# The gas inside of a container exerts 9 Pa of pressure and is at a temperature of 35 ^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?

Jun 5, 2017

$5$ $\text{Pa}$

#### Explanation:

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

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

Let's rearrange this equation to solve for the final pressure, ${P}_{2}$:

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

Plugging in known values, we have

${P}_{2} = \left(\left(9 \text{Pa")(20cancel("K")))/(35cancel("K}\right)\right) = \textcolor{red}{5}$ color(red)("Pa"

rounded to one significant figure, the lowest amount given in the problem.