Question

CK-12 Boyle and Charles's Laws if Mrs. Pa pe prepares 12.8 L of laughing gas at 100.0 k Pa and -108 °C and then she force s the gas into an 855 mL cylinder whe re it warms to room temperature at 22 °C, NC what is the pressure of this gas in kilopascals

Answer 1

Answer:

The answer to your question is   P2 = 2676.6 kPa

Explanation:

Data

Volume 1 = V1 = 12.8 L                        Volume 2 = V2 = 855 ml

Temperature 1 = T1 = -108°C               Temperature 2 = 22°C

Pressure 1 = P1 = 100 kPa                    Pressure 2 = P2 =  ?

Process

- To solve this problem use the Combined gas law.

                     P1V1/T1 = P2V2/T2

-Solve for P2

                     P2 = P1V1T2 / T1V2

- Convert temperature to °K

T1 = -108 + 273 = 165°K

T2 = 22 + 273 = 295°K

- Convert volume 2 to liters

                       1000 ml -------------------- 1 l

                         855 ml --------------------  x

                         x = (855 x 1) / 1000

                         x = 0.855 l

-Substitution

                    P2 = (12.8 x 100 x 295) / (165 x 0.855)

-Simplification

                    P2 = 377600 / 141.075

-Result

                   P2 = 2676.6 kPa