# The gas inside of a container exerts 27 Pa of pressure and is at a temperature of 150 ^o K. If the pressure in the container changes to 30 Pa with no change in the container's volume, what is the new temperature of the gas?

Jul 28, 2017

${T}_{2} = 170$ $\text{K}$

#### Explanation:

We'r asked to find the new temperature of a gas,after it is subjected to a pressure change with no change in volume.

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

$\frac{{P}_{1}}{{T}_{1}} = \frac{{P}_{2}}{{T}_{2}}$ (constant volume)

Here,

• ${P}_{1} = 27$ $\text{Pa}$

• ${T}_{1} = 150$ $\text{K}$

• ${P}_{2} = 30$ $\text{Pa}$

• T_2 = ?

Plugging in known values and solving for ${T}_{2}$, we have

T_2 = (P_2T_1)/(P_1) = ((30cancel("Pa"))(150color(white)(l)"K"))/(27cancel("Pa")) = color(red)(170 color(red)("K"

rounded to $2$ significant figures.