A tank contains 15 kg of air at 165 kPa and 62 0C The air is

A tank contains 15 kg of air at 165 kPa and 62 0C. The air is now heated until the pressure increases by 3.5 times. Determine: (a) Volume of tank (b) Amount of heat added (c) Final temperature of air.

I found the volume of the tank, V = 8.74m^3, but no clue for heat added and final temp. Please help. Thank you.

Solution

Mass of air m= 15 kg = 15 x10 ³ g

Molecular weight of air M = 28.9645 g.mol¹

Number of moles n = m/M

                            = 517.87 mol

Pressure P = 165 kPa

                 = 165000 Pa

Temprature T = 62 ^o C

                    = 62 + 273

                    = 335 K

From the relation PV = nRT

Volume V = nRT / P

Where R = Gas constant = 8.314 J / mol K

Substitute values you get , V = 8.74 m ³

(b).Final pressure P \' = 3.5 P

In this process, volume remains constant.

So, heat added Q = n Cv (T \' - T )

Where T \' = Final temprature.

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

                                        = 3.5 P / P

                                        = 3.5

                                    T \' = 3.5 T

                                        = 3.5 x 335

                                        = 1172.5 K

C v = Specific heat at constant volume.

      = (5/2)R

      = 2.5 R

      = 2.5 x8.314 J / mol K

Substitute values you get Q = 517.87x(2.5 x8.314) x(1172.5 -335)

                                          = 9.014 x10 ⁶ J

(c). Final temprature T \' = 1172.5 K