# The gas inside of a container exerts 27 Pa of pressure and is at a temperature of 180 ^o K. If the pressure in the container changes to 40 Pa with no change in the container's volume, what is the new temperature of the gas?

Apr 8, 2018

Approximately $267$ kelvin.

#### Explanation:

At a constant number of moles and volume, we can use Gay-Lussac's law, which states that

$P \propto T$

or

${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$

So, we can plug in our given values into the equation, and get,

$\frac{27 \setminus \text{Pa")/(180 \ "K")=(40 \ "Pa}}{T} _ 2$

=>T_2=(180 \ "K")/(27color(red)cancelcolor(black)"Pa")*40color(red)cancelcolor(black)"Pa"

$= 266. \overline{6} \setminus \text{K}$

$\approx 267 \setminus \text{K}$