A gas occupies a volume of 0.560 L in a flexible container at a temperature of 393 K. To what temperature must the gas be lowered if we desire it to occupy a space of 0.400 L?

The answer to your question is  T2 = 280.7 °K

Explanation:

Data

Volume 1 = V1 = 0.56 l

Temperature 1 = T1 = 393°K

Volume 2 = V2 = 0.40 l

Temperature 2 = T2 = ?

Process

To solve this problem use Charles law.

             V1/T1 = V2/T2

-Solve for T2

             T2 = V2T1 / V1

-Substitution

              T2 = (0.40 x 393) / 0.56

-Simplification

               T2 = 157.2/0.56

-Result

              T2 = 280.7 °K

while cooking a piece of 1/2 pound beef steak, a cast iron pan (2.4kg in weight) was heated 260 celsius.