A rigid tank contains 228 kg of R134a at 30 degree C The vol

A rigid tank contains 2.28 kg of R134a at -30 degree C. The volume of the tank is 0.33069 m^3. The R134a is heated until its pressure reaches 200 kPa. Determine the initial pressure and final temperature.

Solution

Mass m = 2.28 kg

Molar mass of R134 a is M = 102.03 g / mol

                                        = 0.10203 kg / mol

Number of moles n = m/ M

                            = 22.34 mol

Initial emprature T = -30 ^oC

                           = (-30 + 273 )K

                           = 243 K

Initial volume V = 0.33069 m³

Initial pressure P = nRT / V

Where R = Gas constant = 8.314 / mol K

Substitute values you get P = (22.34)(8.314)(243) /(0.33069)

                                         = 136.52x10 3 Pa

                                         = 136.52 kPa

Final pressure P \' = 200 kPa

At constant volume , P \' / P = T \' / T

                                     T \' = ( P \' / P ) T

                                         = (200 kPa/136.52 kPa) x243K

                                         = 1.46 x243 K

                                          = 356 K

                                          = (356-273) ^oC

                                          = 83 ^o C