Page 43 - BASIC CONCEPTUAL OF THERMOFLUID
P. 43

THE IDEAL GAS LAW

                PV = nRT

                 P = Air Pressure in ATM
                 V = Volume in (m )
                                3
                 n = Number of Moles
                 R = Constant (8.314)
                 T = Absolute Temp in Kelvin

                 The characteristic equation is usually written;
                                                            PV = mRT

                                                          m = mass of gas (kg)
                                                          V = volume (m )
                                                                       3
                If gas changes from state 1 to state 2 during a process, then




                                                                 =
                The gas constant, R is different for each gas and is determined from;


                                                               =

                                                                ( kJ/kg.K)

                 Experiment  has shown that the universal gas constant, Ro = 8314.4 J/ mole K



                 There for the value of   =

                                               .
                                        =


                                                          = 259.8


               EXAMPLE 9

                                                                                   2
                      3
               0.046 m of gas are contained in a sealed cylinder at a pressure of 300 kN/m and a temperature of 45
               o C. The gas is compressed until the pressure reaches 1.27 MN/m . If the gas is assumed to be a perfect
                                                                        2
               gas, determine:
                   a) the mass of gas (kg)
                                                o
                   b) the final temperature of gas ( C)
                Given:R = 0.29 kJ/kg K
               SOLUTION : EXAMPLE 9 10

                V1 = 0.046 m 3          V = ? m 3
                                         2
                P1 = 300 kN/m 2         P 2  =1.27 MN/m =1.27 ´10 kN/m 2
                                                       2
                                                                 3
               T1 = 45 + 273 = 318 K
                                                 b) sealed cylinder = volume constant
               a)     =                                      =



                     =
                                                       =

                      (300)(0.046)
                   =                                       318   1.27 10
                       0.29(318)                        =
                   = 0.150 kg                                    300
                                                        = 1346.2 k                                             39
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