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?

1 Answer

317.867 K

Explanation:

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

P_1/T_1 = P_2/T_2P1T1=P2T2

If we rearrange and get T_2T2 by itself, we get

T_2= (P_2T_1)/ P_1T2=P2T1P1

Since temperature needs to be in Kelvin for all gas law calculations, T_1T1 = 298. That gives us the following

T_2= (16 * 298)/ 15T2=1629815

That answer = 317.866666666666667