A gas at 110.0 kPa and 30.0˚C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0˚C and the pressure increased to 440. KPa. What is the new volume?

The answer to your question is   V2 = 0.58 l

Explanation:

Data

Pressure 1 = P1 = 110 kPa                      Pressure 2 = P2 = 440 kPa

Temperature 1 = T1 = 30°C                   Temperature 2 = T2 = 80°C

Volume 1 = V1 = 2 L                               Volume 2 = V2 = ?

Process

- Use the combined gas law to solve this problem

                                P1V1/T1 = P2V2/T2

-Solve for V2

                                V2 = P1V1T2 / T1P2

-Substitution (temperature must be in °K)

                                V2 = (110 x 2 x (80 + 273)) / (30 + 273)(440)

-Simplification

                                V2 = (220 x 353) / (303 x 440)

                                V2 = 77660 / 133320

-Result

                                V2 = 0.58 l

the product of the pressure and volume is also constant