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
Aug 4, 2017

Answer:

#"new pressure" = 8.0# #"Pa"#

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"" ")|)#

(#2# significant figures)

The final pressure is thus #color(red)(8.0color(white)(l)"pascals"#.