Given:
P1 = 13.0 atm
T1 = 20 °C
T2 = 102 °C
Required:
P2 of oxygen
Solution:
At constant volume, we can apply Gay-Lussac’s law of pressure and temperature relationship
P1/T1=P2/T2
(13.0 atm) / (20 °C) = P2 / (102 °C)
P2 = 66.3 atm
The answer is not in the choices given.