2K#

Answer:

#220^@K#

Explanation:

Using the combined Gas Law, #(P_1V_1)/T_1=(P_2V_2)/T_2#, you can simplify it to #P_1/T_1=P_2/T_2# since volume is constant.

Since to want to find #T_2#(Temperature 2), you rearrange the formula to become #T_2=(T_1*P_2)/P_1#

Then you can plug and chug
#T_2=(150^@K*40Pa)/(27Pa)#
#=222.222222222...^@K#
#=220^@K# (with Significant Figures)