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

Here is the formula that will help to solve this task: P1/T1 = P2/T2 

Keep in mind that value 20 C is an equivalent to 293 K. 
If you have to make new pressure 12.4 atm, the given temperature must be changed intor 284K.
Therefore: 
The correct answer is  B)At 15.4 atm, T2 would have to be 353 K (which is 120 C) 
As you can see, these options are not suitable: 
At 16.6 atm, T2 would have to be 381 K 
At 18.6 atm, T2 would have to be 426 K

Gay-Lussac explained the relation between the temperature and the pressure of the gas.
Gay-Lussac's law states that: "At constant volume, the pressure of the gas is directly proportional to its temperature in kelvin"
This means that:
P/T = constant

Based on this:
P1/T1 = P2/T2 where:
P1 is the initial pressure = 13 atm
T1 is the initial temperature = 20 + 273 = 293 degree kelvin
P2 is the final pressure that we need to calculate
T2 is the final temperature = 102 + 273 = 375 degree kelvin

Substitute with the givens to get the final pressure as follows:
12/293 = P2/375
4500 = 293 P2
P2 = 15.4 atm

Based on the above calculations, the correct choice is: B.15.4 atm

Is there any options, for this