# If a gas in a closed container is pressurized from 15.0 atm to 16.0 atm and its original temperature was 25°C, what would be the final temperature of the gas in Kelvin?

317.867 K

#### Explanation:

Gay-Lussac's Law deals with pressure and temperature.

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

If we rearrange and get ${T}_{2}$ by itself, we get

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

Since temperature needs to be in Kelvin for all gas law calculations, ${T}_{1}$ = 298. That gives us the following

${T}_{2} = \frac{16 \cdot 298}{15}$