# If a gas in a closed container is pressurized from 1.5 atm to 1.6 atm and its original temperature was 298 K, what would be the final temperature of the gas?

Dec 4, 2016

The answer is $= 317.9 K$

#### Explanation:

We use Gay Lussac's Law

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

${P}_{1} = 1.5 a t m$

${P}_{2} = 1.6 a t m$

${T}_{1} = 298 K$

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

$= \frac{1.6 \cdot 298}{1.5} = 317.9 K$