A cylinder has a volume of 30 cubic centimeters. The gas inside is at a temperature of 420 K and a pressure of 110 kPa. Find the temperature of the gas which has a volume of 40 cubic centimeters at a pressure of 120 kPa.

The answer to your question is T2 = 610.9 °K

Explanation:

Data

Volume 1 = V1 = 30 cm³

Temperature 1 = T1 = 420 °K

Pressure 1 = P1 = 110 kPa

Volume 2 = V2 = 40 cm³

Temperature 2 = T2 = ?

Pressure 2 = P2 = 120 kPa

Process

To solve this problem use the combined gas law.

P1V1/T1 = P2V2/T2

-Solve for T2

T2 = P2V2T1 / P1V1

-Substitution

T2 = (120 x 40 x 420) / (110 x 30)

-Simplification

T2 = 2016000 / 3300

-Result

T2 = 610.9 °K