The gas inside of a container exerts #12 Pa# of pressure and is at a temperature of #30 ^o K#. If the temperature of the gas changes to #20 ^oK# with no change in the container's volume, what is the new pressure of the gas?
1 Answer
Explanation:
We're asked to find the final pressure of a gas, given its initial pressure, initial temperature, and final temperature.
To do this, we can use the pressure-temperature relationship of gases under constant volume (and quantity), illustrated by Gay-Lussac's law:
#ulbar(|stackrel(" ")(" "(P_1)/(T_1) = (P_2)/(T_2)" ")|)#
where
-
#P_1# is the initial pressure -
#T_1# is the initial absolute temperature -
#P_2# is the final pressure (what we're trying to find) -
#T_2# is the final absolute temperature
The temperature values must ALWAYS be in units of Kelvin for any gas law.
We're given:
#P_1 = 12# #"Pa"#
#T_1 = 30# #"K"#
#P_2 = ?#
#T_2 = 20# #"K"#
Let's rearrange the equation to solve for the final pressure:
#P_2 = (T_2P_1)/(T_1)#
Plugging in values:
#P_2 = ((20cancel("K"))(12color(white)(l)"Pa"))/(30cancel("K")) = color(red)(ulbar(|stackrel(" ")(" "8.0color(white)(l)"Pa"" ")|)#
(
The final pressure is thus
