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?

Jun 22, 2017

${T}_{2} = 375$ $\text{K}$

Explanation:

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

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

Our known values are

• ${P}_{1} = 30$ $\text{Pa}$

• ${T}_{1} = 450$ $\text{K}$

• ${P}_{2} = 25$ $\text{Pa}$

We're asked to solve for the new temperature, so let's rearrange the equation to solve for ${T}_{2}$:

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

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".