The gas inside of a container exerts #24 Pa# of pressure and is at a temperature of #210 ^o K#. If the pressure in the container changes to #60 Pa# with no change in the container's volume, what is the new temperature of the gas?

1 Answer
Aug 21, 2016

Answer:

The new temperature is #525 K#

Explanation:

The answer can be determined by using Gay-Lussac's law:

#P_1/T_1 = P_2/T_2#

The number 1 represents the initial conditions and the number 2 represents the final conditions.

Identify your known and unknown variables:

#color(indigo)("Knowns:"#
#P_1#= 24 Pa
#P_2#= 60 Pa
#T_1#= 210K

#color(gold) ("Unknowns:"#
#T_2#

Rearrange the equation to solve #T_2#

#T_2=(P_2xxT_1)/P_1#

Plug in your given values to obtain the final temperature:

#T_2=(60cancel"Pa"xx210K)/(24\cancel"Pa")# = #525K#

I should also mention that when you are dealing with the Kelvin scale, don't write the degrees symbol. Just write the letter K.