# A gas is held at 3.8 atm and 500 K. If the pressure is then decreased to 1.2 atm, what will the new temperature be?

Mar 18, 2017

The final temperature will be $\approx 200 \textcolor{w h i t e}{.} \text{K}$.

#### Explanation:

This question involves Gay-Lussac's law to answer, which states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.

This means that as the pressure goes up, the temperature also goes up, and vice-versa.

The equation to use is:

${P}_{1} / \left({T}_{1}\right) = {P}_{2} / \left({T}_{2}\right)$,

where $P$ is pressure in atm, and $T$ is temperature in Kelvins.

Given/Known Information
${P}_{1} = \text{3.8 atm}$
${T}_{2} = \text{500 K}$
${P}_{2} = \text{1.2 atm}$

Unknown: ${T}_{2}$

Solution
Rearrange the equation to isolate ${T}_{2}$.

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

T_2=(1.2color(red)cancel(color(black)("atm"))xx500 "K")/(3.8color(red)cancel(color(black)("atm")))~~"200 K" rounded to 1 significant figure