Acontainer of oxygen with a fixed volume has a pressure of 13.0 atm at a temperature of 20 °c. what will the pressure of the oxygen be if the container is heated to 102 °c? a. 12.4 atm b. 15.4 atm c. 16.6 atm d. 18.6 atm

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.

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