If 850 L of gas at a gauge pressure of 720 kPa and a tempera

If 8.50 L of gas at a gauge pressure of 720 kPa and a temperature of 0 degree C is heated until the gauge pressure is 1500 kPa and the temperature is 250 degree C, what is the final volume? Assume the atmospheric pressure is 101 kPa.^H

Solution

Initial volume V = 8.5 L = 8.5 x10 ^-3 m ³

Initial gauge pressure p = 720 kPa

Initial absolute pressure of the gas P = p + 101 kPa

                                                     = 720 kPa +101 kPa

                                                     = 821 kPa = 821 x10 ³ Pa

Initial temprature T = 0 ^o C = 0 + 273 = 273 K

R = gas constant = 8.314 J / mol K

Final gauge pressure p \' = 1500 kPa

Final absolute pressure P \' = p \' + 101 kPa

                                        = 1500 kPa +101 kPa

                                        = 1601 kPa

                                        = 1601 x10 ³ Pa

Final temprature T \' = 250 ^o C = 250 + 273 = 523 K

weKnow PV = nRT

From this PV /T = constant

    P \'V \'/ T \' = PV /T

Final volume V \' = (T \'/T ) (PV /P \')

                        = (523/273) (821kPa)(8.5 L) /(1601 kPa)

                        = 8.35 L