A gas occupies 12.3 liters at a temperature of 40 K. What is the volume when the temperture is increased to 60 K?

The answer to your question is  V2 = 18.45 l

Explanation:

Data

Volume 1 = V1 = 12.3 l

Temperature 1 = T1 = 40°K

Volume 2 = V2 = ?

Temperature 2 = T2 = 60°K

Process

To solve this problem use Charles' law

              V1/T1 = V2/T2

-Solve for V2

              V2 = V1T2 / T1

-Substitution

              V2 = (12.3 x 60) / 40

-Simplification

              V2 = 738 / 40

-Result

               V2 = 18.45 l

i think that the answer is c