# The gas inside of a container exerts 12 Pa of pressure and is at a temperature of 360 ^o K. If the pressure in the container changes to 64 Pa with no change in the container's volume, what is the new temperature of the gas?

Aug 18, 2017

${T}_{2} = 1920$ $\text{K}$

#### Explanation:

We're asked to find the final temperature of a gas, given some pressure and temperature information.

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

$\frac{{P}_{1}}{{T}_{1}} = \frac{{P}_{2}}{{T}_{2}} \text{ }$ (constant volume and quantity of gas)

where

• ${P}_{1}$ and ${P}_{2}$ are the initial and final pressures of the gas

• ${T}_{1}$ and ${T}_{2}$ are the initial and final absolute temperatures of the gas (which must be in Kelvin)

We know:

• ${P}_{1} = 12$ $\text{Pa}$

• ${P}_{2} = 64$ $\text{Pa}$

• ${T}_{1} = 360$ $\text{K}$

• T_2 = ?

Plugging these in:

$\frac{12 \textcolor{w h i t e}{l} \text{Pa")/(360color(white)(l)"K") = (64color(white)(l)"Pa}}{{T}_{2}}$

color(red)(ulbar(|stackrel(" ")(" "T_2 = 1920color(white)(l)"K"" ")|)