# 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?

Aug 4, 2017

$\text{new pressure} = 8.0$ $\text{Pa}$

#### 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:

$\underline{\overline{| \stackrel{\text{ ")(" "(P_1)/(T_1) = (P_2)/(T_2)" }}{|}}}$

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:

${P}_{1} = 12$ $\text{Pa}$

${T}_{1} = 30$ $\text{K}$

P_2 = ?

${T}_{2} = 20$ $\text{K}$

Let's rearrange the equation to solve for the final pressure:

${P}_{2} = \frac{{T}_{2} {P}_{1}}{{T}_{1}}$

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".