The gas inside of a container exerts #30 Pa# of pressure and is at a temperature of #450 ^o K#. If the pressure in the container changes to #25 Pa# with no change in the container's volume, what is the new temperature of the gas?

1 Answer
Jun 22, 2017

Answer:

#T_2 = 375# #"K"#

Explanation:

To solve this problem, we can use the pressure-temperature relationship of gases, illustrated by Gay-Lussac's law:

#(P_1)/(T_1) = (P_2)/(T_2)#

Our known values are

  • #P_1 = 30# #"Pa"#

  • #T_1 = 450# #"K"#

  • #P_2 = 25# #"Pa"#

We're asked to solve for the new temperature, so let's rearrange the equation to solve for #T_2#:

#T_2 = (P_2T_1)/(P_1)#

Plugging in the known variables, we have

#T_2 = ((25cancel("Pa"))(450color(white)(l)"K"))/(30cancel("Pa")) = color(red)(375# #color(red)("K"#

The new temperature is thus #color(red)(375# #color(red)("K"#.